MCQ Questions for Class 12 Commerce Maths Three Dimensional Geometry Quiz 7

The acute angle between two lines such that the direction cosines l,m,n of each of them satisfy the equations

$l+m+n=0$ and

${ l }^{ 2 }+{ m }^{ 2 }-{ n }^{ 2 }=0$ is :-

0%
$30$
0%
$45$
0%
$60$
0%
$15$

Explanation

Lines are

$l+m+n=0\Rightarrow -l=(m+n)$
and

${ l }^{ 2 }-{ m }^{ 2 }+{ n }^{ 2 }=0\Rightarrow { l }^{ 2 }={ m }^{ 2 }-{ n }^{ 2 }$
Solving them gives,

${ \left( -\left( { m }{ +n } \right) \right) }^{ 2 }={ m }^{ 2 }+{ n }^{ 2 }+2mn={ m }^{ 2 }-{ n }^{ 2 }\Rightarrow 2n(n+m)=0$
Now, for

$n=0$

$m=-l$ , so d.c's

$\left( \cfrac { 1 }{ \sqrt { 2 } } ,-\cfrac { 1 }{ \sqrt { 2 } } ,0 \right)$
And for

$n=-m$

$l=0$ so d.c's

$\left( 0,\cfrac { 1 }{ \sqrt { 2 } } ,-\cfrac { 1 }{ \sqrt { 2 } } \right)$
Angle between the lines

$\left| \cos { \theta } \right| =\left| \cfrac { 1 }{ 2 } \right| \Rightarrow \theta =\cfrac { \pi }{ 3 } =60$

Find in a symmetrical form, the equations of the line formed by the planes

$x+y+z+1=0, 4x+y-2z+2=0$ and find its direction-cosines.

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$\dfrac{x-\frac{1}{3}}{1}=\dfrac{y+\frac{2}{3}}{-2}=\dfrac{z-0}{1}; -\dfrac{1}{\sqrt 6},\dfrac{2}{\sqrt{6}},\dfrac{1}{\sqrt 6}$
0%
$\dfrac{x-\frac{1}{3}}{-1}=\dfrac{y-\frac{2}{3}}{2}=\dfrac{z-0}{1}; \dfrac{1}{\sqrt 6},\dfrac{2}{\sqrt{6}},-\dfrac{1}{\sqrt 6}$
0%
$\dfrac{x+\frac{1}{3}}{-1}=\dfrac{y+\frac{2}{3}}{2}=\dfrac{z+0}{-1}; -\dfrac{1}{\sqrt 6},\dfrac{2}{\sqrt{6}},-\dfrac{1}{\sqrt 6}$
0%
$\dfrac{x+\frac{1}{3}}{1}=\dfrac{y-\frac{2}{3}}{2}=\dfrac{z+0}{1}; \dfrac{1}{\sqrt 6}, -\dfrac{2}{\sqrt{6}},\dfrac{1}{\sqrt 6}$

Explanation

The lines should be

$x + y + z = 0$

and,

$4x + y - 2z = (-2)$

Now, equation should lie on both planes.

So, the equation of line should be perpendicular to both planes.

Let the DRs of the line

$(a, b, c)$

So,

$(a, b, c) \perp (1,1, 1)$

and,

$(a, b, c) \perp (4, ,1, -2)$

$\therefore a + b + c = 0$

$4a + b - 2c = 0$

On solving,

$\dfrac{a}{-2-1} = \dfrac{-b}{-2-4} = \dfrac{c}{1-4}$

$\Rightarrow a = -3, b = 6, c = -3$

$\Rightarrow a = -1, b = 2, c = -1$

Let the coordinates of the point lying in

$xy$ plane

$(x, y, 0)$

on subtracting

$x + y = -1 -(1)$

and

$4x + y = -2$

$\Rightarrow -3x = 1$

$\Rightarrow x = \dfrac{-1}{3}$

Putting vbalue of

$x$ in

$(1)$

$\dfrac{-1}{3} + y = -1$

$y = -1 + \dfrac{-1}{3}$

$=\dfrac{-2}{3}$

Coordinate

$(\dfrac{-1}{3} , \dfrac{-2}{3}, 0)$

$\therefore Equation = \dfrac{x + \dfrac{1}{3}}{-1} = \dfrac{y + \dfrac{2}{3}}{2} =\dfrac{z - 0}{-1}$

DC =

$\dfrac{-1}{\sqrt{(-1)^2 + (2)^2 +(1)^2}} ,\dfrac{2}{\sqrt{(-1)^2 + (2)^2 +(1)^2}} , \dfrac{-1}{\sqrt{(-1)^2 + (2)^2 +(1)^2}}$

$= (\dfrac{-1}{\sqrt{6}}, \dfrac{2}{\sqrt{6}}, \dfrac{-1}{\sqrt{6}})$

The equation of plane passing through

$(4, 5, -1)$ having normal

$3\hat{i}-\hat{j}+\hat{k}$ is ___________.

0%
$4x-5y+z=6$
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$3x-y+z=6$
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$3x+y+z=6$
0%
$4x+5y-z=6$

Explanation

$A(\bar{a})=(4, 5, -1)$ and

$\bar{n}=(3, -1, 1)$

$\therefore$ Required equation of plane

$\bar{r}\cdot \bar{n}=\bar{a}\cdot \bar{n}$

$\therefore (x, y, z)(3, -1, 1)=(4, 5, -1)\cdot (3, -1, 1)$

$\therefore 3x-y+z=12-5-1$

$\therefore 3x-y+z=6$

Vector equation of line

$\dfrac{3-x}{3}=\dfrac{2y-3}{5}=\dfrac{z}{2}$ is __________

$k\in R$ .

0%
$\bar{r}=(3, 5, 2)+k(3, 3, 0)$
0%
$\bar{r}=\left(3, \dfrac{3}{2}, 0\right)+k(-6, 5, 4)$
0%
$\bar{r}=(3, 3, 0)+k(3, 5, 2)$
0%
$\bar{r}=(-6, 5, 4)+$ $k\left( 3, \dfrac{3}{2}, 0\right)$

Explanation

$\dfrac{3-x}{3}=\dfrac{2y-3}{5}=\dfrac{z}{2}$

$\therefore\dfrac{x-3}{-3}=\dfrac{y-\dfrac{3}{2}}{\dfrac{5}{2}}=\dfrac{z-0}{2}$
Here

$A(\bar{a})=\left(3, \dfrac{3}{2}, 0\right)$ and

$(\bar{l})=\left(-3, \dfrac{5}{2}, 2\right)$
Vector equation of line :

$\bar{r}=\bar{a}+k(\bar{l})$

$\therefore r=\left(3, \dfrac{3}{2}, 0\right)+k'\left(-3, \dfrac{5}{2}, 2\right)$

$=\left(3, \dfrac{3}{2}, 0\right)+\dfrac{k'}{2}(-6, 5, 4)$

$=\left(3, \dfrac{3}{2}, 0\right)+k(-6, 5, 4)$

$\therefore k=\dfrac{k'}{2}\in R$ .

The vector equation of the plane which is at distance of

$10$ unit from the origin and perpendicular to the vector

$4i+4j-2k$ is

0%
$r.(4i+4j-2k)=10$
0%
$r.(4i+4j-2k)=20$
0%
$r.(4i+4j-2k)=30$
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$r.(4i+4j-2k)=60$

Explanation

Consider the problem

Let the Normal vector be

$N=4\hat i + 4\hat j - 2\hat k$

Then unit vector in the direction of

$\vec N$ is

$=\frac{\vec N}{|\vec N|}$

$= \frac{{4\hat i + 4\hat j - 2\hat k}}{{\sqrt {{4^2} + {4^2} + {{( - 2)}^2}} }}$

$= \frac{1}{6}\left( {4\hat i + 4\hat j - 2\hat k} \right)$

As we know that the equation of the plane with the position vector

$\vec r$

$\vec r. \hat n=d$

this imlies,

$\vec r. \frac{1}{6}\left( {4\hat i + 4\hat j - 2\hat k} \right)=10$

Hence,

the required equation

$r.\left( {4i + 4j - 2k} \right) = 60$

Hence option

$D$ is the correct answer.

A line making angles

$45^o$ and

$60^o$ with the positive direction of

$x-$ axis and

$y-$ axis respectively. Then the angle made by the line with positive direction of

$z-$ axis is

0%
$60^o$
0%
$120^o$
0%
$60^o$ or $120^o$
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$None\ of\ these$

Explanation

$cos^2\alpha + cos^2\beta + cos^2\gamma = 1$

$\alpha = 45^{\circ}$

$\beta = 60^{\circ}$

$\therefore cos^245^{\circ} + cos^260^{\circ} + cos^2\gamma = 1$

$\dfrac{1}{2} + \dfrac{1}{4} + cos^2\gamma = 1$

$cos^2\gamma = 1 - \dfrac{3}{4}$

$cos\gamma = \pm\dfrac{1}{2}$

$cos\gamma = \dfrac{1}{2}, cos\gamma = -\dfrac{1}{2}$

$\gamma = 60^{\circ}, \gamma = 120^{\circ}$

If the direction cosine of a directed line be

$a, 3a, 7a$ then

$a =$

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$\pm 1/\sqrt{59}$
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$\pm 1/9$
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$\pm 2/7$
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None of these

Explanation

Give

$a,3a,7a$ be the direction cosines of a directed line.

