If $$L_1 = 0$$ is the reflected ray, then its equation is

$$\frac{x + 10}{4} = \frac{y + 15}{5} = \frac{z + 14}{3}$$

$$\frac{x + 10}{4} = \frac{y - 5}{4} = \frac{z + 2}{3}$$

MCQ Questions for Class 12 Commerce Maths Three Dimensional Geometry Quiz 9

A line with direction cosines proportional to 2 , 1, 2 meets each of the lines x = y + a = z and x + a = 2y = 2z. The co-ordinates of each of the point of intersection are given by ______________.

0%
(3a, 3a, 3a), (a,a,a)
0%
(3a,2a, 3a), (a,a,a)
0%
(3a, 2a, 3a), (a, a, 2a)
0%
(2a, 3a, 3a), (2a, a, a)

the points

$(\alpha ,\beta )(\gamma ,\delta ),(\alpha ,\delta )and (\gamma ,\beta )$ where

$\alpha ,\beta ,\gamma ,\delta$ are different real numbers, are

0%
Collinear
0%
vertices of square
0%
vertices of rhomubs
0%
concyclic

The angle between the lines

$2x=3y=-z$ and

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

0%
$0^o$
0%
$45^o$
0%
$90^o$
0%
$30^o$

Explanation

$\dfrac{x}{1/2}=\dfrac{y}{1/3}=\dfrac{x}{-1}$ ;

$\dfrac{x}{1/6}=\dfrac{y}{-1}=\dfrac{x}{-1/4}$

Here:

$\dfrac{1}{12}-\dfrac{1}{3}+\dfrac{1}{4}=\dfrac{1-4+3}{12}=0$

Lines are perpendicular.

A plane passes through the point

$(0, -1, 0)$ and

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

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

$y-z=0$ then the point which satisfies the desired plane is?

0%
$(\sqrt{2}, -1, 4)$
0%
$(\sqrt{2}, 1, 2)$
0%
$(\sqrt{2}, 1, 4)$
0%
$(\sqrt{2}, 2, 4)$

Explanation

$ax+by+cz=1$ ,

$-b=1$ ,

$c=1$

$ax-y+z=1$

$\dfrac{1}{\sqrt{2}}=\cos\dfrac{\pi}{4}=\dfrac{-1-1}{\sqrt{2}\sqrt{a^2+2}}$

$\Rightarrow a^2+2=4\Rightarrow a=-\sqrt{2}$

$\Rightarrow$ Plane is

$-\sqrt{2}x-y+z=1$
Clearly

$(\sqrt{2}, 1, 4)$ satisfy the plane.

The angle between two adjacent sides

$\vec { a }$ and

$\vec { b }$ of parallelogram is

$\cfrac{\pi}{6}$ . If

$\vec { a } =\left( 2,-2,1 \right)$ and

$\left| \vec { b } \right| =2\left| \vec { a } \right|$ , then area of this parallelogram is ______

0%
$9$
0%
$18$
0%
$\cfrac{9}{2}$
0%
$\cfrac{3}{4}$

Explanation

$\vec { a } =2\hat { i } -2\hat { j } +\hat { k } ,\left| \vec { a } \right| =3$

$\sin { { 30 }^{ o } } =\cfrac { x }{ \left| \vec { b } \right| } \Rightarrow x=3\quad \quad$
area

$=3\times 3=9$
1342946_1646721_ans_809da32cf4234602a5c13f159dbc2444.png

The equation of the plane passing through the point

$(-1, 2, 1)$ and perpendicular to the line joining the points

$(-3, 1, 2)$ and

$(2, 3, 4)$ is ________.

0%
$\bar{r}\cdot (5\hat{i}+2\hat{j}+2\hat{k})=1$
0%
$\bar{r}\cdot (5\hat{i}+2\hat{j}+2\hat{k})=-1$
0%
$\bar{r}\cdot (5\hat{i}-2\hat{j}+2\hat{k})=-5$
0%
$\bar{r}\cdot (5\hat{i}-2\hat{j}-2\hat{k})=1$

Explanation

The plane is perpendicular to the line joining the points

$(-3, 1, 2)$ and

$(2, 3, 4).$ So, its normal vector is

$(5, 2, 2).$

Therefore, the equation of the plane is

$\bar{r} . (5\hat{i} + 2\hat{j} + 2\hat{k}) = a$ for some constant

$a.$

It is given that the plane passes through the point

$(-1, 2, 1).$ So, we substitute

$\bar{r} = (-1, 2, 1)$ in the above equation and get

$a = -5 + 4 + 2 = 1.$

So the equation of the plane is

$\bar{r}. (5\hat{i} + 2\hat{j} + 2\hat{k}) = 1.$

The direction ratios of the line perpendicular to the lines

$\dfrac {x - 7}{2} = \dfrac {y + 17}{-3} = \dfrac {z - 6}{1}$ and,

$\dfrac {x + 5}{1} = \dfrac {y + 3}{2} = \dfrac {z - 4}{-2}$ are proportional to

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$4, 5, 7$
0%
$4, -5, 7$
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$4, -5, -7$
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$-4, 5, 7$

Explanation

We have,

$\dfrac{x-7}{2}=\dfrac{y+17}{-3}=\dfrac{z-6}{1}$

and

$\dfrac{x+5}{1}=\dfrac{y+3}{2}=\dfrac{z-4}{-2}$

The direction ratios of the given lines are proportional to

$2,-3,1$ and

$1,2,-2$

The vectors parallel to the given vectors are

$\vec{{b}_{1}}=2\hat{i}-3\hat{j}+\hat{k}$ and

$\vec{{b}_{2}}=\hat{i}+2\hat{j}-2\hat{k}$

Vector perpendicular to the given two lines is

$\vec{b}=\vec{{b}_{1}}\times\vec{{b}_{2}}$

$=\left| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & 1 \\ 1 & 2 & -2\end{matrix} \right|$

