MCQ Questions for Class 12 Commerce Maths Three Dimensional Geometry Quiz 3

Find the direction cosines of vector

$\overrightarrow { r }$ which is equally inclined to

$OX,OY$ and

$OZ$ . Find total number of such vectors.

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$\displaystyle \dfrac { 1 }{ \sqrt { 3 } } ,\dfrac { 1 }{ \sqrt { 3 } } ,\dfrac { 1 }{ \sqrt { 3 } } ;6$
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$\displaystyle \dfrac { 1 }{ \sqrt { 3 } } ,\pm \dfrac { 1 }{ \sqrt { 3 } } ,\dfrac { 1 }{ \sqrt { 3 } } ;8$
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$\displaystyle \pm \dfrac { 1 }{ \sqrt { 3 } } ,\pm \dfrac { 1 }{ \sqrt { 3 } } ,\pm \dfrac { 1 }{ \sqrt { 3 } } ;8$
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None of these

Explanation

Let $l,m,n$ be the direction cosines of $\overrightarrow { r }$ .

Since $\overrightarrow { r }$ is equally inclined with $x,y$ and $z$ axis, $l=m=n$

$\displaystyle \therefore { l }^{ 2 }+{ m }^{ 2 }+{ n }^{ 2 }\Rightarrow 3{ l }^{ 3 }=1\Rightarrow l=\pm \dfrac { 1 }{ \sqrt { 3 } }$

$\therefore$ direction cosines of $\overrightarrow { r }$ are $\displaystyle \pm \dfrac { 1 }{ \sqrt { 3 } } ,\pm \dfrac { 1 }{ \sqrt { 3 } } ,\pm \dfrac { 1 }{ \sqrt { 3 } }$

Now, $\displaystyle \overrightarrow { r } =\left| \overrightarrow { r } \right| \left( li+mj+nk \right) =\overrightarrow { r } =\left| \overrightarrow { r } \right| \left( \pm \dfrac { 1 }{ \sqrt { 3 } } i\pm \dfrac { 1 }{ \sqrt { 3 } } j\pm \dfrac { 1 }{ \sqrt { 3 } } k \right)$

Since $+$ and $-$ signs can be arranged at three places,

$\Rightarrow$ there are eight vectors, i.e $2\times 2\times 2$ which are equally inclined to axes.

$\theta$ is the angle between two lines whose d.c.s are

$l_{1},m_{1},n_{1}$ and

$l_{2}, m_{2}, n_{2}$ , then the d.cs of one of the angular bisectors of the two lines are

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$\displaystyle \dfrac{l_{1}+l_{2}}{2}, \displaystyle \dfrac{m_{1}+m_{2}}{2}, \displaystyle \dfrac{n_{1}+n_{2}}{2}$
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$\displaystyle \dfrac{l_{1}+l_{2}}{2\cos(\dfrac{\theta}{2})},\dfrac{m_{1}+m_{2}}{2\cos(\dfrac{\theta}{2})},\dfrac{n_{1}+n_{2}}{2\cos(\dfrac{\theta}{2})}$
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$\displaystyle \dfrac{l_{1}+l_{2}}{\cos (\dfrac{\theta}{2})},\dfrac{m_{1}+m_{2}}{\cos(\dfrac{\theta}{2})},\dfrac{n_{1}+n_{2}}{\cos(\dfrac{\theta}{2})}$
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$\displaystyle \dfrac{l_{\mathrm{I}}+l_{2}}{2\sin(\dfrac{\theta}{2})}\dfrac{m_{1}+m_{2}}{2\sin(\dfrac{\theta}{2})}\dfrac{n_{]}+n_{2}}{2\sin(\dfrac{\theta}{2})}$

Explanation

Let the direction cosines of the line be $l_{1},m_{1},n_{1}$ .

Hence, $l_{1}=\cos\alpha$ , $m_{1}=\cos\beta$ , $n_{1}=\cos\gamma$

Similarly, for another line

$l_{2}=\cos\alpha'$ , $m_{2}=\cos\beta'$ , $n_{2}=\cos\gamma'$

Now the direction cosines of the angle bisector will be,

$\displaystyle \cos\left(\dfrac{\alpha-\alpha'}{2}+\alpha'\right),\cos \left(\dfrac{\beta-\beta'}{2}+\beta' \right),\cos\left(\dfrac{\gamma-\gamma'}{2}+\gamma'\right)$

$=\cos\left(\dfrac{\alpha+\alpha'}{2}\right),\cos\left(\dfrac{\beta+\beta'}{2}\right),\cos\left(\dfrac{\gamma+\gamma'}{2}\right)$

$=l,m,n$

Now consider $\cos\left(\dfrac{\alpha+\alpha'}{2}\right)$

Multiplying and dividing by $\displaystyle 2\cos\left(\dfrac{\alpha-\alpha'}{2}\right)$ , we get

$\displaystyle \dfrac{2\cos\left(\dfrac{\alpha-\alpha'}{2}\right)\cos\left(\dfrac{\alpha+\alpha'}{2}\right)}{2\cos\left(\dfrac{\alpha-\alpha'}{2}\right)}$

$=\displaystyle \dfrac{\cos\alpha+\cos\alpha'}{2\cos(\dfrac{\alpha-\alpha'}{2})}$

$=\displaystyle \dfrac{l_{1}+l_{2}}{2\cos(\dfrac{\alpha-\alpha'}{2})}$

$=l$

Hence, $l = \dfrac{l_1+l_2}{2 \cos \left(\dfrac{\theta}{2} \right) }$

Similarly,

$m=\displaystyle \dfrac{m_{1}+m_{2}}{2\cos(\dfrac{\beta-\beta'}{2})}$

$n=\displaystyle \dfrac{n_{1}+n_{2}}{2\cos(\dfrac{\gamma-\gamma'}{2})}$

Hence, option B is correct.

$l_{1},m_{1},n_{1}$ and

$1_{2},m_{2},n_{2}$ are the direction cosines of two lines, then

$(l_{1}l_{2}+m_{1}m_{2}+n_{1}n_{2})^{2}+\displaystyle\sum(m_{1}n_{2}-m_{2}n_{1})^{2}=$

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$0$
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$-1$
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$1$
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$2$

Explanation

Dot product of dc's of two lines gives,

$(l_{ 1 }\hat i+m_{ 1 }\hat j+n_{ 1 }\hat k).(l_{ 2 }\hat i+m_{ 2 }\hat j+n_{ 2 }\hat k)=\sqrt { { l_{ 1 } }^{ 2 }+{ m_{ 1 } }^{ 2 }+{ n_{ 1 } }^{ 2 } } \times \sqrt { { l_{ 2 } }^{ 2 }+{ m_{ 2 } }^{ 2 }+{ n_{ 2 } }^{ 2 } } \times \cos\theta$
We know

${ l_{ 1 } }^{ 2 }+{ m_{ 1 } }^{ 2 }+{ n_{ 1 } }^{ 2 }=1,\quad { l_{ 2 } }^{ 2 }+{ m_{ 2 } }^{ 2 }+{ n_{ 2 } }^{ 2 }=1$

$\Rightarrow l_{ 1 }l_{ 2 }+m_{ 1 }m_{ 2 }+n_{ 1 }n_{ 2 }=1.1.\cos\theta \Rightarrow { (l_{ 1 }l_{ 2 }+m_{ 1 }m_{ 2 }+n_{ 1 }n_{ 2 }) }^{ 2 }={ \cos }^{ 2 }\theta$
Cross product of dc's of two lines