Then from the property of direction cosines we get

$a^2+(3a)^2+(7a)^2=1$

or,

$59a^2=1$

or,

$a=\pm\dfrac{1}{\sqrt{59}}$ .

The direction ratios of the line

$6x - 2 = 3y + 1 = 2z - 2$ are

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$\dfrac{1}{\sqrt{3}} , \dfrac{1}{\sqrt{3}} ,\dfrac{1}{\sqrt{3}}$
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$\dfrac{1}{\sqrt{14}} , \dfrac{2}{\sqrt{14}} , \dfrac{3}{\sqrt{14}}$
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$1, 2, 3$
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None of these

Explanation

$6x-2=3y+1=2x-2$

$6(x-\dfrac26)=3(y+\dfrac13)=2(x-\dfrac22)$

$\dfrac{(x-\dfrac13)}1=\dfrac{(y+\dfrac13)}2=\dfrac{(x-1)}3$

Line will be passing through the poits,

$(\dfrac13,-\dfrac13,1)$

and parallel to the line having direction ratios is

$1,2,3$

option

$C$ will be the correct answer

If O is the origin and the coordinates of P is

$(1, 2, -3)$ , then find the equation of the plane passing through P and perpendicular to OP.

0%
$x-2y-3z=-15$
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$x+2y-3z=14$
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$x-2y+3z=15$
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$x-2y-3z=15$

Explanation

Consider the problem

Origin

$O(0,0,0)$ and

$P(1,2,-3)$

then direction ratios of

$OP$

$\begin{array}{l} ({ x_{ 2 } }-{ x_{ 1 } }),({ y_{ 2 } }-{ y_{ 1 } }),({ z_{ 2 } }-{ z_{ 1 } }) \\ (1-0),(2-0),(-3-0) \\ \Rightarrow (1,2,-3) \end{array}$

Equation of plane passing through

$({x_1},{y_1},{z_1})$

$a(x - {x_1}) + b(y - {y_1}) + c(z - {z_1}) = 0$

where

$a,b,c$ are direction ratios of normal are

$1,2,-3$

And the point

$P(1,2,-3)$

therefore equation of required plane

$1(x - 1) + 2(y - 2) - 3(z - ( - 3)) = 0$

$x - 1 + 2y - 4 - 3z - 9 = 0$

$x + 2y - 3z - 14 = 0$

$x + 2y - 3z = 14$

The direction cosines of two lines are related by

$l+m+n=0$ and

$al^2+bm^2+cn^2=0$ . The lines are parallel if

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$a+b+c=0$
0%
$a^{-1}+b^{-1}+c^{-1}=0$
0%
$a=b=c$
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$None\ of\ these$

Explanation

Let

${ l }_{ 1 }{ m }_{ 1 }{ n }_{ 1 }$ and

${ l }_{ 2 }{ m }_{ 2 }{ n }_{ 2 }$ be the direction cosines.

$l+m+m=0\Rightarrow l=-m-n\\ a{ l }^{ 2 }+b{ m }^{ 2 }+c{ n }^{ 2 }=0\\ a({ m }^{ 2 }+{ n }^{ 2 }+2mn)+b{ m }^{ 2 }+c{ n }^{ 2 }=0\\ (a+b){ m }^{ 2 }+(a+c){ n }^{ 2 }+2amn=0$

$(a+b){ (\cfrac { m }{ n } ) }^{ 2 }+2a(\cfrac { m }{ n } )+(a+c)=0$ -Quadratic in

$\cfrac { m }{ n }$

$\therefore$ product of roots

$=\cfrac { { m }_{ 1 } }{ { n }_{ 1 } } \times \cfrac { { m }_{ 2 } }{ { n }_{ 2 } } =\cfrac { a+c }{ a+b }$

Similarly,

$m=-l-n\\ a{ l }^{ 2 }+b({ l }^{ 2 }+{ n }^{ 2 }+2ln)+c{ n }^{ 2 }=0\\ (a+b){ l }^{ 2 }+(b+c){ n }^{ 2 }+2bln=0$

$(a+b){ (\cfrac { l }{ n } ) }^{ 2 }+2b(\cfrac { l }{ n } )+(b+c)=0-$ Quadratic in

$\cfrac { l }{ n }$

Product of roots

$=\cfrac { { l }_{ 1 } }{ { n }_{ 1 } } \times \cfrac { { l }_{ 2 } }{ { n }_{ 2 } } =\cfrac { b+c }{ a+b }$

Since the two lines are perpendicular,

$\therefore { l }_{ 1 }{ l }_{ 2 }+{ m }_{ 1 }{ m }_{ 2 }+{ n }_{ 1 }{ n }_{ 2 }=cos{ 90 }\\ { l }_{ 1 }{ l }_{ 2 }+{ m }_{ 1 }{ m }_{ 2 }+{ n }_{ 1 }{ n }_{ 2 }=0\\ \cfrac { { l }_{ 1 }{ l }_{ 2 } }{ { n }_{ 1 }{ n }_{ 2 } } +\cfrac { { m }_{ 1 }{ m }_{ 2 } }{ { n }_{ 1 }{ n }_{ 2 } } +1=\cfrac { 0 }{ { n }_{ 1 }{ n }_{ 2 } } \\ \cfrac { a+c }{ a+b } +\cfrac { b+c }{ a+b } +1=0\\ 2(a+b+c)=0\\ a+b+c=0$

The direction cosines of a line segment AB are

$- \dfrac{2}{{\sqrt {17} }},\dfrac{3}{{\sqrt {17} }}, - \dfrac{2}{{\sqrt {17} }}$ . If

$AB=\sqrt {17}$ and the coordinates of A are

$(3,-6,10)$ , then the coordinates of B are

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$(1,-2,4)$
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$(2,5,8)$
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$(-1,3,-8)$
0%
$(1,-3,8)$

Explanation

DC of the line

$AB <\dfrac{-2}{\sqrt{17}}, \dfrac{3}{\sqrt{17}}, \dfrac{-2}{\sqrt{17}}>$

Since

$AB = \sqrt{17}$

$\therefore$ DR of the line AB

$=<-12, 3, -2>$

DR of the line

$AB$

$=$ Co-ordinates of

$B$ - co-ordinates

$A$

Let co-ordinates of

$B$ be

$(x, y, z)$

$\Rightarrow \therefore <-2, 3, -2> = <x - 3, y + 6, z - 10>$

Co-ordinates of

$A$ :

$x - 3 = -2$

$\therefore x = 1$

$y + 6 = 3$

$\therefore y = -3$

$z - 10 = -2$

$\therefore z = 8$

$\therefore <x, y, z> = <1, -3, 8>$

option D is the correct option

The foot of the perpendicular drawn from the origin to a plane is

$(1, 2, -3)$ . Find the equation of the plane.

0%
$x-2y-3z=14$
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$x-2y+3z=14$
0%
$x+2y-3z=14$
0%
$x+2y+3z=14$

Explanation

Given:

Let the equation of plane passing through $\left ( 1, 2, 3 \right )$ and perpendicular to $OA$ .

$A = \hat{i}+2\hat j-3\hat k$

$\vec{n} = \overrightarrow{OA}=\hat i+2\hat j-3\hat k$

$a=|\hat i+2\hat j-3\hat k|=\sqrt{1+4+9}=\sqrt{14}$

$\hat{n}=\dfrac{\hat i+2\hat j-3\hat k}{\sqrt{14}}$

The equation of the required plane is $\vec{r}\cdot \vec{n}=a$

$\Rightarrow \vec{r}\cdot \left (\dfrac{\hat i+2\hat j-3\hat k}{\sqrt{14}} \right ) =\sqrt{14}$

$\Rightarrow \vec{r} \cdot \left (\hat i+2\hat j-3\hat k \right ) =14$

$\Rightarrow x + 2y - 3z = 14$

Thus, the equation of the plane is $x + 2y - 3z = 14$ .

$l,m,n$ are d.c's of vector

$\overline {OP}$ then maximum value of

$lmn$ is

0%
$\dfrac{1}{{\sqrt 3 }}$
0%
$\dfrac{1}{{2\sqrt 3 }}$
0%
$\dfrac{1}{{3\sqrt 3 }}$
0%
$\dfrac{2}{{\sqrt 3 }}$

Explanation

Let the

$3$ number be

$l^2, m^2, n^2$

we know that

$AM \geq GM$

i.e. Arithemetic Mean

$\geq$ Geometric Mean

$\therefore \dfrac{l^2 + m^2 + n^2}{3} \geq \sqrt[3]{(lmn)^2}$

$(lmn)^\dfrac{2}{3} \leq \dfrac{1}{3}$

$(l^2 + m^2 + n^2 =1)$

$lmn \leq \dfrac{1}{3\sqrt{3}}$

$\therefore$ option C is the correct answer

If a line has the direction ratios

$4, -12,18$ then find its direction cosines.

0%
$-\dfrac{2}{11},-\dfrac{6}{11},-\dfrac{9}{11}$
0%
$-\dfrac{2}{11},\dfrac{6}{11},-\dfrac{9}{11}$
0%
$\dfrac{2}{11},-\dfrac{6}{11},\dfrac{9}{11}$
0%
$\dfrac{2}{11},\dfrac{6}{11},\dfrac{9}{11}$

Explanation

DR of a line

$<4, -12, 18>$

$=\dfrac{4}{\sqrt{4^2 + (-12)^2 + (18)^2}}, \dfrac{-12}{\sqrt{4^2 + (-12)^2 + (18)^2}}, \dfrac{18}{\sqrt{4^2 + (-12)^2 + (18)^2}}$

$= \dfrac{4}{22}, \dfrac{-12}{22}, \dfrac{18}{22}$

$= <\dfrac{2}{11}, \dfrac{-6}{11}, \dfrac{9}{11}>$

A mirror and a source of light are situated at the origin

${\rm O}$ and at a point on

${\rm O}X$ , respectively. A ray of light from the source strikes the mirror and is reflected. If the direction ratios of the normal to the plane are

$1,\, - 1,\,1$ , then find the DCs of the reflected ray.