$=\left(6-2\right)\hat{i}-\left(-4-1\right)\hat{j}+\left(4+3\right)\hat{k}$

$=4\hat{i}+5\hat{j}+7\hat{k}$

Hence, the direction ratios of the line perpendicular to the given two lines are proportional to

$4,5,7$

The angle between the pair of lines

$\dfrac{x-2}{2} = \dfrac{y-1}{5} = \dfrac{z+3}{-3}$ and

$\dfrac{x+2}{-1} = \dfrac{y-4}{8} = \dfrac{z-5}{4}$ is

0%
$\cos^{-1} \left(\dfrac{21}{9\sqrt{38}}\right)$
0%
$\cos^{-1} \left(\dfrac{23}{9\sqrt{38}}\right)$
0%
$\cos^{-1} \left(\dfrac{24}{9\sqrt{38}}\right)$
0%
$\cos^{-1} \left(\dfrac{26}{9\sqrt{38}}\right)$

Explanation

Given lines are

$\dfrac{x-2}{2} = \dfrac{y - 1}{5} = \dfrac{z+3}{-3}$ and

$\dfrac{x+2}{-1} = \dfrac{y-4}{8} = \dfrac{x-5}{4}$

$dr'$ of above lines are

$< 2,5,-3 >$ and

$< -1, 8, 4 >$ respectively.

$\therefore \cos \theta = \dfrac{a_1a_2 + b_1b_2 + c_1c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$

$= \dfrac{(2)(-1)+(5)(8)+(-3)(4)}{\sqrt{(2)^2+ (5)^2 + (-3)^2} \sqrt{(-1)^2 + (8)^2 + (4)^2}}$

$= \dfrac{-2 + 40 - 12}{\sqrt{4+25+9} \sqrt{1+64+16}}$

$=\dfrac{26}{\sqrt{38} \sqrt{81}}$

$= \dfrac{26}{9\sqrt{38}}$

$\therefore \theta = \cos^{-1} \left(\dfrac{26}{9\sqrt{38}}\right)$

The projections of a line segment on

$X, Y$ and

$Z$ axes are

$12, 4$ and

$3$ respectively. The length and direction cosines of the line segment are

0%
$13; \dfrac {12}{13}, \dfrac {4}{13}, \dfrac {3}{13}$
0%
$19; \dfrac {12}{19}, \dfrac {4}{19}, \dfrac {3}{19}$
0%
$11; \dfrac {12}{11}, \dfrac {4}{11}, \dfrac {3}{11}$
0%
None of these

Explanation

If a line makes angles

$\alpha,\,\beta$ and

$\gamma$ with the axes, then

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

Let

$r$ be the length of the line segment, then,

$r\cos\alpha=12,\,r\cos\beta=4,\,r\cos\gamma=3$ ---- ( 2 )

$\Rightarrow$

$(r\cos\alpha)^2+(r\cos\beta)^2+(r\cos\gamma)^2=(12)^2+(4)^2+(3)^2$

$\Rightarrow$

$r^2\cos^2\alpha+r^2\cos^2\beta+r^2\cos^2\gamma=144+16+9$

$\Rightarrow$

$r^2(\cos^2\alpha+\cos^2\beta+\cos^2\gamma)=169$

$\Rightarrow$

$r^2(1)=169$ [ From ( 1 ) ]

$\Rightarrow$

$r=\sqrt{169}$

$\Rightarrow$

$r=\pm13$

Since, the length cannot be negative

$\Rightarrow$

$r=13$

Substituting

$r=13$ in ( 2 )

we get,

$\cos\alpha=\dfrac{12}{13},$

$\cos\beta=\dfrac{4}{13},$

$\cos\gamma=\dfrac{3}{13}$

$\therefore$ The direction cosines of the line are

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

The Cartesian equation of a line are

$\cfrac{x-2}{2}=\cfrac{y+1}{3}=\cfrac{z-3}{-2}$ . What is its vector equation?

0%
$\vec { r } =\left(2 \hat { i } +3\hat { j } -2\hat { k } \right) +\lambda \left( 2\hat { i } -\hat { j } +3\hat { k } \right)$
0%
$\vec { r } =\left(2 \hat { i } -\hat { j } +3\hat { k } \right) +\lambda \left( 2\hat { i } +3\hat { j } -2\hat { k } \right)$
0%
$\vec { r } =\left(2 \hat { i } +3\hat { j } -2\hat { k } \right)$
0%
none of these

A line passes through the point

$A(-2,4,-5)$ and is parallel to the line

$\cfrac{x+3}{3}=\cfrac{y-4}{5}=\cfrac{z+8}{6}$ . The vector equation of the line is

0%
$\vec { r } =\left( -3\hat { i } +4\hat { j } -8\hat { k } \right) +\lambda \left( -2\hat { i } +4\hat { j } -5\hat { k } \right)$
0%
$\vec { r } =\left( -2\hat { i } +4\hat { j } -5\hat { k } \right) +\lambda \left( -3\hat { i } +5\hat { j } +6\hat { k } \right)$
0%
$\vec { r } =\left( 3\hat { i } +5\hat { j } +6\hat { k } \right) +\lambda \left( -2\hat { i } +4\hat { j } -5\hat { k } \right)$
0%
$\vec { r } =\left( -2\hat { i } +4\hat { j } -5\hat { k } \right) +\lambda \left( 3\hat { i } +5\hat { j } +6\hat { k } \right)$