$\left| (l_{ 1 }\hat i+m_{ 1 }\hat j+n_{ 1 }\hat k)\times (l_{ 2 }\hat i+m_{ 2 }\hat j+n_{ 2 }\hat k) \right| =\left| \sqrt { { l_{ 1 } }^{ 2 }+{ m_{ 1 } }^{ 2 }+{ n_{ 1 } }^{ 2 } } \times \sqrt { { l_{ 2 } }^{ 2 }+{ m_{ 2 } }^{ 2 }+{ n_{ 2 } }^{ 2 } } \times \sin\theta \right|$

$\Rightarrow \sqrt { \Sigma (m_{ 1 }n_{ 2 }-m_{ 2 }n_{ 1 })^{ 2 } } =\sin\theta \Rightarrow \Sigma (m_{ 1 }n_{ 2 }-m_{ 2 }n_{ 1 })^{ 2 }={ \sin }^{ 2 }\theta$

${ (l_{ 1 }l_{ 2 }+m_{ 1 }m_{ 2 }+n_{ 1 }n_{ 2 }) }^{ 2 }+\Sigma (m_{ 1 }n_{ 2 }-m_{ 2 }n_{ 1 })^{ 2 }={ \cos }^{ 2 }\theta +{ \sin }^{ 2 }\theta =1$

The coordinates of a point P are

$(3,12,4)$ w.r.t origin O, then the direction cosines of

$OP$ are

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$3,12,4$
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$\displaystyle \dfrac{1}{4}, \dfrac{1}{3}, \dfrac{1}{2}$
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$\displaystyle \dfrac{3}{\sqrt{13}},\dfrac{1}{\sqrt{13}},\dfrac{2}{\sqrt{13}}$
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$\displaystyle \dfrac{3}{13},\dfrac{12}{13},\dfrac{4}{13}$

Explanation

Given points are

$O(0,0,0)=(x_1,y_1,z_1)$ and

$P(3,12,4)=(x_2,y_2,z_2)$

Direction cosines of

$OP$ is given by

$\cos \alpha=\dfrac{x_2-x_1}{\sqrt{x_2^2+y_2^2+z_2^2}}$

$\cos \beta=\dfrac{y_2-y_1}{\sqrt{x_2^2+y_2^2+z_2^2}}$

$\cos \gamma=\dfrac{z_2-z_1}{\sqrt{x_2^2+y_2^2+z_2^2}}$

$\implies$

$\cos \alpha=\dfrac{3-0}{\sqrt{3^2+12^2+4^2}}=\dfrac{3}{\sqrt{169}}=\dfrac{3}{13}$

$\implies$

$\cos \beta=\dfrac{12-0}{\sqrt{3^2+12^2+4^2}}=\dfrac{12}{\sqrt{169}}=\dfrac{12}{13}$

$\implies$

$\cos \gamma=\dfrac{4-0}{\sqrt{3^2+12^2+4^2}}=\dfrac{4}{\sqrt{169}}=\dfrac{4}{13}$

Hence, the direction cosines are

$\dfrac{3}{13},\dfrac{12}{13},\dfrac{4}{13}$ .

$ox, oy$ are positive

$\mathrm{x}$ -axis, positive

${y}$ -axis respectively where

${O}=(0, 0,0)$ . The

${d.c.}$ s of the llne which bisects

$\angle xoy$ are

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$1, 1, 0$
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$\displaystyle \dfrac{1}{\sqrt{2}}, \displaystyle \dfrac{1}{\sqrt{2}},0$
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$\displaystyle \dfrac{1}{\sqrt{2}},0, \displaystyle \dfrac{1}{\sqrt{2}}$
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$0, 0, 1$

Explanation

Equation of line bisecting

$XOY$ is

$x=y$
Therefore, d.r. s are

$(1,1,0)$
And thus d.c. s are

$\left( \dfrac { 1 }{ \sqrt { 2 } } ,\dfrac { 1 }{ \sqrt { 2 } } ,0 \right)$
Ans: B

$(2, 1, 3)$ and

$(-1, 2, 4)$ are the extremities of a diagonal of a rhombus then the d.rs of the other diagonal are

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$2, 3, -9$
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$-2, 3, -9$
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$1, -2, 4$
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$2, 3,1$

Explanation

$(2, 1, 3)$ and

$(-1, 2, 4)$ are extremities of a diagonal of a rhombus.
Hence, the direction ratios of that diagonal are equal to

$2-(-1), 1-2, 3-4$ .
So, the direction ratios are

$3, -1, -1$ .
In rhombus the diagonals are perpendicular to each other. Suppose the direction ratios of the other diagonal are

$l. m , n$ . So,

$3l -m - n$ should be equal to zero.

$3l-m-n = 0$
Only option B satisfies this condition. Hence the direction ratios are

$-2,\; 3\;, -9\;$ .

$A = (1, 2, 3), B = (4, 5, 7), C = (-4, 3, -6), D =(2, k, 2)$ are four points. If the lines

$AB$ and

$CD$ are parallel, then

$k =$

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$0$
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$-9$
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$9$
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$2$

Explanation

Given,

$A=(1,2,3), B=(4,5,7), C=(-4,3,-6), D=(2,k,2)$
For

$AB$ to be parallel to

$CD$ , their direction cosines should be equal.
Therefore, Dcs of

$AB$ will be

$(\dfrac{3}{\sqrt{34}},\dfrac{ 3}{\sqrt{34}},\dfrac{4}{\sqrt{34}})$

D.C.s of

$CD$ will be

$\left (\dfrac{6}{\sqrt{100+(k-3)^{2}}},\dfrac{k-3}{\sqrt{100+(k-3)^{2}}},\dfrac{8}{\sqrt{100+(k-3)^{2}}}\right)$
Now

$\dfrac{6}{\sqrt{100+(k-3)^{2}}}=\dfrac{3}{\sqrt{34}}$

$\dfrac{k-3}{\sqrt{100+(k-3)^{2}}}=\dfrac{ 3}{\sqrt{34}}$

$\dfrac{8}{\sqrt{100+(k-3)^{2}}}=\dfrac{4}{\sqrt{34}}$

Solving any of the above can give us the value of

$k$ :

$\dfrac{6}{\sqrt{100+(k-3)^{2}}}=\dfrac{3}{\sqrt{34}}$

$4(34)=100+(k-3)^{2}$

$136=100+(k-3)^{2}$

$36=(k-3)^{2}$

$k-3=\pm6$

$k=9$ and

$k=-3$

lf a line makes angles

$\alpha, \beta,\gamma$ with

$OX, OY, OZ$ respectively where

${O}=(0,0,0)$ , then

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

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$1$
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$0$
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$2$
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$-1$

Explanation

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

$2 \times [(\cos\alpha )^2+(\cos\beta )^2+(\cos\gamma )^2]=2$

$2 \times [(\cos\alpha )^2+(\cos\beta )^2+(\cos\gamma )^2]-3=-1$

$2(\cos\alpha )^2-1+2(\cos\beta )^2-1+2(\cos\gamma )^2-1]=-1$

$(\cos2\alpha )+(\cos2\beta )+(\cos2\gamma )=-1$

The direction cosines of the line passing through

$\mathrm{P}(2,3,-1)$ and the origin are

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$\displaystyle \dfrac{2}{\sqrt{14}},\dfrac{3}{\sqrt{14}},\dfrac{1}{\sqrt{14}}$
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$\displaystyle \dfrac{2}{\sqrt{14}},\dfrac{-3}{\sqrt{14}},\dfrac{1}{\sqrt{14}}$
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$\displaystyle \dfrac{-2}{\sqrt{14}},\dfrac{-3}{\sqrt{14}},\dfrac{1}{\sqrt{14}}$
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$\displaystyle \dfrac{2}{\sqrt{14}},\dfrac{-3}{\sqrt{14}},\dfrac{-1}{\sqrt{14}}$