0%
$\dfrac {1}{3},\dfrac {2}{3},\dfrac {2}{3}$
0%
$-\dfrac {1}{3},\dfrac {2}{3},\dfrac {2}{3}$
0%
$-\dfrac {1}{3},-\dfrac {2}{3},-\dfrac {2}{3}$
0%
$-\dfrac {1}{3},-\dfrac {2}{3},\dfrac {2}{3}$

Explanation

Suppose the source of light be situated at

$A(a,0,0)$

where

$a\neq 0$ ,

$AO=$ incident ray,

$OB=$ reflected ray.

$ON=$ normal to the mirror at

$0$ .

$\therefore \angle AON=\angle NOB=\dfrac{\theta}{2}$ (say)

Direction ratios of normal place are

$(1,-1,1)$

$\therefore$ Direction cosine of normal :

$\left(\dfrac{1}{\sqrt{1^2+(-1)^2+1^2}},\dfrac{1}{\sqrt{1^2+(-1)^2+1^2}},\dfrac{1}{\sqrt{1^1(-1)^2+1^2}}\right)$

$=\left(\dfrac{1}{\sqrt{3}},\dfrac{-1}{\sqrt{3}},\dfrac{1}{\sqrt{3}}\right)$

Hence,

$\cos \left(\dfrac{\theta}{2}\right)=\dfrac{1}{\sqrt{3}}$

Direction ratio of

$AO$ are

$(\alpha,0,0)$ and direction cosines are

$(1,0,0)$

$\therefore$ Now let

$l, m, n$ be direction cosines of reflected ray

$OB$ .

$\therefore \dfrac{1+1}{2\cos\dfrac{\theta}{2}}=\dfrac{1}{\sqrt{3}}$

$\Rightarrow 1+1=\dfrac{2}{\sqrt{3}}\cos \dfrac{\theta}{2}=\dfrac{2}{3}$

$\Rightarrow 1=\dfrac{2}{3}-1=\dfrac{-1}{3}$

$\dfrac{m+o}{2\cos\left(\dfrac{\theta}{2}\right)}=\dfrac{-1}{\sqrt{3}}$

$\Rightarrow m=\dfrac{-2}{\sqrt{3}}\cos\dfrac{\theta}{2}=\dfrac{-2}{3}$

$\dfrac{n+o}{2\cos\left(\dfrac{\theta}{2}\right)}=\dfrac{1}{\sqrt{3}}$

$\Rightarrow n=\dfrac{2}{\sqrt{3}}\cos\dfrac{\theta}{2}=\dfrac{2}{3}$

$\therefore$ Direction cosines of reflected ray are

$\left(\dfrac{-1}{3},\dfrac{-2}{3},\dfrac{2}{3}\right)$

1427667_1047102_ans_a7c25c51850d4a0d94cde67513a21d08.png

The direction`cosines of a line equally inclined to three mutually perpendicular lines having D.C.'s as

${\ell _1}{m_1}{n_1}:{\ell _2}{m_2}{n_2}:{\ell _3}{m_3}{n_3}\,\,$ are

0%
${l _1} + {l _2} + {l _3},\,{m_1} + {m_2} + {m_3},\,{n_1} + {n_2} + {n_3}$
0%
$\left( \pm \dfrac{1}{\sqrt{3}},\pm \dfrac{1}{\sqrt{3}},\pm \dfrac{1}{\sqrt{3}} \right)$
0%
$\left( \pm \dfrac{1}{\sqrt{2}},\pm \dfrac{1}{\sqrt{3}},\pm \dfrac{1}{\sqrt{4}} \right)$
0%
none of these

Explanation

We know that, ${{\cos }^{2}}\alpha +{{\cos }^{2}}\beta +{{\cos }^{2}}\gamma =1$

But given that,

$\alpha$ = $\beta$ = $\gamma$

${{\cos }^{2}}\alpha +{{\cos }^{2}}\alpha +{{\cos }^{2}}\alpha =1$

$3{{\cos }^{2}}\alpha =1$

$\cos \alpha =\pm \dfrac{1}{\sqrt{3}}$

Hence, direction cosine are $\left( \pm \dfrac{1}{\sqrt{3}},\pm \dfrac{1}{\sqrt{3}},\pm \dfrac{1}{\sqrt{3}} \right)$

$\overrightarrow a ,\overrightarrow b ,\overrightarrow c$ are three non-zero vectors, no two of which are collinear and the vector

$\overrightarrow a + \overrightarrow b$ is collinear with

$\overrightarrow c$ ,

$\overrightarrow b + \overrightarrow a$ is collinear with

$\overrightarrow a$ , then

$\overrightarrow a + \overrightarrow b + \overrightarrow c$ is equal to -

0%
$\overrightarrow a$
0%
$\overrightarrow b$
0%
$\overrightarrow c$
0%
none

Explanation

${\textbf{Step -1: Write equation using given information and assume with constant.}}$

${\text{Given,}}$

$\bar a + \bar b$

${\text{is collinear with}}$

$\bar c$

$\Rightarrow \bar a + \bar b = \lambda \bar c$

$..... (\lambda$

${\text{is constant)}}$

$\bar b + \bar c$

${\text{is collinear with}}$

$\bar a$

$\Rightarrow \bar b + \bar c = \mu \bar c$

$..... (\mu$

${\text{is constant)}}$

${\text{Then,}}$

$\Rightarrow \bar a + \bar b + \bar c =\lambda \bar c + \bar c ..(1)$

${\text{and}}$

$\Rightarrow \bar a + \bar b + \bar c =\mu \bar a + \bar a ....(2)$

${\textbf{Step -2: Compare both equations and find values of}}$

${\mathbf{ \lambda and \mu.}}$

$\Rightarrow \lambda \bar c + \bar c = \mu \bar a + \bar a$

$\Rightarrow 0 \bar a + \lambda \bar c + \bar c = \mu \bar a + \bar a + 0 \bar c$

$\Rightarrow 0 \bar a + ( \lambda +1 ) \bar c = ( \mu + 1) \bar a+ 0 \bar c$

${\text{So,}}$

$\Rightarrow 1+ \mu =0$

${\text{and}}$

$\Rightarrow 1+ \lambda = 0$

$\Rightarrow \lambda = -1, \mu = -1$

${\textbf{Step -3: Put the values in equations (1) and (2).}}$

$\Rightarrow \bar a + \bar b + \bar c =\lambda \bar c + \bar c$

$\Rightarrow \bar a + \bar b + \bar c = (\lambda + 1) \bar c$

$\Rightarrow \bar a + \bar b + \bar c = ( -1 + 1) \bar c$

$\Rightarrow \bar a + \bar b + \bar c = \bar 0$

${\text{and}}$

$\Rightarrow \bar a + \bar b + \bar c =\mu \bar a + \bar a$

$\Rightarrow \bar a + \bar b + \bar c = (\mu + 1) \bar a$

$\Rightarrow \bar a + \bar b + \bar c = ( -1 + 1) \bar a$

$\Rightarrow \bar a + \bar b + \bar c = \bar 0$

${\textbf{ Hence, the option (D) is correct.}}$

$\dfrac{2}{\sqrt{3}}, \dfrac{-2}{\sqrt{3}}, \dfrac{-1}{\sqrt{3}}$ can be the direction ratios of a directed line.

0%
True
0%
False

Explanation

For direction ratios

$(\dfrac{2}{\sqrt{3}})^2 + (\dfrac{-2}{\sqrt{3}})^2 + (\dfrac{-1}{\sqrt{3}})^2$

$=\dfrac{4}{3} + \dfrac{4}{3} + \dfrac{1}{3}$

$= 3 \neq 1$ (hence they are not direction ration)

A line passes through the point

$(6,-7, -1)$ and

$(2,-3,1)$ . if the angle

$\alpha$ which the line makes with the positive direction of x-axis is acute, the direction cosines of the line are,

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$2/3, -2/3 , -1/3$
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$2/3 , 2/3 , -1/3$
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$2/3, -2/3 , 1/3$
0%
$2/3 , 2/3, 1/3$

Explanation

$P(6,-7,-1)$ and

$Q(2,-3,1)$

$\overrightarrow { PQ } =-4\hat { i } +4\hat { j } +2\hat { k }$

Direction of

$\overrightarrow { PQ } =-2\hat { i } +2\hat { j } +\hat { k }$

$\hat { PQ } =\cfrac { -2 }{ 3 } \hat { i } +\cfrac { 2 }{ 3 } \hat { j } +\cfrac { 1 }{ 3 } \hat { k } =\cos { \alpha } \hat { i } +\cos { \beta } \hat { j } +\cos { \beta } \hat { k }$

Since

$\alpha$ is acute.