A line passes through the point

$A(5,-2,4)$ and it is parallel to the vector

$\left(2 \hat { i } -\hat { j } +3\hat { k } \right)$ . The vector equation of the line is

0%
$\vec { r } =\left( 2\hat { i } -\hat { j } +3\hat { k } \right) +\lambda \left( 5\hat { i } -2\hat { j } +4\hat { k } \right)$
0%
$\vec { r } =\left( 5\hat { i } -2\hat { j } +4\hat { k } \right) +\lambda \left( 2\hat { i } -\hat { j } +3\hat { k } \right)$
0%
$\vec { r } .\left( 5\hat { i } -2\hat { j } +4\hat { k } \right) =\sqrt{14}$
0%
none of these

The Cartesian equations of a line are

$\cfrac{x-1}{2}=\cfrac{y+2}{3}=\cfrac{z-5}{-1}$ . Its vector equation is

0%
$\vec { r } =\left( -\hat { i } +2\hat { j } -5\hat { k } \right) +\lambda \left( 2\hat { i } +3\hat { j } -\hat { k } \right)$
0%
$\vec { r } =\left(2 \hat { i } +3\hat { j } -\hat { k } \right) +\lambda \left( \hat { i } -2\hat { j } +5\hat { k } \right)$
0%
$\vec { r } =\left( \hat { i } -2\hat { j } +5\hat { k } \right) +\lambda \left( 2\hat { i } +3\hat { j } -\hat { k } \right)$
0%
none of these

If the points

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

$C(5,5,\lambda)$ are collinear then the value of

$\lambda$ is

0%
$5$
0%
$7$
0%
$8$
0%
$10$

The direction cosines of the perpendicular from the origin to the plane

$\vec{r}\cdot (6\hat{i}-3\hat{j}+2\hat{k})+1=0$ are?

0%
$\dfrac{6}{7}, \dfrac{3}{7}, \dfrac{-2}{7}$
0%
$\dfrac{6}{7}, \dfrac{-3}{7}, \dfrac{2}{7}$
0%
$\dfrac{-6}{7}, \dfrac{3}{7}, \dfrac{2}{7}$
0%
None of these

The direction consines of the line drawn from

$P\left ( -5,3,1 \right )\,to\,Q\left ( 1,5,-2 \right )$ is

0%
$\left ( 6,2,-3 \right )$
0%
$\left ( 2,-4,1 \right )$
0%
$\left ( -4,8,-1 \right )$
0%
$\left ( \dfrac {6}{7},\dfrac {2}{7},-\dfrac {3}{7} \right )$

If a straight line makes an angle of

$60^\circ$ with each of the X and Y axes, the angle which it makes with the Z axis is

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

The points

$(p+1, 1), (2p+1, 3)$ and

$(2p+2,2p)$ are collinear if

0%
$p=-1$
0%
$p=1/2$
0%
$p=2$
0%
$p=-\dfrac{1}{2}$

The equation of the plane which passes through the x-axis and perpendicular to the line

$\dfrac {(x - 1)}{cos\theta} = \dfrac {(y + 2)}{sin\theta} = \dfrac {(z - 3)}{0}$ is

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

The direction cosines of the normal to the plane

$5y+4=0$ are?

0%
$0, \dfrac{-4}{5}, 0$
0%
$0, 1, 0$
0%
$0, -1, 0$
0%
None of these

Explanation

$5y+4=0$

$\Rightarrow -5y=4$

$\Rightarrow -y=\dfrac{4}{5}$

$\Rightarrow 0x-1\cdot y+0\cdot z=\dfrac{4}{5}$

D.r.'s of the normal to this plane are

$0, -1, 0$ and

$\sqrt{0^2+(-1)^2+0^2}=\sqrt{1}=1$

$\therefore$ d.c.'s of the normal to the given plane are

$0, -1, 0$ .

If O is the origin and

$P(1, 2, -3)$ is a given point, then the equation of the plane through P and perpendicular to OP is?

0%
$x+2y-3z=14$
0%
$x-2y+3z=12$
0%
$x-2y-3z=14$
0%
None of these

Explanation

Let the required equation of the plane through

$P(1, 2, -3)$ be

$a(x-1)+b(y-2)+c(z+3)=0$ ........(i)

D.r.'s of OP are

$(1-0), (2-0), (-3, 0)$ i.e.,

$1, 2, -3$

$\therefore a=1, b=2, c=-3$

Hence, the required equation of the plane is

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

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

If the directions cosines of a line are

$k, k, k,$ then

0%
$k>0$
0%
$0< k< 1$
0%
$k=1$
0%
$k=\dfrac{1}{\sqrt{3}}$ or $-\dfrac{1}{\sqrt{3}}$

Explanation

since, direction cosines of a line are

$k, k$ and

$k$

$\therefore l=k, m=k$ and

$n=k$

We know that,

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

$\Rightarrow k^{2}+k^{2}+k^{2}=1$

$\Rightarrow k^{2}=\dfrac{1}{3}$

$\therefore k=\pm \dfrac{1}{\sqrt{3}}$

What is the equation of the plane which passes through the z-axis and is 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%
$y + x \tan \theta = 0$
0%
$x \cos \theta - y \sin \theta = 0$
0%
$x \sin \theta - y \cos \theta = 0$

Explanation

The plane is perpendicular to the line

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

Hence, the direction ratios of the normal of the plane are

$\cos \theta, \sin \theta$ and

$0$ . (i)

Now, the required plane passes through the z-axis. Hence the point

$\left (0, 0, 0 \right )$ lies on the plane.