Explanation

Given points are

$P(2,3,-1)=(x_1,y_1,z_1)$ and

$(0,0,0)=(x_2,y_2,z_2)$

Direction cosines of a line passing through these two points is given by

$\cos \alpha=\dfrac{x_2-x_1}{\sqrt{x_1^2+y_1^2+z_1^2}}$

$\cos \beta=\dfrac{y_2-y_1}{\sqrt{x_1^2+y_1^2+z_1^2}}$

$\cos \gamma=\dfrac{z_2-z_1}{\sqrt{x_1^2+y_1^2+z_1^2}}$

$\implies$

$\cos \alpha=\dfrac{0-2}{\sqrt{2^2+3^2+(-1)^2}}=\dfrac{-2}{\sqrt{14}}$

$\cos \beta=\dfrac{0-3}{\sqrt{2^2+3^2+(-1)^2}}=\dfrac{-3}{\sqrt{14}}$

$\cos \gamma=\dfrac{0-(-1)}{\sqrt{2^2+3^2+(-1)^2}}=\dfrac{1}{\sqrt{14}}$

Hence, the direction cosines are

$\dfrac{-2}{\sqrt{14}},\dfrac{-3}{\sqrt{14}},\dfrac{1}{\sqrt{14}}$ .

$A$ is

$(2, 4, 5),$ and

$B$ is

$(-7, -2, 8)$ , then which of the following is collinear with

$A$ and

$B$ is

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$(1, 2, 6)$
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$(2, -1,6)$
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$(-1, 2, 6)$
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$(2, 6, -1)$

Explanation

Line passing through

$A(2,4,5)$ and

$B(-7,-2,8)$ is

$L: \dfrac{x-2}{9} = \dfrac{y-4}{6} = \dfrac{z-5}{-3}$
Lets suppose

$O$ is collinear with

$A$ and

$B$ , then

$L = k$

$\therefore$ Coordinates of

$O$ are

$(2+9k,4+6k,5-3k)$
Looking at options, only option C satisfies the coordinates of

$O$ for

$k = -\dfrac{1}{3}$

lf a line makes

$\dfrac{\pi }{3},\dfrac{\pi }{4}$ with the

$x$ -axis,

${y}$ -axis respectively, then the angle made by that line with the

$z$ - axis is

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$\displaystyle \frac{\pi}{2}$
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$\displaystyle \frac{\pi}{3}$
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$\displaystyle \frac{\pi}{4}$
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$\displaystyle \frac{5\pi}{12}$

Explanation

We have, $\left (\cos\left (\dfrac {\pi}{3}\right)\right)^{2}+\left (\cos\left (\dfrac {\pi}{4}\right)\right)^{2}+(\cos(\gamma))^{2}=1$
$\Rightarrow \cos^{2}\gamma = \dfrac {1}{4}$

$\displaystyle \Rightarrow \cos \gamma = \pm \frac{1}{2}$
$\Rightarrow \gamma=\dfrac {\pi}{3}$ or $\dfrac {2 \pi}{3}$

$l, m, n$ are the d.cs of the line joining

$(5, -3, 8)$ and

$(6, -1, 6)$ then

$l + m + n=$

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$1$
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$\displaystyle \dfrac{1}{3}$
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$-1$
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$\displaystyle \dfrac{5}{3}$

Explanation

The line joining

$(5,-3,8)$ and

$(6,-1,6)$ is given by the vector

$-i+2j-2k.$

Hence, the direction cosines are given by:

$l=\cfrac{1}{\sqrt{1^2+2^2+(-2)^2} }=\cfrac{1}{3}$ ,

$m=\cfrac{2}{\sqrt{1^2+2^2+(-2)^2}}=\cfrac{2}{3}$ ,

$n=\cfrac{-2}{\sqrt{1^2+2^2+(-2)^2}}=\cfrac{-2}{3}$

$\Rightarrow l+m+n=\cfrac{1}{3}$

$\dfrac{1}{2}, \displaystyle \dfrac{1}{2}$ ,

$n(n<0)$ are the dcs of a line, then the angle made by that line with

$OZ$ where

$O=(0,0,0)$ is

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$\displaystyle \dfrac{-1}{\sqrt{2}}$
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$45^{o}$
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$60^{o}$
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$135^{o}$

Explanation

Given

$l=\dfrac{1}{2}, m=\dfrac{1}{2} , n$ as the d.c's of a line.
Let

$\gamma$ be the angle which the line makes with

$OZ$ axis i.e.

$\cos\gamma=n$
We know that

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

$\Rightarrow n^2=\dfrac{1}{2}$

$\Rightarrow n=\pm \dfrac{1}{\sqrt{2}}$
Since,

$n<0$

$\Rightarrow n=-\dfrac{1}{\sqrt{2}}$

$\Rightarrow \cos \gamma =-\dfrac{1}{\sqrt{2}}$

$\Rightarrow \gamma =-\cos {45^0}$

$\Rightarrow \gamma =135^{0}$

If the direction ratios of two lines are given by

$3lm-4ln+mn=0$ and

$l+2m+3n=0$ , then the angle between the lines is

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$\displaystyle \dfrac { \pi }{ 6 }$
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$\displaystyle \dfrac { \pi }{ 4 }$
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$\displaystyle \dfrac { \pi }{ 3 }$
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$\displaystyle \dfrac { \pi }{ 2 }$

Explanation

Given $3lm-4ln+mn=0$ ... $(1)$

and $l+2m+3n=0$ .... $(2)$

From equation $(2)$ , $l=-(2m+3n)$ , putting in equation $(1)$ , we get

$-3\left( 2m+3n \right) m+4\left( 2m+3n \right) n+mn=0\\ \Rightarrow -6{ m }^{ 2 }+12{ n }^{ 2 }=0\Rightarrow m=\pm \sqrt { 2 } n$

Now, $m=\pm \sqrt { 2 } n\Rightarrow l=-\left( 2\sqrt { 2 } n+3n \right) =-\left( 2\sqrt { 2 } +3 \right) n$

$\therefore l:m:n=-\left( 3+2\sqrt { 2 } \right) n:\sqrt { 2 } n:n=-\left( 3+2\sqrt { 2 } \right) :-\sqrt { 2 } :1$

Also, $m=-\sqrt { 2 } n\Rightarrow l=-\left( -2\sqrt { 2 } +3 \right) n$

$\therefore l:m:n=-\left( 3-2\sqrt { 2 } \right) n:-\sqrt { 2 } n:n=-\left( 3-2\sqrt { 2 } \right) :-\sqrt { 2 } :1$

$\displaystyle \Rightarrow \cos { \theta } =\dfrac { \left( 3+2\sqrt { 2 } \right) \left( 3-2\sqrt { 2 } \right) +\left( \sqrt { 2 } \right) \left( -\sqrt { 2 } \right) +1.1 }{ \sqrt { { \left( 3+2\sqrt { 2 } \right) }^{ 2 }+{ \left( \sqrt { 2 } \right) }^{ 2 }+{ 1 }^{ 2 } } \sqrt { { \left( 3-2\sqrt { 2 } \right) }^{ 2 }+{ \left( -\sqrt { 2 } \right) }^{ 2 }+{ 1 }^{ 2 } } }$

$\displaystyle \Rightarrow \theta =\dfrac { \pi }{ 2 }$

$P = (3, 4, 5)$ ,

$Q= (4, 6, 3)$ ,

$R= (-1, 2, 4)$ and

$S=(1, 0, 5)$ are four points, then the projection of

$RS$ on

$PQ$ is

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$\displaystyle \dfrac{8}{3}$
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$\displaystyle \dfrac{4}{3}$
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$4$
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$0$

Explanation

Let $P(3,4,5)$ $Q(4,6,3)$ $R(-1,2,4)$ and $S(1,0,5)$ be any four points.