$\therefore \cos { \alpha } >0\\ \therefore \hat { PQ } =\cfrac { 2 }{ 3 } \hat { i } -\cfrac { 2 }{ 3 } \hat { j } -\cfrac { 1 }{ 3 } \hat { k }$

$\therefore$ Direction of cosines

$(\cfrac { 2 }{ 3 } ,\cfrac { -2 }{ 3 } ,\cfrac { -1 }{ 3 } )$

In a line

$OP$ through the origin

$O$ makes angles of

${90^ \circ },\,{60^ \circ }\,and\,{60^ \circ }$ with

$x, y$ and

$z$ axis respectively then the direction cosines of

$OP$ are

0%
$\left( A \right)\,\,\dfrac{1}{2},\dfrac{1}{{\sqrt 2 }},\dfrac{{\sqrt 3 }}{2}$
0%
$\left( B \right)\,\,\sqrt 2 ,\,2\,\sqrt 6$
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$\left( C \right)\,\,\dfrac{{\sqrt 3 }}{2},\,\dfrac{1}{{\sqrt 2 }},\,\dfrac{1}{2}$
0%
None of these

Explanation

We know that ,

The direction cosines of a line making angle $\alpha$ with x-axis, $\beta$ with y- axis and $\gamma$ with z-axis are $l,m,n.$

Then,

$l=\cos \alpha ,\,\,\,m=\cos \beta ,\,\,\,n=\cos \gamma$

Given that $\alpha ={{90}^{0}},\,\,\,\,\beta ={{60}^{0}}\,and\,\,\,\,\gamma ={{60}^{0}}$

So, direction cosines of a line

$l=\cos {{90}^{0}},\,\,\,m=\cos {{60}^{0}},\,\,\,n=\cos {{60}^{0}}$

$l=0,\,\,\,m=\dfrac{1}{2},\,\,\,n=\dfrac{1}{2}$

Therefore direction cosines are $\left( 0,\dfrac{1}{2},\dfrac{1}{2} \right)$ .

Option (D) is correct answer.

The direction cosines of the line which is perpendicular to the lines with direction cosines proportional to

$(1, -2, -2)$ &

$(0, 2, 1)$ are

0%
$\left( {\dfrac{2}{3}, - \dfrac{1}{3},\dfrac{2}{3}} \right)$
0%
$\left( {\dfrac{2}{3},\dfrac{1}{3},\dfrac{2}{3}} \right)$
0%
$\left( {\dfrac{2}{3},\dfrac{1}{3},\dfrac{{ - 2}}{3}} \right)$
0%
$\left( {\dfrac{{ - 2}}{3},\dfrac{1}{3},\dfrac{2}{3}} \right)$

Explanation

$\begin{array}{l} According\, to\, \, the\, \, question: \\ direction\, ratio\, equ:a,b,c\, \, and\, \, two\, line\, is{ { } }perpendicular: \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, (1,-2,-2,)\, \, \& \left( { 0,2,1 } \right) \\ relation\, of\, direction\, ratio\, \, with\, perpendicular\, line\, is: \\ \, \Rightarrow { a_{ 1 } }{ a_{ 2 } }+{ b_{ 1 } }{ b_{ 2 } }+{ c_{ 1 } }{ c_{ 2 } }=0 \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, a-2b-2c=0 \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \underline { \, \, 0a+2b+c=0 } \\ the\, ratio\, of\, 3direction\, po{ { int } }\, is: \\ \frac { a }{ 2 } =\frac { { -b } }{ 0 } =\frac { c }{ 2 } \\ so\, the\, ratio\, is\, \left( { \frac { 2 }{ 3 } ,\frac { { -1 } }{ 3 } ,\frac { 2 }{ 3 } } \right) \\ and\, the\, \, { { correct } }\, option\, is\, A. \end{array}$

The projection of the join of the points

$(3,4,2),(5,1,8)$ on the line whose d.c's are

$\left( {\frac{2}{7},\frac{3}{7},\frac{6}{7}} \right)$ is

0%
$7$
0%
$\frac{{31}}{{7}}$
0%
$\frac{{42}}{{13}}$
0%
$\frac{{38}}{{13}}$

Explanation

Let

$A\equiv(3,4,2);\quad B\equiv (5,1,8)$

$\bar { AB } =2\widehat { i} -3\widehat { j } +6\widehat { k}$

$\bar{a}=dc=\cfrac{2}{7}\widehat{i}+\cfrac{3}{7}\widehat{j}+\cfrac{6}{7}\widehat{k}$

$Projection=\bar{AB}\times \bar{a}=\cfrac{4}{7}-\cfrac{9}{7}+\cfrac{36}{7}=\cfrac{31}{7}$

Direction ratios of the normal to the plane passing through the points

$(0, 1, 1),(1, 1, 2)$ and

$(-1, 2, -2)$ are

0%
$(1, 1, 1)$
0%
$(2, 1, -1)$
0%
$(1, 2, -1)$
0%
$(1, -2, -1)$

Explanation

$\overrightarrow { a } =0\hat { i } +\hat { j } +\hat { k } ,\overrightarrow { b } =\hat { i } +\hat { j } +2\hat { k } ,\overrightarrow { c } =-\hat { i } +2\hat { j } -2\hat { k }$

$\therefore$ Direction of line

$\overrightarrow { AB }$ and

$\overrightarrow { BC } \quad$

Directions of normal is perpendicular to that of lines in plane.

$\therefore \overrightarrow { n } =\overrightarrow { AB } \times \overrightarrow { BC } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 1 & 0 & 1 \\ -2 & 1 & -4 \end{matrix} \right| =\hat { -i } +2\hat { j } +\hat { k }$

$\therefore$ Directions of normal is

$(1,-2,-1)$

A lines makes angles

$\dfrac{\alpha }{2},\dfrac{\beta }{2},\dfrac{\gamma }{2}$ with positive direction of coordinate axes, then

$\cos \alpha + \cos \beta + \cos \gamma$ is equal to

0%
$-1$
0%
$1$
0%
$2$
0%
$3$

Explanation

angles made with

$x,y,z$ axis are

$\cfrac { \alpha }{ 2 } ,\cfrac { \beta }{ 2 } ,\cfrac { \gamma }{ 2 }$

$\therefore \cos { ^{ 2 }\cfrac { \alpha }{ 2 } } +\cos { ^{ 2 }\cfrac { \beta }{ 2 } } +\cos { ^{ 2 }\cfrac { \gamma }{ 2 } } =1\\ \cos { 2\theta } =2\cos { ^{ 2 }\theta } -1\quad \quad \Rightarrow 1+\cos { 2\theta } \\ \Rightarrow 2\cos { ^{ 2 }\cfrac { \alpha }{ 2 } } +2\cos { ^{ 2 }\cfrac { \beta }{ 2 } } +2\cos { ^{ 2 }\cfrac { \gamma }{ 2 } } =2\\ 1+\cos { \cfrac { 2\alpha }{ 2 } } +1+\cos { \cfrac { 2\beta }{ 2 } } +1+\cos { \cfrac { 2\gamma }{ 2 } } =2\\ \cos { \alpha } +\cos { \beta } +\cos { \gamma } =-1$

State whether the following statement is true or false.

If l, m, n are the direction cosines of a line, then

$l^2+m^2+n^2=1$ .

0%
True
0%
False

A vector

$\vec{V}$ is inclined at equal angles to axes OX, OY and OZ. If the magnitude of

$\vec{V}$ is

$6$ units, then

$\vec{V}$ is?

0%
$2\sqrt{3}(\hat{i}+\hat{j}+\hat{k})$
0%
$2\sqrt{3}(\hat{i}-\hat{j}+\hat{k})$
0%
$\sqrt{2}(\hat{i}+\hat{j}+\hat{k})$
0%
$2\sqrt{3}(\hat{i}+\hat{j}-\hat{k})$

Explanation

Let the angle subtended with each of the coordinate axis be Q

then

${l} = {m} = {n} = \cos \theta$

${l^2} = {m^2} = {n^2} = 1$

${\cos ^2}\theta + {\cos ^2}\theta + {\cos ^2}\theta = 1$

$3{\cos ^2}\theta = 1$

${\cos ^2} = \dfrac{1}{3}$

$\cos {\theta ^{}} = \pm \dfrac{1}{{\sqrt 3 }}$

therefore l = m = n =

$\pm \dfrac{1}{{\sqrt 3 }}$

then vector

$\begin{array}{l} \vec { v } =6\times \left( { \dfrac { 1 }{ { \sqrt { 3 } } } \hat { i } +\dfrac { 1 }{ { \sqrt { 3 } } } \hat { j } +\dfrac { 1 }{ { \sqrt { 3 } } } \hat { k } } \right) \end{array}$

$\begin{array}{l} \dfrac { { 6\sqrt { 3 } } }{ 3 } (\widehat { i } +\widehat { j } +\widehat { k } ) \\ 2\sqrt { 3 } (\widehat { i } +\widehat { j } +\widehat { k } ) \end{array}$

The points with position vectors

$\vec {a}=\hat {i}-2\hat {j}+3\hat {k}, \vec {b}=2\hat {i}+3\hat {j}-4\hat {k}$ &

$-7\hat {j}+10\hat {k}$ are collinear.