From Eqs. (i) and (ii), we get equation of the plane as

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

$\cos \theta \space x + \sin \theta \space y = 0$

$x + y \tan \theta = 0$

$P_1 : \overrightarrow{r} \cdot \overrightarrow{n_1} - d_1 = 0, P_2 : \overrightarrow{r} \cdot \overrightarrow{n_2} - d_2 = 0$ and

$P_3 : \overrightarrow{r} \cdot \overrightarrow{n_3} - d_3 = 0$ are three planes and

$\overrightarrow{n_1}, \overrightarrow{n_2}$ and

$\overrightarrow{n_3}$ are three non-copllanar vectors, then three lines

$P_1 = 0, P_2 = 0; P_2 = 0, P_3 = 0$ and

$P_3 = 0, P_1 = 0$ are

0%
parallel lines
0%
coplanar lines
0%
coincident lines
0%
concurrent lines

Explanation

$P_1 = P_2 = 0, P_2 = P_3 = 0$ and

$P_3 = P_1 = 0$ are lines of intersection of the three planes

$P_1, P_2$ and

$P_3$ . As

$\overrightarrow{n_1}, \overrightarrow{n_2}$ and

$\overrightarrow{n_3}$ are non-coplanar, planes

$P_1, P_2$ and

$P_3$ will intersect at unique point. So the given lines will pass through a fixed point.

$L_1 = 0$ is the reflected ray, then its equation is

0%
$\frac{x + 10}{4} = \frac{y - 5}{4} = \frac{z + 2}{3}$
0%
$\frac{x + 10}{3} = \frac{y + 15}{5} = \frac{z + 14}{5}$
0%
$\frac{x + 10}{4} = \frac{y + 15}{5} = \frac{z + 14}{3}$
0%
none of these

Explanation

The equation of the reflected ray

$L_1 = 0$ is the line joining

$Q(x_2, y_2, z_2)$ and

$B(-10, -15, -14)$ .

$\frac{x + 10}{16} = \frac{y + 15}{20} = \frac{z + 14}{12}$

$\frac{x + 10}{3} = \frac{y + 15}{5} = \frac{z + 14}{5}$

State true or false.

The equation of a line, which is parallel to

$2 \hat{i}+\hat{j}+3 \hat{k}$ and which passes through the point (5,-2,4) is

$\dfrac{x-5}{2}=\dfrac{y+2}{-1}=\dfrac{z-4}{3}$

0%
True
0%
False

Explanation

The equation of the line passing through

$(x_1,y_1,z_1)$ and parallel to line with direction ratio

$a,b,c$ is given by

$\Rightarrow \dfrac{x-x_1}{a}=\dfrac{y-y_1}{b}=\dfrac{z-z_1}{c}$

Here,

$x_{1}=5, y_{1}=-2, z_{1}=4$

And

$a=2, b=1, c=3$

$\Rightarrow \dfrac{x-5}{2}=\dfrac{y+2}{1}=\dfrac{z-4}{3}$

Which is not similar to given equation.

The given statement is false.

State true or false.

The vector equation of the line

$\dfrac{x-5}{3}=\dfrac{y-4}{7}=\dfrac{z-6}{2}$ is

$\vec{r}=5 \hat{i}-4 \hat{j}+6 \hat{k}+\lambda(3 \hat{i}+7 \hat{j}+2 \hat{k})$

0%
True
0%
False

Explanation

The vector equation of the line is given by

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

Where

$\vec{a}=(x_1 \hat{i}+y_1 \hat{j}+z_1\hat{k}),\vec{b}=(a \hat{i}+b \hat{j}+c \hat{k})$

We have,

$x_1=5, y_1=-4, z_1=6$

And

$a=3, b=7, c=2$

$\vec{r}=(5 \hat{i}-4 \hat{j}+6 \hat{k})+\lambda(3 \hat{i}+7 \hat{j}+2 \hat{k})$

State true or false.

The unit vector normal to the plane

$x+2 y+3 z-6=0$ is

$\dfrac{1}{\sqrt{14}} \hat{i}+\dfrac{2}{\sqrt{14}} \hat{j}+\dfrac{3}{\sqrt{14}} \hat{k}$

0%
True
0%
False

Explanation

We have,

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

$\therefore \hat{n}=\dfrac{\hat{i}+2 \hat{j}+3 \hat{k}}{\sqrt{1^{2}+2^{2}+3^{2}}}=\dfrac{\hat{i}}{\sqrt{14}}+\dfrac{2 \hat{j}}{\sqrt{14}}+\dfrac{3 \hat{k}}{\sqrt{14}}$

$\alpha, \ \beta,\ \gamma$ are direction angles of a line and

$\alpha = 60^{o},\ \beta=45^{o},\ \gamma =$ ____.