The dr's of the line $PQ$ are $(1,2,-2)$

The dr's of the line $RS$ are $(2,-2,1)$

The projection of $RS$ on $PQ$ will be $RS \times \cos \theta$

$RS \cos\theta$ $=$ $\displaystyle \dfrac{PQ \times RS}{|PQ|}$

$=$ $\displaystyle \dfrac{(i+2j-2k).(2i-2j+k)}{3}$ $=$ $\displaystyle \dfrac{4}{3}$

If the projections of the line segment

$AB$ on the coordinate axes are

$2, 3, 6$ , then the square of the sine of the angle made by

$AB$ with

$x=0$ , is

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$\displaystyle \dfrac{3}{7}$
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$\displaystyle \dfrac{3}{49}$
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$\displaystyle \dfrac{4}{7}$
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$\displaystyle \dfrac{40}{49}$

Explanation

Projections of the line segment

$AB$ on coordinate axes are

$2,3,$ and

$6$

$\therefore$ the line can be

$L: 2\hat i + 3\hat j + 6\hat k$
Angle of line

$L$ with

$x = 0\>i.e.\>\hat j$ is:

$\cos\theta = \dfrac{L\cdot\hat j}{|L||j|} = \dfrac{(2\hat i + 3\hat j + 6\hat k)\cdot\hat j}{|(2\hat i + 3\hat j + 6\hat k)||j|} = \dfrac{3}{7}$

$\sin^2\theta = 1 - \cos^2\theta = 1 - \left(\dfrac{3}{7}\right)^2 = \dfrac{40}{49}$

If the d.rs of

$OA$ and

$OB$ are

$1, -1, -1$ and

$2, -1, 1$ , then the d.cs of the line perpendicular to both

$OA$ and

$OB$ are

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$0,1, -1$
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$-2, -3,1$
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$\displaystyle \dfrac{-2}{\sqrt{14}},\dfrac{-3}{\sqrt{14}},\dfrac{1}{\sqrt{14}}$
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$\displaystyle \dfrac{2}{\sqrt{41}},\dfrac{3}{\sqrt{41}},\dfrac{-1}{\sqrt{41}}$

Explanation

The Directionsratios of

$OA$ will be,

$(\dfrac{1}{\sqrt{3}},\dfrac{-1}{\sqrt{3}}, \dfrac{-1}{\sqrt{3}})$
The Direction cosines of

$OB$ will be,

$(\dfrac{2}{\sqrt{6}},\dfrac{-1}{\sqrt{6}}, \dfrac{1}{\sqrt{6}})$
Now the lines should be such that the product of Dcs will be

$0$ .
This is only satisfied by Option C

The projection of the line segment joining

$(0, 0, 0)$ and

$(5, 2, 4)$ on the line whose direction ratios are

$2, -3, 6$ is

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$28$
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$4$
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$\displaystyle \dfrac{40}{7}$
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$\sqrt{45}$

Explanation

The line segment passing through the points

$(0,0,0)$ and

$(5,2,4)$ is parallel to

$\vec{a}=5i+2j+4k$ . The dr's of the line (as given ) is

$\vec{l}=2i-3j+6k$
Hence the required projection is

$=\dfrac{\vec{a}.\vec{l}}{|\vec{l}|}$

$=\dfrac{10-6+24}{\sqrt{4+36+9}}$

$=\dfrac{28}{7}$

$=4$ .

${P}(x,y,z)$ is a point on the line segment joining

${A}(2,2,4)$ and

${B}(3,5,6)$ such that projection of

$\overline{OP}$ on axes are

$\displaystyle \dfrac{13}{5},\dfrac{19}{5},\dfrac{26}{5}$ respectively, then

${P}$ divide

${A}{B}$ in the ratio

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$3: 2$
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$2 : 3$
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$1 : 2$
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$1 : 3$

Explanation

$P\left( x,y,z \right)$ lies on line joining

$A\left( 2,2,4 \right) ,B\left( 3,5,6 \right)$

Equation of

$AB:\dfrac { x-2 }{ 3-2 } =\dfrac { y-2 }{ 5-2 } =\dfrac { z-4 }{ 6-4 }$

$AB:\dfrac { x-2 }{ 1 } =\dfrac { y-2 }{ 3 } =\dfrac { z-4 }{ 2 } =t$

$\Rightarrow x=t+2,\quad y=3t+2,\quad z=2t+4$

$\overrightarrow { OP } =\left( t+2 \right) \hat { i } +\left( 3t+2 \right) \hat { j } +\left( 2t+4 \right) \hat { k }$

Projection of

$OP$ on axes are

$\dfrac { 13 }{ 5 } ,\dfrac { 19 }{ 5 } ,\dfrac { 26 }{ 5 }$

$\overrightarrow { OP } .\hat { i } =t+2=\dfrac { 13 }{ 5 }$

$\Rightarrow t=\dfrac { 3 }{ 5 }$

$\overrightarrow { OP } =\left( \dfrac { 13 }{ 5 } ,\dfrac { 19 }{ 5 } ,\dfrac { 26 }{ 5 } \right)$

$P$ divides

$AB$ in

$m:n$ ratio

$\Rightarrow \dfrac { 13 }{ 5 } =\dfrac { m(3)+n(2) }{ m+n }$

$13m+13n=15m+10n$

$3n=2m$

$m:n=3:2$

The projections of a line segment on

$x,y\ and\ z$ axes are respectively

$\sqrt{2},3,5$ . The length of the line segment is

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$6$
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$11$
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$8$
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$5$

Explanation

Let the line segment be represented in vector form as

$\vec a = a_1\vec i+a_2\vec j+a_3\vec k$

Vector along coordinate axes are

$\vec i,\vec j,\vec k$

Given that projection of

$\vec a$ on x axis is

$\sqrt3$ , that with y axis is

$3$ and that with z axis is

$5$ .

$\Longrightarrow \vec a . \vec i$ =

$\sqrt2$ ,

$\therefore \ a_1=\sqrt2$

similarily,

$a_2=3,a_3=5$

$\therefore \vec a=\sqrt2\vec i +3\vec j+5\vec k$

the length of the line segment=

$|\vec a| =\sqrt{2+9+25}=\sqrt36=6$

If the d.rs of two lines are

$1, -2, 3$ and

$2, 0, 1$ , then the d.rs of the line perpendicular to both the given lines is

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$-2,5, 4$
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$2,-5,4$
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$2,5,-4$
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$1,5,-4$

Explanation

$OA$ and

$OB$ are given by

$(1,-2,3), (2,0,1)$ .
A line that will be perpendicular to both

$OA$ and

$OB$ can be obtained by doing the cross product of

$OA$ with

$OB$ .

Then,

$n= OA\times OB$
Hence,

$n=-2i+5j+4k$
Hence, the dr's are

$(-2,5,4)$ .