0%
True
0%
False

A point at a distance of

$\sqrt6$ from the origin which lies on the straight line

$\frac{x-1}{1}=\frac{y-2}{2}=\frac{z+1}{3}$ will be

0%
$(1, -1, 2)$
0%
$(1, 2, -1)$
0%
$\left( \frac{5}{7}, \frac{10}{7}, \frac{-13}{7}\right)$
0%
$\left( \frac{5}{7}, \frac{2}{7}, \frac{-6}{7}\right)$

Explanation

$L:\cfrac { x-1 }{ 1 } =\cfrac { y-2 }{ 2 } =\cfrac { z+1 }{ 3 } =k\\ \therefore (k+1,2k+3,3k-1)\\ \sqrt { { (k+1) }^{ 2 }+{ (2k+2) }^{ 2 }+{ (3k-1) }^{ 2 } } =\sqrt { 6 } \\ { (k+1) }^{ 2 }+{ (2k+2) }^{ 2 }+{ (3k-1) }^{ 2 }=6\\ { k }^{ 2 }+1+2k+4{ k }^{ 2 }+4+8k+9{ k }^{ 2 }+1-6k=6\\ 14{ k }^{ 2 }+4k=0\\ \therefore \Rightarrow k=0,\cfrac { -2 }{ 7 }$

$\therefore$ point

$(x,y,z)=(1,2,-1)\quad (\cfrac { 5 }{ 7 } ,\cfrac { 10 }{ 7 } ,\cfrac { -13 }{ 7 } )$

The angle between the lines whose direction cosines satisfy the equations

$l+m+n=0$ and

$l^{2}+m^{2}+n^{2}$ is

0%
$\dfrac{\pi}{2}$
0%
$\dfrac{\pi}{3}$
0%
$\dfrac{\pi}{4}$
0%
$\dfrac{\pi}{6}$

Explanation

According to question.................

$\begin{array}{l} \Rightarrow l+m+n=0----(i) \\ \, \, \, \, \, \, l=-\, (m+n) \\ \, \Rightarrow { l^{ 2 } }={ m^{ 2 } }+{ n^{ 2 } }---(ii) \\ \, \, \, \, \, \, \, \, \Rightarrow \, { \left( { -\, (m+n) } \right) ^{ 2 } }={ m^{ 2 } }+{ n^{ 2 } } \\ \, \, \, \, \, \, \, \, \, \Rightarrow { (m+n)^{ 2 } }={ m^{ 2 } }+{ n^{ 2 } } \\ \, \, \, \, \, \, \Rightarrow { m^{ 2 } }+{ n^{ 2 } }+2mn={ m^{ 2 } }+{ n^{ 2 } } \\ \, \, \, \, \, \, \, \, \, \, \, \Rightarrow 2mn=0 \\ \, \, \, \, \, \, \, \, \, \, \, \therefore \, \, \, \, mn=0 \\ there\, \, are\, \, two\, \, \, case:\, \, \, m=0,n=0 \\ case\, \, 1:\, \, (m=0) \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, l=-n\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \left| { u{ { sing } }\, equ\, :l+m+n=0----(i) } \right. \\ \, suppose: \\ \, l=k,\, \, m=o,\, \, n=-k \\ \Rightarrow { l^{ 2 } }+{ m^{ 2 } }+{ n^{ 2 } }=1 \\ \Rightarrow { k^{ 2 } }+0+{ k^{ 2 } }=1 \\ \Rightarrow k=\dfrac { 1 }{ { \sqrt { 2 } } } \, \, \\ Apply, \\ \Rightarrow l=\dfrac { 1 }{ { \sqrt { 2 } } } \, ,\, \, \, \, m=o,\, \, \, \, n=\dfrac { { -1 } }{ { \sqrt { 2 } } } \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, (we\, Asume:\, { l_{ 1 } }=\dfrac { 1 }{ { \sqrt { 2 } } } \, ,\, \, \, \, { m_{ 1 } }=o,\, \, \, \, { n_{ 1 } }=\dfrac { { -1 } }{ { \sqrt { 2 } } } \, \\ case\, \, 2:\, \, \, (n=0) \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, l=-m\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \left| { u{ { sing } } } \right. \, \, equ:l+m+n=0----(i) \\suppose:\, \, \\ \, \, \Rightarrow \, \, l=k,\, \, \, \, \, m=-k,\, \, \, n=0\, \, \, \, \, \, \, ( \\ \, then, \\ { l^{ 2 } }+{ m^{ 2 } }+{ n^{ 2 } }=1 \\ { k^{ 2 } }+{ k^{ 2 } }+0=1 \\ k=\dfrac { 1 }{ { \sqrt { 2 } } } \\ apply, \\ \Rightarrow l=\dfrac { 1 }{ { \sqrt { 2 } } } \, ,\, \, m=\dfrac { { -1 } }{ { \sqrt { 2 } } } ,\, \, n=0\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, (we\, Assume:{ l_{ 2 } }=\dfrac { 1 }{ { \sqrt { 2 } } } \, ,\, \, { m_{ 2 } }=\dfrac { { -1 } }{ { \sqrt { 2 } } } ,\, \, { n_{ 2 } }=0\, \, \\ Angle\, \, between\, \, \, 2\, lines=\theta \\ \cos \theta =\dfrac { { \, \, \, \, \, \, \, ({ l_{ 1 } }+{ m_{ 1 } }+{ n_{ 1 } })\, .({ l_{ 2 } }+{ m_{ 2 } }+{ n_{ 2 } })\, \, \, \, } }{ { \sqrt { ({ l_{ 1 } }^{ 2 }+{ m_{ 1 } }^{ 2 }+{ n_{ 1 } }^{ 2 })\, .\, \, ({ l_{ 2 } }^{ 2 }+{ m_{ 2 } }^{ 2 }+{ n_{ 2 } }^{ 2 })\, } } } \\ \, \, \, =\dfrac { { { l_{ 1 } }\, { l_{ 2 } }+{ m_{ 1 } }{ m_{ 2 } }+{ n_{ 1 } }{ n_{ 2 } } } }{ { \sqrt { ({ l_{ 1 } }^{ 2 }+{ m_{ 1 } }^{ 2 }+{ n_{ 1 } }^{ 2 })\, \, (\, { l_{ 2 } }^{ 2 }+{ m_{ 2 } }^{ 2 }+{ n_{ 2 } }^{ 2 }) } } } \\ \, =\dfrac { { \dfrac { 1 }{ { \sqrt { 2 } } } . \dfrac { 1 }{ { \sqrt { 2 } } } + 0 .\dfrac { { -1 } }{ { \sqrt { 2 } } } +\dfrac { { -1 } }{ { \sqrt { 2 } } } . 0 } }{ { \sqrt { \left( { \dfrac { 1 }{ 2 } +\dfrac { 1 }{ 2 } +0} \right) \, \, \left( { \dfrac { 1 }{ 2 } +\dfrac { 1 }{ 2 } \ +0 } \right) } } } =\dfrac { { \dfrac { 1 }{ 2 } } }{ { \sqrt { 1 } } } \\ \cos \theta =\dfrac { 1 }{ 2 } ,\, \, \, \, \, \, \, \, \therefore \, \, \, \theta =\dfrac { \pi }{ 3 } \end{array}$

So,that the correct option is B.

A line passes through the points

$(6,-7,-1)$ and

$(2,-3, 1)$ . If the angle a which the line makes with the positive direction of x-axis is acute, the direction cosines of the line are.

0%
$(2/3),(-2/3),(-1/3)$
0%
$(2/3),(2/3), (-1/3)$
0%
$(2/3), (-7/3), (1/3)$
0%
$(8/3), (2/3), (1/3)$

Explanation

Direction Ratios of line joining

$(6,-7,-1)$ and

$(2,-3,1)$ are

$\rightarrow (6,-2,-7-(-3)-(-1))$

$\rightarrow (4,-4,-2)$

if it make acute angle with positive x-axis then

$D.R(x)>0$

D.C are

$\left(\dfrac{4}{\sqrt{4^2+4^2+2^2}},\dfrac{-4}{4^2+4^2+2^2},\dfrac{-2}{4^2+4^2+2^2}\right)$

$\left(\dfrac{4}{6},\dfrac{-4}{6},\dfrac{-2}{6}\right)$

$\left(\dfrac{2}{3},\dfrac{-2}{3},\dfrac{-1}{3}\right)$

$A$ is correct.

The equation to the altitude of the altitude triangle formed by

$\left( {1,1,1} \right).\left( {1,2,3} \right),\left( {2, - 1,1} \right)$ through

$\left( {1,1,1} \right)$ is

0%
$\bar r = \left( {\bar i + \bar j + \bar k} \right) + t\left( {\bar i - \bar j - 2\bar k} \right)$
0%
$\bar r = \left( {\bar i - \bar j + \bar k} \right) + t\left( {\bar i + \bar j - 2\bar k} \right)$
0%
$\bar r = \left( {\bar i + \bar j + \bar k} \right) + t\left( {\bar i - \bar j + 2\bar k} \right)$
0%
$\bar r = \left( {\bar i - \bar j - \bar k} \right) + t\left( {\bar i + \bar j - 2\bar k} \right)$

Explanation

drs of

$BC=(x_{b}-x_{c}), (y_{b}-y_{c}), (z_{b}-z_{c})$

$=(-1, 3, 2)$

$\therefore$ drt of

$AD$ will be of the form

$-1(l)+3(m)+2(n)=0$

$(l_{1}l_{2}+m_{1}m_{2}+n_{1}n_{2})=0$

$l-3m-2n=0$

$\vec{r}=\vec{a}+t\vec{b}$

$\vec{r}=a_{x}\hat{i}+a_{y}\hat{j}+a_{z}\hat{k}+t(x_{d}-x_{a})\hat{i}+(y_{d}-y_{a})\hat{j}+(z_{d}-z_{a})\hat{k})$

when

$l-3m-2n=0$

$\therefore \vec{r}=(\hat{i}+\hat{j}+\hat{k})+t(l\hat{i}+m\hat{j}+n\hat{k})$

$l=1$ and

$m=-1$

$\Rightarrow 1-3(-1)-2n=0$

$2n\Rightarrow n=2$

$(1, -1,2)$ is a dutim

$\therefore \vec{r}=(\hat{i}+\hat{j}+\hat{k})+t(\hat{i}.\hat{j}+2\hat{k})$

Equation of pair of lines passing through origin and making and angle

${\tan ^{ - 1}}2$ with the lines

$4x-3y+7=0$ .