0%
$30^o$ or $90^o$
0%
$45^o$ or $60^o$
0%
$90^o$ or $30^o$
0%
$60^o$ or $120^o$

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

$\sin^{2}\alpha + \sin^{2}\beta + \sin^{2}\gamma$ is

0%
$1$
0%
$2$
0%
$3$
0%
$4$

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

$\sin^{2}\alpha + \sin^{2}\beta + \sin^{2}\gamma$ is

0%
$1$
0%
$2$
0%
$3$
0%
$4$

The direction ratios of the line which is perpendicular to the two lines

$\dfrac{x-7}{2}=\dfrac{y+17}{-3}=\dfrac{z-6}{1}and\dfrac{x+5}{1}=\dfrac{y+3}{2}=\dfrac{z-6}{-2}$ are

0%
$4 , 5 , 7$
0%
$4 , -5 , 7$
0%
$4 , -5 , -7$
0%
$-4 , 5 , 8$

The angle between the lines 2x = 3 y = - z and 6 x = -y = -4 z is

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

Te direction ratios of the line

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

0%
$2 , 1 , 6$
0%
$2 , 1 , -6$
0%
$2, -1 , 6$
0%
$-2 , 1, 6$

Position vectors of two points are

$P(2\hat i+\hat j+3\hat k)$ and

$Q(-4\hat i-2\hat j+\hat k)$
Equation of plane passing through

$Q$ and perpendicular of

$PQ$ is

0%
$\vec r.(6\hat i+3\hat j+2\hat k)=28$
0%
$\vec r.(6\hat i+3\hat j+2\hat k)=32$
0%
$\vec r.(6\hat i+3\hat j+2\hat k)+28=0$
0%
$\vec r.(6\hat i+3\hat j+2\hat k)+32=0$

Explanation

Let position vector of point

$P$ .

$\vec a=2\hat i+\hat j+3\hat k$
and position vector of point

$Q$ .

$\vec b=-4\hat i-2\hat j+\hat k$
then

$\hat PQ=$ position vector of

$Q-$ position of vector of

$P$

$\vec n=-4\hat i-2\hat j+\hat k-(2\hat j+\hat k-/(2\hat i+\hat j+3\hat k)$

$\vec n=-6\hat i-3\hat j-2\hat k$ $

$\therefore$ Equation of plane passing through point

$Q(\vec b)$ perpendicular to

$PQ$ is

$(\vec r-\vec b).\vec n=0$

$\vec r.\vec n=\vec b.\vec n$

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

$=24+6-2=28$

$\Rightarrow \hat r(-6\hat i-3\hat j-2\hat k)=28$

$\Rightarrow -\hat r(6\hat i+3\hat j-2\hat k)=28$

$\Rightarrow \hat r(6\hat i+3\hat j-2\hat k)+28=0$

which of the following group is not direction cosines of a line:

0%
$1,1,1$
0%
$0,0,-1$
0%
$-1,0,0$
0%
$0,-1,0$

Explanation

Direction cosines of a line are proportional to direction ratio's.

Let

$a,b$ and

$c$ are direction ratio's, then according to equation

$1\alpha \dfrac{1}{a}$

$\rightarrow =\dfrac{l}{a}$ where

$k$ is any constant

Similarly

$k=\dfrac{m}{b}$

$k=\dfrac{n}{c}$

$k=\dfrac{l}{a}=\dfrac{m}{b}=\dfrac{n}{c}=\dfrac{\sqrt{l^2+m^2+n^2}}{\sqrt{a^2+b^2+c^2}}$

$k=\dfrac{1}{\sqrt{a^2+b^2+c^2}}$

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

But

$\sqrt{a^2+b^2+c^2}\ne 1$

$l=\dfrac{a}{\sqrt{a^2+b^2+c^2}}\ne 1$

$m=\dfrac{a}{\sqrt{a^2+b^2+c^2}}\ne 1$

$n=\dfrac{a}{\sqrt{a^2+b^2+c^2}}\ne 1$

$\therefore$ Group

$(1,1,1)$ is not dc's.

Direction cosines of

$3i$ be

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

Explanation

Given vector

$\vec a=3\hat i+0\hat j+0\hat k$
Whose direction ratio's are

$3,0,0$ .
Direction cosines of

$\vec a$ are

$\dfrac{3}{\sqrt{(3)^2+0+0}},\dfrac{0}{\sqrt{(3)^2+0+0}},\dfrac{0}{\sqrt{(3)^2+0+0}}$
or

$\dfrac{3}{3},\dfrac{0}{3},\dfrac{0}{3}$
or

$1,0,0$

$A(1, 0,0), B(0, 2, 0), C(0, 0, 3)$ form the triangle

$ABC$ . Then the direction ratios of the line joining orthocenter and circumcentre of

$ABC$ are

0%
$58, 43, 36$
0%
$59, - 44, - 37$
0%
$59, - 44, - 111$
0%
None of these

Explanation

Let H(a,b,c) be the orthocenter of

$\Delta ABC$
AH is perpendicular to BC

$\Rightarrow (a-1)0+b(2-0)+c(0-3)=0$

$2b=3c\Rightarrow b=\dfrac{3}{2}c$
Also BH is perpendicular to CA

$\Rightarrow (a-0)(0-1)+(b-2)0+(c-0)3=0$

$\Rightarrow -a+3c=0$

$\Rightarrow a=3c$
Now (a,b,c) lies on the plane

$\dfrac{x}{1}+\dfrac{y}{2}+\dfrac{z}{3}=1$

$3c+\dfrac{1}{2}\dfrac{3}{2}c+\dfrac{1}{3}c=1$

$\Rightarrow c=\dfrac{12}{49}$

$\therefore a=\dfrac{36}{49},b=\dfrac{18}{49}$

$\therefore$ orthocenter =

$\left ( \dfrac{36}{49} ,\dfrac{18}{49},\dfrac{12}{49}\right )$
Centroid of

$\Delta ABC$ is

$\left ( \dfrac{1}{3} ,\dfrac{2}{3},1\right )$
Dr s of the line joining orthocentre and crirumcentre

$=$ Dr s of the line joining orthocenter and centroid .