${A}=(3,1, -2), {B}=(-1,0,1)$ and

$l,m$ are the projections of

${A}{B}$ on the

${y}$ -axis,

$zx$ -plane respectively, then

$3l^{2}-m+1=$

0%
$-1$
0%
$0$
0%
$1$
0%
$9$

Explanation

Given

$A=\left( 3,1,-2 \right) ,B=\left( -1,0,1 \right)$

DR's of

$AB$ are

$\left( 3-(-1),1-0,-2-1 \right)$

$\overrightarrow { AB } =\left( 4,1,-3 \right)$ or

$\left( 4\hat { i } +\hat { j } -3\hat { k } \right)$

Projection of

$AB$ on

$x-$ axis is

$\overrightarrow { AB } .\hat { i }$

${ a }_{ x }=4$

Projection of

$AB$ on

$y-$ axis is

$\overrightarrow { AB } .\hat { j }$

$l={ a }_{ y }=1$

Projection of

$AB$ on

$z-$ axis is

$\overrightarrow { AB } .\hat { k }$

${ a }_{ z }=-3$

Projections of

$\overrightarrow { AB }$ on

$zx$ plane

$=\sqrt { { { a }_{ x } }^{ 2 }+{ { a }_{ z } }^{ 2 } }$

$=\sqrt { { 4 }^{ 2 }+{ 3 }^{ 2 } } =5$

$3{ l }^{ 2 }-m+1=3\times 1-5+1=-1$

If the projections of the line segment

$AB$ on the coordinate axes are

$12, 3, k$ and

$AB = 13$ , then

$k^{2}-2k+3$ is equal to:

0%
$0$
0%
$1$
0%
$11$
0%
$27$

Explanation

Let a,b,c be the projection of a line on the coordinate axes.

Then the length of the line given by

$\sqrt{a^2+b^2+c^2}$ .

Here we have

$12^2+3^2+k^2=169$

$\Rightarrow k=\pm 4$ .

Thus

$k^2-2k+3=11$ or

$27$ .

The d.rs of the lines

$x= ay + b$ ,

$z= cy + d$ are:

0%
$1, a, c$
0%
$a, 1, c$
0%
$b, 1, c$
0%
$c, a, 1$

Explanation

Given

$x=ay+b$ and

$z=cy+d$

$\Rightarrow \dfrac{x-b}{a}=y$ and

$\dfrac{z-d}{c}=y$

$\Rightarrow \dfrac{x-b}{a}=\dfrac{y}{1}=\dfrac{z-d}{c}$

Therefore Drs of given line is

$a,1,c$

Find the angles between the lines, whose direction cosines are give by the equation

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

0%
$0$
0%
$\displaystyle \frac{\pi}{6}$
0%
$\displaystyle \frac{\pi}{4}$
0%
$\displaystyle \frac{\pi}{3}$

Explanation

Lines are $l+m+n=0\Rightarrow -l=\left( m+n \right)$

and ${ l }^{ 2 }-{ m }^{ 2 }+{ n }^{ 2 }=0\Rightarrow { l }^{ 2 }={ m }^{ 2 }-{ n }^{ 2 }$

Solving them gives,

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

Now, For $n=0$

$m=-l$ , so d.c's $\displaystyle \left( \dfrac { 1 }{ \sqrt { 2 } } ,-\dfrac { 1 }{ \sqrt { 2 } } ,0 \right)$

And for $n=-m$

$l=0$ so d.c's $\displaystyle \left( 0,\dfrac { 1 }{ \sqrt { 2 } } ,-\dfrac { 1 }{ \sqrt { 2 } } \right)$

Angle between the lines $\displaystyle \left| \cos { \theta } \right| =\left| \dfrac { 1 }{ 2 } \right| \Rightarrow \theta =\dfrac { \pi }{ 3 }$

$AB \perp BC$ , then the value of

$\lambda$ equal, where

$A(2k,2,3), B(k,1,5),C(3+k,2,1)$

0%
$3$
0%
$\displaystyle \dfrac{1}{3}$
0%
$-3$
0%
$-\displaystyle \dfrac{1}{3}$

Explanation

The dr's of

$AB$ are

$(k,1,-2)$
The dr's of

$BC$ are

$(3,1,-4)$
Since, they are perpendicular,

$AB.BC = 0$

$\Rightarrow 3k + 1 +8 = 0$

$\Rightarrow k = -3$

If the projections of the line segment

$AB$ on the coordinate axes are

$2, 3, 6$ , then the sum of the d.cs of the line

$AB$ is

0%
$11$
0%
$1$
0%
$\displaystyle \frac{11}{49}$
0%
$\displaystyle \frac{11}{7}$

Explanation

Let

$\overrightarrow { AB } =l\hat { i } +m\hat { j } +n\hat { k }$

Projection of

$AB$ on

$x,y,z$ axis are

$l,m,n$ respectively

$\overrightarrow { AB } .\hat { i } =l=2$

$\overrightarrow { AB } .\hat { j } =m=3$

$\overrightarrow { AB } .\hat { k } =n=6$

$\overrightarrow { AB } =2\hat { i } +3\hat { j } +6\hat { k }$

$\left| \overrightarrow { AB } \right| =\sqrt { { 2 }^{ 2 }+{ 3 }^{ 2 }+{ 6 }^{ 2 } } =7$

DR's of

$\overrightarrow { AB }$ are

$(2,3,6)$

DR's of

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

Sum of DR's

$\dfrac { 2+3+6 }{ 7 } =\dfrac { 11 }{ 7 }$

lf the

$\mathrm{D.R}$ s of two lines are given by the equations

$l+m+n=0$ and

$l^{2}+m^{2}-n^{2}=0$ , then the angle between the two lines is:

0%
$60^{o}$
0%
$30^{o}$
0%
$45^{o}$
0%
$90^{o}$

Explanation

Given that

$l+m=-n$
Substituting

$n$ in

$l^2+m^2-n^2=0$ , we get

$l^2+m^2-(l+m)^2=0$

$\Rightarrow lm=0$

$\Rightarrow l=0$ or

$m=0$
Substituting in

$l+m=-n$ , we get

$m=-n$ or

$l=-n$
Now,

$(l,m,n)=(0,-n,n)$ or

$(-n,0,n)$
Therefore, d.r.s are

$(0,-1,1)$ or

$(-1,0,1)$
Now, angle between two lines

$\cos { \theta } =\dfrac { \left( 0,-1,1 \right) .\left( -1,0.1 \right) }{ \sqrt { { \left( -1 \right) }^{ 2 }+{ 1 }^{ 2 } } \sqrt { { \left( -1 \right) }^{ 2 }+{ 1 }^{ 2 } } } =\dfrac { 1 }{ 2 }$

$\Rightarrow \theta= \cos ^{-1}\left (\dfrac {1}{2}\right)$

$\Rightarrow \theta =\dfrac { \pi }{ 3 }$
Ans: A

lf the projections ofthe line segment

${A}{B}$ on the

$yz$ -plane,

$zx$ -plane,

$xy$ -plane are

$\sqrt{160}, \sqrt{153},5$ respectively, then the projection of

${A}{B}$ on the

${z}$ -axis is

0%
$\sqrt{12}$
0%
$\sqrt{13}$
0%
$12$
0%
$144$

Explanation

Let

$\overrightarrow { AB } =l\hat { i } +m\hat { j } +n\hat { k }$

$l,m,n$ are projections of

$\overrightarrow { AB }$ on

$x,y,z$ respectively

Projection of

$\overrightarrow { AB }$ on

$yz$ plane

$=\sqrt { { m }^{ 2 }+{ n }^{ 2 } } =\sqrt { 160 }$

Projection of

$\overrightarrow { AB }$ on

$zx$ plane

$=\sqrt { { n }^{ 2 }+{ l }^{ 2 } } =\sqrt { 153 }$

Projection of

$\overrightarrow { AB }$ on

$xy$ plane

$=\sqrt { { l }^{ 2 }+{ m }^{ 2 } } =5$

$\left( { m }^{ 2 }+{ n }^{ 2 } \right) -\left( { n }^{ 2 }+{ l }^{ 2 } \right) =160-153$

$\left( m-l \right) \left( m+l \right) =7$

$\Rightarrow m=4,\quad l=3,\quad n=12$

Projection of

$AB$ on

$z-$ axis

$=n=12$

A line OP where O

$=$

$(0, 0, 0)$ makes equal angles with ox, oy, oz. The point on OP, which is at a distance of

$6$ units from O is:

0%

$(\displaystyle \dfrac{6}{\sqrt{3}}, \dfrac{6}{\sqrt{3}}, \dfrac{6}{\sqrt{3}})$
0%
$(2\sqrt{3}, -2\sqrt{3},2\sqrt{3})$
0%
$-(2\sqrt{3},2\sqrt{3},2\sqrt{3})$
0%
$(6\sqrt{3},6\sqrt{3},6\sqrt{3})$

Explanation

Let

$l,m,n$ be the direction cosines.