0%
${\left( {4x - 3y} \right)^2} - 4{\left( {3x + 4y} \right)^2} = 0$
0%
${\left( {4x - 3y} \right)^2} - {\left( {3x + 4y} \right)^2} = 0$
0%
${\left( {4x - 3y} \right)^2} - 3{\left( {3x + 4y} \right)^2} = 0$
0%
$4{\left( {4x - 3y} \right)^2} - {\left( {3x + 4y} \right)^2} = 0$

Explanation

Slope of the line

$4x-3y+7=0$ is

${m}_{1}=-\dfrac{coeff.\,of\,x}{coeff.\,of\,y}=\dfrac{4}{3}$

$\tan{\theta}=\left|\dfrac{m-{m}_{1}}{1+m{m}_{1}}\right|$

Given:

${\tan}^{-1}{2}=\theta\Rightarrow\,\tan{\theta}=2$

$\Rightarrow\,2=\left|\dfrac{m-\dfrac{4}{3}}{1+\dfrac{4m}{3}}\right|$

$\Rightarrow\,\left|\dfrac{3m-4}{3+4m}\right|=2\\$

$\Rightarrow\,\dfrac{3m-4}{3+4m}=\pm\,2\\$

$\Rightarrow\,3m-4=\pm\,2\left(4m+3\right)$

$\Rightarrow\,3m-4=8m+6$ and

$3m-4=-8m-6$

$\Rightarrow\,-5m=10$ and

$11m=-2$

$\Rightarrow\,m=-2$ and

$m=\dfrac{-2}{11}$

Equation is

$y=-2x$ and

$y=\dfrac{-2}{11}x$

$\Rightarrow\,2x+y=0$ and

$2x+11y=0$

Pair of lines is

$\left(2x+y\right)\left(2x+11y\right)=0$

$4{x}^{2}+22xy+2xy+11{y}^{2}=0$

$4{x}^{2}+24xy+11{y}^{2}=0$ is the required equation.

Which is equivalent to equation given in option A

${\left( {4x - 3y} \right)^2} - 4{\left( {3x + 4y} \right)^2} = 0$

Prove that the points

$A=(1,2,3),B(3,4,7),C(-3,-2,-5)$ are collinear & find the ratio in which

$B$ divides

$AC$ .

0%
$2:5$
0%
$2:3$
0%
$2:8$
0%
$2:7$

Explanation

$A(1,2,3),B(3,4,7),C(-3,-2,-5)$

$\overrightarrow { AB } =\,\,2\hat{i}+2\hat{j}+4\hat{k}$

$\overrightarrow {BC}\,\,-6\hat{i}-6\hat{j}-12\hat{k}$

If points are collinear then

$|(\overrightarrow {AB } ).(\overrightarrow { BC } )|=|\overrightarrow {AB } ||\overrightarrow {BC } |$

$\overrightarrow {AB } .\overrightarrow {AC } =-12-12-24=-48$

$|\overrightarrow { AB } .\overrightarrow {AC } |=48$

$|\overrightarrow { AB } |.|\overrightarrow {AC } |=\sqrt{24}\sqrt{216}=2\sqrt{24}\sqrt{24}=48$

Points are collinear

$|\overrightarrow {AC } |=|-4\hat{i}-4\hat{j}-8\hat{k}|=4\sqrt{6}$

$|\overrightarrow {AB } |=2\sqrt{6}$

$B$ divides

$AC$ in the ration

$2:3$

1063349_1116830_ans_7f7ff9bc711a4c1cbba938e1a935ba61.PNG

$\vec { a } ,\vec { b } ,\vec { c }$ are three non-zero vectors, no two of which are collinear and the vector

$\vec { a } +\vec { b }$ is collinear with

$\vec { c }, \vec { b } +\vec { c }$ is collinear with

$\vec {a},$ then

$\vec { a } +\vec { b } +\vec { c }$ is equal to -

0%
$\vec {a}$
0%
$\vec {b}$
0%
$\vec {c}$
0%
$none\ of\ these$

Explanation

$\bar a+\bar b=K_{1}\bar c$

$\bar b+\bar c=K_{2}\bar a$

$\bar a- \bar c=K_{1}c-K_{2}\bar a$

$(k_{2}+1) \bar a-\bar c(1+k_{1})$ =0

$k_{2}=-1$ and

$k_{1}=-1$

$\bar{a}+\bar{b}+\bar{c}=0$

Find the equation of the plane if the foot of the perpendicular from origin to the plane is

$(2, 3, -5 )$

0%
$2x+3y+5y=38$
0%
$2x+3y-5y=38$
0%
$2x-3y-5y=38$
0%
None of these

Explanation

$\textbf{Step -1: Find a,b,c}$

$\text{The line is passing through }(2,3,-5)\text{ and let the direction ratios of the line be a,b,c}$

$\text{Then the equation of the line is}$

$\dfrac{x-2}{a}=\dfrac{y-3}{b}=\dfrac{z-(-5)}{c}\Rightarrow{\dfrac{x-2}{a}=\dfrac{y-3}{b}=\dfrac{z+4}{c}=k~(\text{let})}$

$\dfrac{x-2}{a}=k\Rightarrow{x-2=ak}\Rightarrow{x=ak+2}$

$\dfrac{y-3}{b}=k\Rightarrow{y-3=bk}\Rightarrow{y=bk+3}$

$\dfrac{z+5}{c}=k\Rightarrow{z+5=ck}\Rightarrow{z=ck-5}$

$\text{Since the line passes through the origin i.e. }(0,0,0)\text{ then,}$

$\text{ak+2=0}\Rightarrow{a=-\dfrac{2}{k}}$

$\text{bk+3=0}\Rightarrow{b=-\dfrac{3}{k}}$

$\text{ck-5=0}\Rightarrow{c=\dfrac{5}{k}}$

$\textbf{Step -2: Find the equation of plane.}$

$\text{Then the equation of required plane is,}$

$ax+by+cz=d\Rightarrow{\left(-\dfrac{2}{k}\right)x+\left(-\dfrac{3}{k}\right)y+\left(\dfrac{5}{k}\right)z=d}\Rightarrow{-2x-3y+5z=dk}$

$\text{The foot of the perpendicular is }(2,3,-5)\text{ then it satisfies the equation of the plane.}$

$\therefore{-2\times2+-3\times3+5\times-5=dk}\Rightarrow{dk=-4-9-25=-38}$

$\text{Then the equation of the required plane is,}$

$-2x-3y+5z=dk\Rightarrow{-2x-3y+5z=-38}$

$\therefore{2x+3y-5z=38}$

$\textbf{Hence, option (B) 2x+3y-5z=38}\textbf{ is correct answer.}$

If the points with position vectors

$10\hat { i } +\lambda \hat { j } ,3\hat { i } -\hat { j }$ and

$4\hat { i } +5\hat { j }$ are collinear then

$\lambda$ is

0%
$41$
0%
$-41$
0%
$42$
0%
$-42$

Explanation

position vectors :

$(10\hat{i}+\lambda \hat{j}) a, (3\hat{i}-\hat{j})b, (4\hat{i}+5\hat{j})c$

since they are collinear, so

$\vec{AB}$ must parallel to

$\vec{BC}$

$\therefore \vec{AB} = b-a = (3-10)\hat{i}+(-1-\lambda)\hat{j})$

$= (-7\hat{i}+(-1-\lambda )\hat{j})$

$\vec{BC} = \bar{C}-a = (4-3)\hat{i}+(5-(-1)) \hat{j}$

$= \hat{i}+6\hat{j}$

$\therefore \dfrac{-7}{1} = \dfrac{-1-\lambda }{6}$

$\therefore -42 = -1-\lambda$

$\lambda = -1+42$

$\lambda = 41$ (option A)

1116230_1139921_ans_f79953a1fe0543778b8612a58aea337c.jpeg

If the points with position vectors

$60\hat{i}+3\hat{j}, 40\hat{i}-8\hat{j}$ and

$a\hat{i}-52j$ are collinear, then

$a=?$

0%
$-40$
0%
$-20$
0%
$20$
0%
$40$

Explanation

Suppose, position vector

$A=60\widehat i+3\widehat j$

position vector

$B=40\widehat i-8\widehat j$

position vector

$C=a\widehat i-52\widehat j$

Now, find vector AB and BC

$AB= -20\widehat i-11\widehat j$

$BC= (a-40)\widehat i-44\widehat j$

To be collinear, angle between the vector AB and BC made by the given position vectors should be 0 or 180 degree.

That’s why the cross product of the vectors should be zero

$ABXBC=(-20\widehat i-11\widehat j)X(a-40)\widehat i-44\widehat j$

$0\widehat i+0\widehat j+(880+11(a-40))=0$

$a-40= -80$

$a=-40$

Therefore, a should be

$-40$ to be the given positions vectors collinear.