$=\dfrac{36}{49}-\dfrac{1}{3},\dfrac{18}{49}-\dfrac{2}{3},\dfrac{12}{49}-1$

$=59,-44-111$

Assertion (

$A$ ):

Three points with position vectors

$\vec{a},\vec{b},\ \vec{c}$ are collinear if

$\vec{a}\times\vec{b}+\vec{b}\times\vec{c}+\vec{c}\times\vec{a}=\vec{0}$

Reason (

$R$ ):

Three points

${A}, {B},\ {C}$ are collinear if

$\vec{AB}={t}\ \vec{BC}$ , where

${t}$ is a scalar quantity.

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

Explanation

Let the position vectors of

$A, B, C$ be

$\vec a, \vec b, \vec c$ respectively.

Given,

$\vec{AB} = t\ \vec{BC}$

$\Rightarrow \vec{AB}\times\vec{BC} = 0$

$\Rightarrow (\vec b - \vec a)\times(\vec c - \vec b) = 0$

$\Rightarrow (\vec a - \vec b)\times(\vec b - \vec c) = 0$

$\Rightarrow \vec a\times \vec b + \vec b\times \vec c + \vec c \times \vec a = 0$

Hence,

$\vec a, \vec b, \vec c$ are collinear.

Hence, option A.

If a ray makes angles

$\alpha, \beta, \gamma$ and

$\delta$ with the four diagonals of a cube and

$\mathrm{A}:\cos^{2}\alpha+\cos^{2}\beta+\cos^{2}\gamma+\cos^{2}\delta$

$\mathrm{B}:\sin^{2}\alpha+\sin^{2}\beta+\sin^{2}\gamma+\sin^{2}\delta$

$\mathrm{C}:\cos 2\alpha+\cos 2\beta+\cos 2\gamma+\cos 2\delta$
Arrange

$A,B,C$ in descending order

0%
$B,A,C$
0%
$A,B,C$
0%
$C,A,B$
0%
$B,C,A$

Explanation

If a ray makes angles $\alpha, \beta, \gamma$ and $\delta$ with the four diagonals of a cube.

Then we know that $\cos^{2}\alpha+\cos^{2}\beta+\cos^{2}\gamma+\cos^{2}\delta=\displaystyle\dfrac{4}{3}$

$\mathrm{A}:$

$\cos^{2}\alpha+\cos^{2}\beta+\cos^{2}\gamma+\cos^{2}\delta=\displaystyle\dfrac{4}{3}$

$\mathrm{B}:$

$\sin^{2}\alpha+\sin^{2}\beta+\sin^{2}\gamma+\sin^{2}\delta$

$=1-\cos^{2}\alpha+1-\cos^{2}\beta+1-\cos^{2}\gamma+1-\cos^{2}\delta=4-\displaystyle\dfrac{4}{3}=\displaystyle\dfrac{8}{3}$

$\mathrm{C}:$

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

$=1-2\sin^{2}\alpha+1-2\sin^{2}\beta+1-2\sin^{2}\gamma+1-2\sin^{2}\delta=4-\displaystyle\dfrac{16}{3}=\displaystyle-\dfrac{4}{3}$

Descending order is $B,A,C$

Hence, option A is correct answer.

Find the angle between the pair of lines

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

$\overrightarrow { r } =5i-2k+\mu \left( 3i+2j+6k \right)$ .

0%
$\displaystyle \cos ^{ -1 }{ \left( \dfrac { 19 }{ 21 } \right) }$
0%
$\displaystyle \sin ^{ -1 }{ \left( \dfrac { 19 }{ 21 } \right) }$
0%
$\displaystyle \cos ^{ -1 }{ \left( \dfrac { 20 }{ 21 } \right) }$
0%
$\displaystyle \sin ^{ -1 }{ \left( \dfrac { 20 }{ 21 } \right) }$

Explanation

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

and $\overrightarrow { r } =5i-2k+\mu \left( 3i+2j+6k \right)$

We know, angle between $\overrightarrow { r } =\overrightarrow { { a }_{ 1 } } +\lambda \overrightarrow { { b }_{ 1 } }$ and $\overrightarrow { r } =\overrightarrow { { a }_{ 2 } } +\lambda \overrightarrow { { b }_{ 2 } }$ is given by,

$\displaystyle \cos { \theta } =\dfrac { \overrightarrow { { b }_{ 1 } } .\overrightarrow { { b }_{ 2 } } }{ \left| \overrightarrow { { b }_{ 1 } } \right| \left| \overrightarrow { { b }_{ 2 } } \right| }$

$\Rightarrow \cos { \theta } =\dfrac { \left( i+2j+2k \right) .\left( 3i+2j+6k \right) }{ \sqrt { { 1 }^{ 2 }+{ 2 }^{ 2 }+{ 2 }^{ 2 } } \sqrt { { 3 }^{ 2 }+{ 2 }^{ 2 }+{ 6 }^{ 2 } } }$

$\displaystyle =\dfrac { 3+4+12 }{ \sqrt { 9 } \sqrt { 49 } } =\dfrac { 19 }{ 21 }$