We have

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

Given that

$l=m=n$

$\Rightarrow 3l^2=1$ .

Solving we get

$l=\dfrac{1}{\sqrt{3}}$ .

Any point on the line is a multiple of direction cosines.

Let the point P is given by

$(\dfrac{k}{\sqrt{3}},\dfrac{k}{\sqrt{3}},\dfrac{k}{\sqrt{3}})$ .

Using the distance formula we get,

$k=6$ .

A point

$P$ lies on a line whose ends are

$A(1,2,3)$ and

$B(2,10,1).$ If

$z$ component of

$P$ is

$7,$ then the coordinates of

$P$ are

0%
$(-1,-14,7)$
0%
$(1,-14,7)$
0%
$(-1,14,7)$
0%
$(1,14,7)$

Explanation

The equation of line passing through

$A$ and

$B$ is

$\displaystyle\frac { x-1 }{ 2-1 } =\frac { y-2 }{ 10-2 } =\frac { z-3 }{ 1-3 } \Rightarrow \frac { x-1 }{ 1 } =\frac { y-2 }{ 8 } =\frac { z-3 }{ -2 } =r$

Substitute

$z=7$ , we get

$\Rightarrow \displaystyle \frac { x-1 }{ 1 } =\frac { y-2 }{ 8 } =\frac { 7-3 }{ -2 } =-2$

Solve it to get

$x=1-2=-1$ ,

$y=2-16=-14$

$P:$

$\left( -1,-14,7 \right)$

$l,m,n$ are the d.cs of a line and

$l=\displaystyle \dfrac{1}{3}$ , then the maximum value of

$l\times m\times n$ is

0%
$4$
0%
$\displaystyle \dfrac{4}{9}$
0%
$\displaystyle \dfrac{27}{4}$
0%
$\displaystyle \dfrac{4}{27}$

Explanation

Given that

$l=\dfrac {1}{3}$
Now,

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

$m^2+n^2=\dfrac {8}{9}$
Using A.M

$\geq$ GM,

$\dfrac { { m }^{ 2 }+{ n }^{ 2 } }{ 2 } \ge \sqrt { { m }^{ 2 }{ n }^{ 2 } }$

$\Rightarrow \dfrac { 4 }{ 9 } \ge mn$
Therefore,

$lmn \leq \dfrac { 4 }{ 27 }$

lf the direction ratios of two lines are given by the equations

$2l+2m-n=0$ and

$ml+nl+lm=0$ , then the angle between the two lines is

0%
$60^{o}$
0%
$30^{o}$
0%
$45^{o}$
0%
$90^{o}$

Explanation

$2l+2m=n ...............(1)$ and

$mn+nl+lm=0......(2)$

$\Rightarrow 2lm+2m^2+2l^2+2lm+lm=0$

$\Rightarrow 2m^2+5lm+2l^2=0$

On factorisation, we get

$(m+2l)(2m+l)=0$

$\Longrightarrow m=−2l$

$\therefore m=\dfrac{-l}{2}$

Put

$l=1$

then

$d_1=(1,-2,-2,)$

and

$d_2=(1,\dfrac{-1}{2},1)$

$d_1.d_2=0$ for right angle

$i.e. 90$

The angle between the lines

$2x=3y=-z$ and

$6x=-y=-4z$ is

0%
$0^{0}$
0%
$90^{0}$
0%
$45^{0}$
0%
$30^{0}$

Explanation

Direction ratios of line

$2x=3y=-z$ which can be written as

$\frac{x}{\frac{1}{2}}=\frac{y}{\frac{1}{3}}=\frac{z}{-1}$ are

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

Direction ratios of line

$6x=-y=-4z$ which can be written as

$\frac{x}{\frac{1}{6}}=\frac{y}{-1}=\frac{z}{-\frac{1}{4}}$ are

$\left(\dfrac{1}{6},-1,-\dfrac{1}{4}\right).$

Dot product of the direction ratio of the two line

$=\dfrac{1}{2}\times\dfrac{1}{6}+\dfrac{1}{3}\times(-1)+(-1)\times(-\dfrac{1}{4})$

$=\dfrac{1}{12}-\dfrac{1}{3}+\dfrac{1}{4}$

$=0$

So, angle between the lines is

$90^\circ.$

Hence, B is the correct option.

If OA is equally inclined to OX, OY and OZ and if A is

$\sqrt{3}$ units from the origin, then A is

0%
$(3, 3, 3)$
0%
$( 1, -1, -1)$
0%
$( -1, 1, -1)$
0%
$(1,1,1)$

Explanation

Let

$\alpha$ be the angle inclined by

$OA$ with

$OX$ ,

$\beta$ be the angle inclined by

$OA$ with

$OY$ and

$\gamma$ be the angle inclined by

$OA$ with

$OZ$

We have

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

Given

$\alpha=\beta=\gamma$

So, we get

$\cos\alpha=\cos\beta=\cos\gamma=\cfrac{1}{\sqrt3}$

Therefore,

$A$ is

$(OA\cos\alpha,OA\cos\beta,OA\cos\gamma) = (1,1,1)$

If the d.rs of two lines are given by the equations

$l+m+n=0$ and

$2lm-mn+2nl=0$ , then the angle between the two lines is

0%
$120^{o}$
0%
$45^{o}$
0%
$90^{\mathrm{o}}$
0%
$30^{\mathrm{o}}$

Explanation

Given,

$l+m+n=0$ and

$2lm-mn+2nl=0$

$2lm+2nl-mn=0$

$2l(m+n)-mn=0$

$-2(m+n)(m+n)-mn=0$

$2(m+n)^{2}+mn=0$

$2m^2+5mn+2n^2=0$
Hence by applying quadratic formula, we get

$m=\dfrac{-5n\pm3n}{4}$

$m=\dfrac{-n}{2}$ and

$m={-2n}$
Therefore, the corresponding values of l are

$n$ and

$\dfrac{-n}{2}$
Hence, the drs of one line is (-2n,n,n)
And d.r.s of other is

$(\dfrac{-n}{2},n,\dfrac{-n}{2})$

Hence, the drs will be in the ration

$(-2,1,1)$ and

$(-1,2,-1)$
Taking dot product, we get

$(-2)(-1)+(1)(2)+(1)(-1)=\sqrt{6}\sqrt{6}cos\theta$

$3=6\cos\theta$

$\theta=60^0$
Hence, angle between the lines will be

$60^0$ and

$180^0-60^0$

$=120^0$

The line passing through the points

$(5,1,a)$ and

$(3,b,1)$ crosses the

$yz$ -plane at the point

$\left (0,\dfrac {17}{2}, \dfrac {-13}{2}\right)$ , then

0%
$a=11,\ b=4$
0%
$a=8,\ b=2$
0%
$a=2,\ b=8$
0%
$a=4,\ b=6$

Explanation

Equation of line passing through

$(5,1,a)$ and

$(3,b,1)$ is

$\displaystyle \dfrac { x-5 }{ 2 } =\dfrac { y-1 }{ 1-b } =\dfrac { z-a }{ a-b } =\lambda$