$A ( 2 \overline{i} - \overline{j} - 3 \overline{k} , B ( 4 \overline {i} + \overline{j} - \overline{k} )$ and

$D( \overline{i} - \overline{j} - 2 \overline{k})$ then the vector equation of the plane parallel to

$\overline{ABC}$ and passing through the centroid of the tetrahedron

$ABCD$ is :

0%
$\overline{r} = ( 2 \overline{i} - \overline{j} - \overline{k} + s ( 2 \overline{i} + 2 \overline{j} + 2 \overline{k} ) + t (\overline{i} - \overline{k} )$
0%
$\overline{r} = ( 2 \overline{i} - \overline{j} -3 \overline{k})+ s ( \overline{i} + \overline{j} + \overline{k} ) + t (\overline{i} - \overline{k} )$
0%
$\overline{r} = ( 2 \overline{i} - \overline{j} - \overline{k} )+ s ( \overline{i} + \overline{j} + \overline{k} ) + t (\overline{i} + \overline{j} -5 \overline{k} )$
0%
$\overline{r} = ( 2 \overline{i} - \overline{j} - \overline{k} )+ s ( \overline{i} + \overline{j} + \overline{k} ) + t (\overline{i} + \overline{j} +5 \overline{k} )$

The distance of the point

$3\hat{i}+5\hat{k}$ from the line parallel to the vector

$6\hat{i}+\hat{j}-2\hat{k}$ and passing through the point

$8\hat{i}+3\hat{j}+\hat{k}$ is

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$1$
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$2$
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$3$
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$4$

Explanation

The distance of the point

$\overrightarrow{c}$ from the line

$\overrightarrow{r}=\overrightarrow{a}+s\overrightarrow{b}$ is

$\dfrac{\left|\left(\overrightarrow{c}-\overrightarrow{a}\right)\times \overrightarrow{b}\right|} {\left|\overrightarrow{b}\right|}$

$\overrightarrow{c}=3\hat{i}+5\hat{k},\overrightarrow{a}=8\hat{i}+3\hat{j}+\hat{k},\overrightarrow{b}=6\hat{i}+\hat{j}-2\hat{k}$

$\left(\overrightarrow{c}-\overrightarrow{a}\right)\times \overrightarrow{b}=\left|\begin{matrix} \hat{i}&\hat{j}&\hat{k} \\-5&-3&4 \\6&1&-2\end{matrix}\right|$

$=\hat{i}\left(6-4\right)-\hat{j}\left(10-24\right)+\hat{k}\left(-5+18\right)$

$=\hat{i}\left(2\right)-\hat{j}\left(-14\right)+\hat{k}\left(13\right)$

$=2\hat{i}+14\hat{j}+13\hat{k}$
Distance

$=\dfrac{\sqrt{{2}^{2}+{14}^{2}+{13}^{2}}} {\sqrt{{6}^{2}+{1}^{2}+{2}^{2}}}=\sqrt{\dfrac{369}{41}}=3$ on simplification

$A=(-1, 2, -3), B=(5, 0, -6), C=(0, 4, -1)$ are the vertices of a triangle. The d.c's of the internal bisector of

$\angle$ BAC are?

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$\left(\dfrac{25}{\sqrt{714}}, \dfrac{-8}{\sqrt{714}}, \dfrac{-5}{\sqrt{714}}\right)$
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$\left(\dfrac{5}{\sqrt{74}}, \dfrac{6}{\sqrt{74}}, \dfrac{8}{\sqrt{74}}\right)$
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$\left(\dfrac{25}{\sqrt{714}}, \dfrac{8}{\sqrt{714}}, \dfrac{5}{\sqrt{714}}\right)$
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$\left(\dfrac{-5}{\sqrt{74}}, \dfrac{6}{\sqrt{74}}, \dfrac{-8}{\sqrt{74}}\right)$

Explanation

Here,

$\begin{array}{l} AB=\sqrt { { { \left( { 5+1 } \right) }^{ 2 } }+{ { \left( { 0-2 } \right) }^{ 2 } }+{ { \left( { -6+3 } \right) }^{ 2 } } } =\sqrt { 49 } =7 \\ AC=\sqrt { { { \left( { 0+1 } \right) }^{ 2 } }+{ { \left( { 4-2 } \right) }^{ 2 } }+{ { \left( { 1+3 } \right) }^{ 2 } } } =\sqrt { 9 } =3 \end{array}$

by geometry , the bisector of

$\angle BAC$

will divide the side BC in the ration AB:AC i.e.,in the ratio

$7:3$ internally .Let the bisector of

$\angle BAC$ , meets the side BC at point D.

Therefore,D divides BC in the ratio

$7:3$

coordinates of D are

$\begin{array}{l} \left( { \dfrac { { 7\times 0+3\times 5 } }{ { 7+3 } } ,\dfrac { { 7\times 4+3\times 0 } }{ { 7+3 } } ,\dfrac { { 7\times \left( { -1 } \right) +3\times \left( { -6 } \right) } }{ { 7+3 } } } \right) \\ \left( { \dfrac { 3 }{ 2 } ,\dfrac { { 14 } }{ 5 } ,-\dfrac { 5 }{ 3 } } \right) \end{array}$

Therefore, direction ratio of the bisector AD are

$\dfrac{3}{2} - \left( { - 1} \right),\dfrac{{14}}{5} - 2,\dfrac{{ - 5}}{2} + 3\,\,\,\,i.e.\frac{5}{2},\dfrac{4}{5},\dfrac{1}{2}$

Hence , direction cosines of the bisector AD are

$\begin{array}{l} \dfrac { { \dfrac { 5 }{ 2 } } }{ { \sqrt { { { \left( { \dfrac { 5 }{ 2 } } \right) }^{ 2 } }+{ { \left( { \dfrac { 4 }{ 5 } } \right) }^{ 2 } }+{ { \left( { \dfrac { 1 }{ 2 } } \right) }^{ 2 } } } } } \\ \dfrac { { \dfrac { 4 }{ 5 } } }{ { \sqrt { { { \left( { \dfrac { 5 }{ 2 } } \right) }^{ 2 } }+{ { \left( { \dfrac { 4 }{ 5 } } \right) }^{ 2 } }+{ { \left( { \dfrac { 1 }{ 2 } } \right) }^{ 2 } } } } } \\ \dfrac { { \dfrac { 1 }{ 2 } } }{ { \sqrt { { { \left( { \dfrac { 5 }{ 2 } } \right) }^{ 2 } }+{ { \left( { \dfrac { 4 }{ 5 } } \right) }^{ 2 } }+{ { \left( { \dfrac { 1 }{ 2 } } \right) }^{ 2 } } } } } \\ \dfrac { { 25 } }{ { \sqrt { 714 } } } ,\dfrac { 8 }{ { \sqrt { 714 } } } ,\dfrac { 5 }{ { \sqrt { 714 } } } \end{array}$

Let

$\overrightarrow{p}=3a{x}^{2}\hat{i}-2\left(x-1\right)\hat{j}$ and

$\overrightarrow{q}=b\left(x-1\right)\hat{i}+x\hat{j}$ . If

$ab<0$ then

$\overrightarrow{p}$ and

$\overrightarrow{q}$ are parallel for

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atleast one $x$ is $\left(0,1\right)$
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atleast one $x$ is $\left(-1,0\right)$
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atleast one $x$ is $\left(1,2\right)$
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none of these

Explanation

$\overrightarrow{p},\overrightarrow{q}$ are parallel if

$\overrightarrow{p}\times \overrightarrow{q}=0$ and

$\overrightarrow{p}\times \overrightarrow{q}$

$=\left|\begin{matrix}\hat{i}& \hat{j} & \hat{k}\\ 3a{x}^{2} &-2\left(x-1\right)&0 \\ b\left(x-1\right) & x & 0 \end{matrix}\right|$

$=\hat{i}\left(0-0\right)-\hat{j}\left(0-0\right)+\hat{k}\left(3a{x}^{3}+2b{\left(x-1\right)}^{2}\right)$

$=\hat{k}\left(3a{x}^{3}+2b{\left(x-1\right)}^{2}\right)$

$\left|\overrightarrow{p}\times\overrightarrow{q}\right|=3a{x}^{3}+2{\left(x-1\right)}^{2}b=f\left(x\right)$

$f\left(0\right)=2b$ ,

$f\left(1\right)=3a$

$f\left(0\right).f\left(1\right)=2b\times 3a=6ab<0$ (given)

$\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are non-coplanar and

$\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\alpha\overrightarrow{d}$ ,

$\overrightarrow{b}+\overrightarrow{c}+\overrightarrow{d}=\beta\overrightarrow{a}$ then

$\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}+\overrightarrow{d}=$

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$0$
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$\alpha\overrightarrow{a}$
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$\beta\overrightarrow{b}$
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$\left(\alpha+\beta\right)\overrightarrow{c}$

Explanation

$\overrightarrow{a} +\overrightarrow{b} +\overrightarrow{c} =\alpha\overrightarrow{d} \left(given\right)$

$\overrightarrow{b} +\overrightarrow{c} +\overrightarrow{d} =\beta\overrightarrow{a} \left(given\right)$
Eliminating

$\overrightarrow{d}$

$\Rightarrow \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\alpha\overrightarrow{d}$

$\Rightarrow \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\alpha\left(\beta\overrightarrow{a}-\overrightarrow{b}-\overrightarrow{c}\right)$