$\Rightarrow \theta =\cos ^{ -1 }{ \left( \dfrac { 19 }{ 21 } \right) }$

Statement-1 : If a line makes acute angles

$\alpha, \beta, \gamma, \delta$ with diagonals of a cube, then

$\displaystyle \cos^2\alpha+\cos^2\beta+\cos^2\gamma+\cos^2\delta=\frac{4}{3}$
Statement 2 : If a line makes equal angle (acute) with the axes, then its direction cosine are

$\displaystyle \frac{1}{\sqrt{3}} , \frac{1}{\sqrt{2}}$ and

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

0%
Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
0%
Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
0%
Statement-1 is True, Statement-2 is False
0%
Statement-1 is False, Statement-2 is True

Explanation

Let l,m,n be the direction ratio of a line making angles ,,, with four diagonals of a cube then as four diagonals have direction ratio, $( a, a, a)$ ; $(a, a, -a)$ ; $( a, -a, a,)$ ; $(-a, a, a)$ (where $a$ is the side of the cube)

Now, $\displaystyle \cos\alpha =\left| \dfrac{al+am+an}{\sqrt{a^2+a^2+a^2}\sqrt{l^2+m^2+n^2}} \right|=\left|\dfrac{l+m+n}{\sqrt3 \sqrt{l^2+m^2+n^2}} \right|$

$\displaystyle \cos\beta =\left| \dfrac{al+am-an}{\sqrt{a^2+a^2+a^2}\sqrt{l^2+m^2+n^2}} \right|=\left| \dfrac{l+m-n}{\sqrt{3} \sqrt{l^2+m^2+n^2}} \right|$

$\displaystyle \cos\gamma = \left|\dfrac{al-am+an}{\sqrt{a^2+a^2+a^2}\sqrt{l^2+m^2+n^2}} \right|=\left|\dfrac{l-m+n}{\sqrt3 \sqrt{l^2+m^2+n^2}} \right|$

$\displaystyle \cos\delta = \left|\dfrac{al+am+an}{\sqrt{a^2+a^2+a^2}\sqrt{l^2+m^2+n^2}} \right|= \left| \dfrac{-l+m+n}{\sqrt{3} \sqrt{l^2+m^2+n^2}} \right|$

and $\cos^2\alpha +\cos^2\beta+\cos^2\gamma+\cos^2\delta$

$\displaystyle =\dfrac{1}{3}\times \dfrac{(4l^2+4m^2+4n^2)}{l^2+m^2+n^2}=\dfrac{4}{3}$

So statement 1 is true and statement-2 is not true .

Hence, option 'C' is correct.

If the lines

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

$\dfrac{x-1}{k}=\dfrac{y-4}{2}=\dfrac{z-5}{1}$ are coplanar, then

$k$ can have

0%
exactly one value
0%
exactly two values
0%
exactly three values
0%
any value

$\alpha,\ \beta,\ \gamma$ are the angles made by a line with the coordinate axes in the positive direction, then the range of

$\sin\alpha\sin\beta+\sin\beta\sin\gamma+\sin\gamma\sin\alpha$ is

0%
$\left[-\dfrac{1}{2},1 \right]$
0%
$\left[-1,2 \right]$
0%
$\left[-\dfrac{1}{2},\infty \right)$
0%
$\left[-1,\infty \right)$

Explanation

$\alpha,\; \beta$ and

$\gamma$ are angles made by a line with the coordinate axes in the positive direction, then

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

$1-\sin^2\alpha+1-\sin^2\beta+1-\sin^2\gamma=1$

$\sin ^{ 2 }{ \alpha } +\sin ^{ 2 }{ \beta } +\sin ^{ 2 }{ \gamma } =2$
Now , using

$\displaystyle { \left( \sum { \sin { \alpha } } \right) }^{ 2 }=\sum { \sin ^{ 2 }{ \alpha } } +2\sum { \sin { \alpha } .\sin { \beta } }$

$\displaystyle { \left( \sum { \sin { \alpha } } \right) }^{ 2 } \ge 0$

$\Rightarrow \displaystyle \sum { \sin ^{ 2 }{ \alpha } } +2\sum { \sin { \alpha } .\sin { \beta } } \ge 0$

$\displaystyle \Rightarrow 2 + 2\sum { \sin { \alpha } .\sin { \beta } } \ge 0$

$\displaystyle \Rightarrow \sum { \sin { \alpha } .\sin { \beta } } \ge -1$ .....(1)
Next use the equation

$\displaystyle \sum { { \left( \sin { \alpha } -\sin { \beta } \right) }^{ 2 } } =2\sum { \sin ^{ 2 }{ \alpha } } -2\sum { \sin { \alpha } .\sin { \beta } }$

$\displaystyle \sum { { \left( \sin { \alpha } -\sin { \beta } \right) }^{ 2 } } \ge 0$

$\displaystyle \Rightarrow 2\sum { \sin ^{ 2 }{ \alpha } } -2\sum { \sin { \alpha } .\sin { \beta } } \ge 0$

$\displaystyle \Rightarrow 4 -2\sum { \sin { \alpha } .\sin { \beta } } \ge 0$

$\displaystyle \Rightarrow \sum { \sin { \alpha } .\sin { \beta } } \le 2$ .....(2)
From (1) and (2), we get the range of

$\displaystyle \sum { \sin { \alpha } .\sin { \beta } }$ is

$[-1,2]$
Hence, option 'B' is correct.