If line crosses

$yz$ -plane

$i.e.,x=0$

$x=2\lambda +5=0$

$\displaystyle \Rightarrow \lambda =-\dfrac { 5 }{ 2 } ,$

Since,

$\displaystyle y=\lambda \left( 1-b \right) +1\Rightarrow \dfrac { 17 }{ 2 } =\dfrac { -5 }{ 2 } \left( 1-b \right) +1\Rightarrow b=4$

Also,

$\displaystyle z=\lambda \left( a-1 \right) +a\Rightarrow-\dfrac { 13 }{ 2 } =\dfrac { -5 }{ 2 } \left( a-1 \right) +a\Rightarrow a=11$

Assertion $({A})$ . The direction ratios of the line joining origin and point $(x,y, z)$ must be $x, y, {z}$

Reason (R): lf $P(x, y, z)$ is a point in space and $|{OP}|={r},$ then the direction cosines of ${O}{P}$ are $\displaystyle \dfrac{x}{r}$ , $\displaystyle \dfrac{y}{r}$ , $\displaystyle \dfrac{z}{r}$

0%
Both A and R are individually true and R is the correct explanation of A
0%
Both A and R individually true but R is not the correct explanation of A
0%
A is true but R is false
0%
A is false but R is true

Explanation

Assertion: Let

$P(x,y,z)$

$\overrightarrow { OP } =x\hat { i } +y\hat { j } +z\hat { k }$

DR of

$\overrightarrow { OP }$ is any vector parallel to

$x\hat { i } +y\hat { j } +z\hat { k }$

So, given assertion is false.

Reason:

$\overrightarrow { OP } =\left( x\hat { i } +y\hat { j } +z\hat { k } \right) \lambda$

$\overrightarrow { OP } =\lambda \sqrt { { x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 } } =r$

$\alpha ,\beta ,\gamma$ be angles made by

$\overrightarrow { OP }$ with

$x,y,z$ axes respectively

$\cos { \alpha } =\dfrac { \overrightarrow { OP } .\hat { i } }{ \left| \overrightarrow { OP } \right| \left| \hat { i } \right| } =\dfrac { x }{ r }$

$\cos { \beta } =\dfrac { \overrightarrow { OP } .\hat { j } }{ \left| \overrightarrow { OP } \right| \left| \hat { j } \right| } =\dfrac { y }{ r }$

$\cos { \gamma } =\dfrac { \overrightarrow { OP } .\hat { k } }{ \left| \overrightarrow { OP } \right| \left| \hat { k } \right| } =\dfrac { z }{ r }$

DC of

$\overrightarrow { OP }$ are

$\dfrac { x }{ r } ,\dfrac { y }{ r } ,\dfrac { z }{ r }$

Reason is true.

The projection of a directed line segment on the co-ordinate axes are

$12,4,3$ , then the direction cosines of the line are

0%
$\displaystyle \dfrac{-12}{13},\dfrac{-4}{13},\dfrac{-3}{13}$
0%
$\displaystyle \dfrac{12}{13},\dfrac{4}{13},\dfrac{3}{13}$
0%
$\displaystyle \dfrac{12}{13},\dfrac{-4}{13},\dfrac{3}{13}$
0%
$\displaystyle \dfrac{12}{13},\dfrac{4}{13},\dfrac{-3}{13}$

Explanation

$x=12,y=4,z=3$

Direction cosines $\displaystyle =\dfrac{x}{\sqrt{x^{2}+y^{2}+z^{2}}} ,\displaystyle \dfrac{y}{\sqrt{x^{2}+y^{2}+z^{2}}},\dfrac{z}{\sqrt{x^{2}+y^{2}+z^{2}}}$

$\displaystyle = \dfrac{12}{13},\dfrac{4}{13},\dfrac{3}{13}$

Hence, option 'B' is correct.

For waht value of $\lambda$ , the three numbers $2\lambda - 1 , \frac{1}{4}, \lambda -\frac{1}{2}$ can be the direction cosines of a straight line?

0%
$\displaystyle \frac{1}{2} \pm \frac{{\sqrt 3 }}{4}$
0%
$\displaystyle \frac{3}{4}$
0%
$\displaystyle \pm \frac{3}{4}$
0%
$\displaystyle \frac{{\sqrt 3 }}{2} \pm \frac{1}{4}$

Explanation

for being direction cosines of straight line,

$(2\lambda -1 )^2 + \left ( \dfrac{1}{4} \right )^2+ \left ( \lambda -\dfrac{1}{2} \right )^2 =1$

$4\lambda^2 -4 \lambda + y+\dfrac{1}{16} + \lambda^2-\lambda +\dfrac{1}{4} =\lambda$

$5\lambda^2 -5 \lambda + \dfrac{5}{16}=0$

$16 \lambda^2 -16 \lambda +1=0$

$\displaystyle \lambda =\dfrac{16 \pm \sqrt{256 - 64}}{32}$

$\displaystyle =\dfrac{16 \pm \sqrt{13}}{32}=\dfrac{1}{2} \pm \dfrac{\sqrt{3}}{4}$

The direction ratios of a normal to the plane through

$(1, 0, 0)$ and

$(0, 1, 0)$ , which makes an angle of

$\dfrac {\pi}{4}$ with the plane

$x+y=3$ , are:

0%
$< 1, \sqrt 2, 1 >$
0%
$< 1, 1, \sqrt 2 >$
0%
$< 1, 1, 2 >$
0%
$< \sqrt 2, 1, 1 >$

Explanation

Let

$A(1,0,0)$ &

$B(0,1,0)$ be two points on the plane and direction ratio of its normal be

$ai+bj+ck$ , then

$(ai+bj+ck). \vec{BA}=0$

$\Rightarrow (ai+bj+ck).(i-j)=0$

$\Rightarrow a=b$ ....(1)
Since,

$ai+bj+ck$ makes an angle

$\dfrac {\pi}{4}$ with

$x+y-3=0$
Therefore,

$\cos { \dfrac { \pi }{ 4 } } =\dfrac { \left( ai+bj+ck \right) .\left( i+j \right) }{ \sqrt { \left( a^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } \right) \left( { 1 }^{ 2 }+{ 1 }^{ 2 } \right) } }$

$\Rightarrow \dfrac { 1 }{ \sqrt { 2 } } =\dfrac { a+b }{ \sqrt { 2\left( a^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } \right) } } =\dfrac { 2a }{ \sqrt { 2\left( 2a^{ 2 }+{ c }^{ 2 } \right) } }$ ....[ from (1) ]

$\Rightarrow 2a^{ 2 }+{ { c }^{ 2 } }=4{ a }^{ 2 }$

$\Rightarrow c=\pm \sqrt { 2 } a$
Therefore, direction ratios are

$\left( a,b,c \right) =\left( a,a,\pm \sqrt { 2 } a \right) =\left( 1,1,\pm \sqrt { 2 } \right)$

Ans: B

What is the equation of the plane which passes through the z-axis and its perpendicular to the line

$\dfrac {x-a}{cos\theta}=\dfrac {y+2}{sin\theta}=\dfrac {z-3}{0} ?$

0%
$x+y tan\theta=0$
0%
$y+xtan\theta=0$
0%
$x cos\theta-y sin\theta=0$
0%
$x sin\theta-y cos\theta=0$