$\Rightarrow \left(1-\alpha\beta\right)\overrightarrow{a}+\left(1+\alpha\right)\overrightarrow{b}+\left(1+\alpha\right)\overrightarrow{c}=0$
since

$\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are non-coplanar

$\Rightarrow 1-\alpha\beta=0, 1+\alpha=0$

$\Rightarrow \alpha\beta=1, \alpha=-1$

$\Rightarrow \alpha=-1$

$\therefore \beta=-1$

$\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\alpha\overrightarrow{d}=-\overrightarrow{d}$

$\therefore \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}+\overrightarrow{d}=0$

If the angle between the line

$x = \dfrac { y - 1 } { 2 } = \dfrac { z - 3 } { \lambda }$ and the plane

$x + 2 y + 3 z = 4 \text { is } \cos ^ { - 1 } \left( \sqrt { \frac { 5 } { 14 } } \right)$ , then

$\lambda$ equals:-

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$\frac{2}{5}$
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$\frac{5}{3}$
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$\frac{2}{3}$
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$\frac{3}{2}$

Explanation

We have,

$x = \dfrac{{y - 1}}{2} = \dfrac{{z - 3}}{\lambda }\,\,\,\,\quad x + 2y + 3z = 4$

$= {\cos ^{ - 1}}\left( {\dfrac{{\sqrt 5 }}{{14}}} \right)\,\,\, \quad \lambda = ?$

$= \cos = \dfrac{{\sqrt 5 }}{{14}}$

$\theta = {\sin ^{ - 1}}\sqrt {1 - \dfrac{5}{{14}}} = {\sin ^{ - 1}}\left( {\dfrac{{\sqrt 9 }}{{14}}} \right)$

$\sin \theta =\sqrt { \dfrac { 9 }{ { 14 } } }$

$\Rightarrow\dfrac { { 1+4+3\lambda } }{ { \sqrt { 5+{ \lambda ^{ 2 } } } \sqrt { 14 } } } = \dfrac { 3 }{ { \sqrt { 14 } } }$ ..................obtained from the cross product relation of two normal vectors of the line and the plane.

$\Rightarrow \dfrac { { 5+3\lambda } }{ { \sqrt { 5+{ \lambda ^{ 2 } } } \sqrt { 14 } } } =\dfrac{3}{\sqrt {14}}$

$\Rightarrow 25+9{ \lambda ^{ 2 } }+30\lambda =9\left( { 5+\lambda^2 } \right)$

$\Rightarrow 25 + 30\lambda =45$

$20 = 30\lambda$

$\lambda = \dfrac{2}{3}$

Then,

Option

$C$ is correct answer.

If the points whose position vectors are

$2i+j+k, 6i-j+2k$ and

$14i-5j+pk$ are collinear, then the value of p is?

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$2$
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$4$
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$6$
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$8$

Explanation

$Positive\, \, vector\, \, are\, \, \, 2i+j+k,6\hat { i } +\hat { j } +2\hat { k } \\ and\, \, 4i-5j+pk\, \, are\, \, collinear\, \, then\, \, P=2 \\ if\, \, three\, \, position\, \, vectors\, \, are\, \, collinear\, \, then,\, \, its\, \, { { determinantsis } }\left( 0 \right) \\ \Rightarrow \left| \begin{matrix} 2\, \, \, \, \, \, \, 1\, \, \, \, \, \, \, 1 \\ 6\, \, \, \, -1\, \, \, \, \, \, 2 \\ 14\, \, -5\, \, \, \, P \\ \end{matrix} \right| =0 \\ \Rightarrow 2\left( { -P+10 } \right) -1\left( { 6P-28 } \right) +\left( { -30+14 } \right) =0 \\ \Rightarrow -2P+20-6P+28-16=0 \\ \Rightarrow -8P+32=0 \\ \therefore P=4\,$

The projection of a vector on three coordinate axes are

$6,-3, 2$ respectively. The direction cosines of the vector are

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$\left(6,-3,2\right)$
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$\left(\dfrac{6}{5},\dfrac{-5}{5},\dfrac{2}{5} \right)$
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$\left(\dfrac{6}{7},\dfrac{-3}{7},\dfrac{2}{7}\right)$
0%
$\left(\dfrac{-6}{7},\dfrac{-3}{7},\dfrac{2}{7}\right)$

Explanation

$\cos{\alpha},\cos{\beta},\cos{\gamma}$ are the direction cosines then

$r\cos{\alpha}=6, r\cos{\beta}=-3, r\cos{\gamma}=2$
Squaring ,

${r}^{2}{\cos}^{2}{\alpha}=36,{r}^{2}{\cos}^{2}{\beta}=9,{r}^{2}{\cos}^{2}{\gamma}=4$

$\Rightarrow {r}^{2}{\cos}^{2}{\alpha}+{r}^{2}{\cos}^{2}{\beta}+{r}^{2}{\cos}^{2}{\gamma}=\left(36+9+4\right)=49$

$\Rightarrow {r}^{2}\left({\cos}^{2}{\alpha}+{\cos}^{2}{\beta}+{\cos}^{2}{\gamma}\right)=49$

$\Rightarrow {r}^{2}=49 , r=7$

$\therefore$ the direction cosines are

$\dfrac{6}{7},\dfrac{-3}{7},\dfrac{2}{7}$

If a line makes angles

$\alpha ,\beta$ &

$\gamma$ with

$OX,OY$ &

$OZ$ respectively then

$\cos ^{ 2 }{ \alpha } +\cos ^{ 2 }{ \beta } +\cos ^{ 2 }{ \gamma } =-1$

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True
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False

The vector form of the equation of the line passing through points

$(3,4, 7)$ and

$(5,1,6)$ is-

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$\vec { r } =(3\hat { i } +4\hat { j } -7\hat { k } )+\lambda (2\hat { i } -3\hat { j } +13\hat { k } )$
0%
$\vec { r } =(3\hat { i } +4\hat { j } -7\hat { k } )+\lambda (8\hat { i } +5\hat { j } -\hat { k } )$
0%
$\vec { r } =(3\hat { i } +4\hat { j } +7\hat { k } )+\lambda (2\hat { i } -3\hat { j } -\hat { k } )$
0%
$\vec { r } =(3\hat { i } +4\hat { j } -7\hat { k } )+\lambda (2\hat { i } -3\hat { j } -13\hat { k } )$

Explanation

We have,

Given points are $\left( 3,4,7 \right)\,\,and\,\,\left( 5,1,6 \right)$ .

We know that,

Vector equation of a line passing through the two points with position vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ is

$\overrightarrow{r}=\overrightarrow{a}+\lambda \left( \overrightarrow{b}-\overrightarrow{a} \right)$

Let the given points are $A\left( 3,4,7 \right)\,\,and\,\,\left( 5,1,6 \right)$

Now,

Equation of vector,

$\overrightarrow{r}=\left( 3\overrightarrow{i}+4\overrightarrow{j}+7\overrightarrow{k} \right)+\lambda \left( 5\overrightarrow{i}+\overrightarrow{j}+6\overrightarrow{k}-3\overrightarrow{i}-4\overrightarrow{j}-7\overrightarrow{k} \right)$

$\overrightarrow{r}=\left( 3\overrightarrow{i}+4\overrightarrow{j}+7\overrightarrow{k} \right)+\lambda \left( 2\overrightarrow{i}-3\overrightarrow{j}-\overrightarrow{k} \right)$

Hence, this is the answer.

The line perpendicular to the plane

$2x-y+5z=4$ passing through the point

$(-1,0,1)$ is ?

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$\dfrac{x+1}{2}=-y=\dfrac{z-1}{-5}$
0%
$\dfrac{x+1}{-2}=y=\dfrac{z-1}{-5}$
0%
$\cfrac{x+1}{2}=-y=\cfrac{z-1}{5}$
0%
$\dfrac{x+1}{2}=y=\dfrac{z-1}{-5}$

Explanation

$\begin{array}{l} Po{ { int } }\, \, A\left( { -1,0,1 } \right) \, and\, \, \, B\left( { 2,-1,5 } \right) \\ So,\, \, equation\, \, of\, \, line. \\ \frac { { x+1 } }{ 2 } =-y=\frac { { z-1 } }{ 5 } \\ Option\, \, \, C\, \, is\, \, correct\, \, answer. \end{array}$

If a straight line makes an angle

${ cos }^{ -1 }\left( \frac { 1 }{ \sqrt { 3 } } \right)$ with each of the positive

$x, y$ and

$z$ -axis, a vector parallel to that line is

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$\overset { - }{ i }$
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$\overline { i } +\overline { j }$
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$\overline { j } +\overline { k }$
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$\overline { i } +\overline { j } +\overline { k }$

A line makes angles

$\alpha, \beta, \gamma$ with the positive direction of the axes of reference. The value of

$\cos{2\alpha}+\cos{2\beta}+\cos{2\gamma}$ is

0%
$1$
0%
$3$
0%
$-1$
0%
$0$

Explanation

$\cos ^{2}\alpha+\cos^{2}\beta+\cos^{2}r=1$

$\cos 2\alpha+\cos ^{2}\beta+\cos 2r=2\cos^{2}\alpha-1+2\cos^{2}\beta-1+2\cos^{2}r-1$

$=2(\cos^{2}\alpha+\cos^{2}\beta+\cos^{2}r)-3$

$=2(1)-3$

$=-1$

CBSE Questions for Class 12 Commerce Maths Three Dimensional Geometry Quiz 7 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Three Dimensional Geometry Quiz 7 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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