Explanation

Given plane is passing through z-axis,

$i.e$ it also passes through origin
Hence equation of plane passing through origin and perpendicular to line

$\dfrac{x-a}{cos\theta}=\dfrac{y+2}{sin\theta}=\dfrac{z-3}{0}$ is given by,

$cos\theta (x)+sin\theta(y)+0(z)=0$

$\Rightarrow x+y tan\theta =0$

If points

$\hat i + \hat j, \hat i - \hat j$ and

$p \hat i + q \hat j + r \hat k$ are collinear, then

0%
$p = 1$
0%
$r = 0$
0%
$q \in R$
0%
$q \neq 1$

Explanation

Points

$A(\hat i + \hat j), B (\hat i - \hat j)$ and

$C(p \hat i + q \hat j + r \hat k)$ are collinear
Now

$\vec{AB} = - 2 \hat j$ and

$\vec{BC} = (p -1) \hat i + (q - 1) \hat j + r k$
Vectors

$\vec{AB}$ and

$\vec{BC}$ must be collinear

$\Rightarrow p = 1, r= 0$ and

$q \neq 1$

If a line makes an angle of

$\dfrac {\pi}{4}$ with the positive direction of each of

$x$ -axis and

$y$ -axis, then the angle that the line makes with the positive direction of

$z$ -axis is-

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

Explanation

let

$\alpha$ ,

$\beta$ and

$\gamma$ be the angle made by the line with co-ordinate axes.
Therefore,

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

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

$\Rightarrow \cos ^{ 2 }{ \gamma } =0$

$\Rightarrow \gamma =\dfrac { \pi }{ 2 }$

Ans: C

The equation of the plane passing through the lines

$\frac {x-4}{1}=\frac {y-3}{1}=\frac {z-2}{2}$ and

$\frac {x-3}{1}=\frac {y-2}{-4}=\frac {z}{5}$ is-

0%
$11x-y-3z=35$
0%
$11x+y-3z=35$
0%
$11x-y+3z=35$
0%
None of these.

Explanation

Normal vector of the plane is given by,

$\vec{n} = \begin{vmatrix} \hat{i}&\hat{j}&\hat{k}\\1&1&2\\1&-4&5 \end{vmatrix}=9\hat{i}-3\hat{j}-5\hat{k}$
Hence equation of plane is,

$(\vec{r} - (3\hat{i}+2\hat{j}))\cdot \vec{n}=0$

$\Rightarrow 9x-3y-5z-21=0$

A line makes an angle

$\theta$ with each of the

$x$ - and

$z$ - axes. If the angle

$\beta$ , which it makes with the

$y$ -axis, is such that

$\sin^2\beta=3 \sin^2\theta$ , then

$\cos^2\theta$ equals-

0%
$\dfrac {2}{3}$
0%
$\dfrac {1}{5}$
0%
$\dfrac {3}{5}$
0%
$\dfrac {2}{5}$

Explanation

Given that

$\sin^2\beta=3 \sin^2\theta$ ...

$(1)$
Let

$\theta$ ,

$\theta$ and

$\beta$ be the angle made by the line with co-ordinate axes.
Therefore,

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

$\Rightarrow 2\cos ^{ 2 }{ \theta }=\sin ^{ 2 }{ \beta }=3\sin ^{ 2 }{ \theta }=3-3\cos ^{ 2 }{ \theta }$ from

$(1)$
Therefore,

$\cos ^{ 2 }{ \theta }=\dfrac {3}{5}$

Ans: C

If the foot of the perpendicular from the origin to a plane is

$P(a, b, c)$ , the equation of the plane is-

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$\dfrac {x}{a}+\dfrac {y}{b}+\dfrac {z}{c}=3$
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$ax+by+cz=3$
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$ax+by+cz=a^2+b^2+c^2$
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$ax+by+cz=a+b+c$

Explanation

${\textbf{Step - 1: Find direction ratio}}$

${\text{Let the foot of the perpendicular be P(a,b,c)}}{\text{.}}$

${\text{Then, direction ratios of OP are }} {\text{a - 0,b - 0,c - 0 }}\;{\text{i}}{\text{.e}}{\text{ a,b,c}}$

${\textbf{Step - 2: Make equation}}$

${\text{So, the equation of the plane passing through P(a,b,c), }}$

${\text{the direction ratios of the normal to which are a,b,c is}}$

${\text{a(x - a) + b(y - b) + c(z - c) = 0}}$

$\Rightarrow {\text{ax + by + cz = }}{{\text{a}}^2}{\text{ + }}{{\text{b}}^2}{\text{ + }}{{\text{c}}^2}{\text{.}}$

${\textbf{Hence,correct answer is option C}}{\text{.}}$

A straight line

$L$ on the

$xy$ -plane bisects the angle between

$OX$ and

$OY$ . What are the direction cosines of

$L$ ?

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$(1/\sqrt 2, 1/\sqrt 2, 0)$
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$( 1/2, \sqrt 3/2, 0 )$
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$(0, 0, 1)$
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$(2/3, 2/3, 1/3)$

Explanation

L makes an angle

$\dfrac{\pi}{4}$ with X and Y axis and

$\dfrac{\pi}{2}$
Therefore, d.cs are

$\displaystyle \left( \cos { \frac { \pi }{ 4 } } ,\cos { \frac { \pi }{ 4 } } ,\cos { \frac { \pi }{ 2 } } \right) =\left( \frac { 1 }{ \sqrt { 2 } } ,\frac { 1 }{ \sqrt { 2 } } ,0 \right)$

If the points

$(-1, 3, 2), (-4, 2, -2)$ and

$(5, 5, \lambda)$ are collinear, then

$\lambda$ is equal to

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$-10$
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$5$
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$-5$
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$10$

Explanation

Let the points be

$A(-1,3,2) B(-4,2,-2) C(5,5,\lambda)$
Direction ratio of

$AB{=}(-3,-1,-4)$
If

$A,B,C$ are collinear points then direction ratio of

$AB$ and

$BC$ must be proportional.
Direction ratio of

$BC{=}(9,3,\lambda+2)$

$\therefore 9{=}\alpha (-3)$

$\therefore 3{=}\alpha(-1)$

$\therefore \lambda+2{=}\alpha(-4)$ and

$\alpha{=} -3$

$\therefore \dfrac{\lambda+2}{-3}{=}-4$

$\therefore \lambda{=} 10$

If the points

$(0, 1, -2), (3, \lambda, -1)$ and

$(\mu, -3, -4)$ are collinear, the point on the same line is

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$(12, 9, 2)$
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$(1, -1, -2)$
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$(5, -3, 4)$
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$(0, 0, 0)$

Explanation

We know condition of collinearity,

$\dfrac{x-x_1}{x_2-x_1}=\dfrac{y-y_1}{y_2-y_1}=\dfrac{z-z_1}{z_2-z_1}$

We have points

$(0,1,-2),(3,\lambda,-1) and (\mu,-3,-4)$

$\dfrac{x-0}{3-0}=\dfrac{y-1}{\lambda-1}=\dfrac{z+2}{-1+2}.........(1)$

Since

$(\mu,-3,-4)$ is collinear , so it satisfy equation (1)

taking y and z coordinates,

$\dfrac{-3-1}{\lambda-1}=\dfrac{-4+2}{-1+2}$

$\lambda=3$

Now equation (1) becomes

$\dfrac{x-0}{3-0}=\dfrac{y-1}{3-1}=\dfrac{z+2}{-1+2}.........(2)$

Option (A) (12,9,2) satisfy equation (2).

Therefore option (A) is correct.

CBSE Questions for Class 12 Commerce Maths Three Dimensional Geometry Quiz 3 - 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 3 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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