If the straight lines $\displaystyle \dfrac{x-1}{2}=\dfrac{y+1}{K}=\dfrac{z}{2}$ and $\displaystyle \dfrac{x+1}{5}=\dfrac{y+1}{2}=\dfrac{z}{K}$ are coplanar, then the plane(s) containing these two lines is(are)

0%
$y+2z=-1$
0%
$y+z =-1$
0%
$y-z=-1$
0%
$y-2z =-1$

Explanation

Two straight lines

$\dfrac {x-1}{2}=\dfrac {y+1}{K}=\dfrac {z}{2}$ and

$\dfrac {x+1}{5}=\dfrac {y+1}{2}=\dfrac {z}{K}$ are co-planar, then we get

$\begin{vmatrix} 2& K & 2 \\5& 2 & K\\ 2 & 0 & 0\end{vmatrix}=0$

$\Rightarrow K^2=4\Rightarrow K=\pm 2$

For

$K=2$ , the plane

$y+1=z$ is common in both lines.

For

$K=-2$ , family of plane containing first line is

$x+y+\lambda (x-z-1)=0$

Point

$(-1, -1, 0)$ must satisfy

$\lambda$ .

$\Rightarrow -2+\lambda(-2)=0\Rightarrow \lambda=-1$

$\Rightarrow y+z+1=0$

$\Rightarrow y+z=-1$

The intercepts made on the axes by the plane which bisects the line joining the points

$(1,2,3)$ and

$(-3,4,5)$ at right angles are

0%
$\left (-\displaystyle \dfrac{9}{2},9,9\right)$
0%
$\left (\displaystyle \dfrac{9}{2},9,9\right)$
0%
$\left (9,-\displaystyle \dfrac{9}{2},9\right)$
0%
$\left (9,\displaystyle \dfrac{9}{2},9\right)$

Explanation

Since, the line joining the

$2$ points is a normal to the plane, the equation of the plane is of the form,

$4x-2y-2z = d$
Since, the midpoint of the

$2$ points lies on the plane,

$-4-6-8 = d$

$4x-2y-2z = -18$
The respective intercepts are,

$\left ( \dfrac{-9}{2} , 9 , 9 \right )$

lf a line makes angles

$60^{o}, 45^{o}, 45^{o}$ and

$\theta$ with the four diagonals of a cube, then

$\sin^{2}\theta =$

0%
$\displaystyle \frac{1}{12}$
0%
$\displaystyle \frac{11}{12}$
0%
$\displaystyle \sqrt{\frac{11}{12}}$
0%
$\displaystyle \frac{31}{12}$

Explanation

Let the length of the cube be

$a$ units.
Let

$AA', BB', CC' , OP$ be the diagonals of the cube.
Coordinates of vertices are

$O(0,0,0),A(a,0,0), A'(0,a,a),B(0,b,0),B'(b,0,b),C(0,0,c),C'(c,c,0), P(a,a,a)$

D.Rs of

$AA'$ are

$(-a,a,a)$ .
D.Rs of

$BB'$ are

$(a,-a,a)$
D.Rs of

$CC'$ are

$(a,a,-a)$
D.Rs of

$OP$ are

$(a,a,a)$

D.c's of

$AA'$ are

$\displaystyle \frac{-a}{\sqrt{a^2+a^2+a^2}},\frac{a}{\sqrt{a^2+a^2+a^2}},\frac{a}{\sqrt{a^2+a^2+a^2}}=-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$

D.c's of

$BB'$ are

$\displaystyle \frac{1}{\sqrt{3}} ,-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$
D.c's of

$CC'$ are

$\displaystyle \frac{1}{\sqrt{3}} ,\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}$
D.c's of

$OP$ are

$\displaystyle \frac{1}{\sqrt{3}} ,\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$

Let the d.c's of given line be

$l,m,n$ such that

$l^2+m^2+n^2=1$
Given , the line makes angles

$60^{o}, 45^{o}, 45^{o}$ and

$\theta$ with the four diagonals of a cube.

$\displaystyle \cos {60}^{0}=\frac{-l+m+n}{\sqrt{3}}$

$\Rightarrow \displaystyle \frac{1}{2}=\frac{-l+m+n}{\sqrt{3}}$ .....(1)

$\displaystyle \cos {45}^{0}=\frac{l-m+n}{\sqrt{3}}$

$\Rightarrow \displaystyle \frac{1}{\sqrt{2}}=\frac{l-m+n}{\sqrt{3}}$ .....(2)

$\displaystyle \cos {45}^{0}=\frac{l+m-n}{\sqrt{3}}$

$\Rightarrow \displaystyle \frac{1}{\sqrt{2}}=\frac{l+m-n}{\sqrt{3}}$ .....(3)

$\displaystyle \cos {\theta}=\frac{l+m+n}{\sqrt{3}}$ ....(4)

Squaring and adding (1),(2),(3),(4), we get

$\Rightarrow \displaystyle \frac{5}{4}+\cos^{2}\theta=\frac{4}{3}$ (

$\because l^2+m^2+n^2=1$ )

$\Rightarrow \cos^{2}\theta=\displaystyle \frac{1}{12}$

$\Rightarrow \displaystyle \sin^{2}\theta=\frac{11}{12}$

The direction ratios of the diagonal of a cube which joins the origin to the opposite corner are (when the three concurrent edges of the cube are coordinate axes)

0%
$\displaystyle \frac { 2 }{ \sqrt { 3 } } ,\frac { 2 }{ \sqrt { 3 } } ,\frac { 2 }{ \sqrt { 3 } }$
0%
$1,1,1$
0%
$2,-2,1$
0%
$1,2,3$

Explanation

Let the length of the sides of cube is a, then coordinates of corner (P) opposite to origin are (a,a,a)

$\therefore$ Direction ratios of diagonal OP are

$(a-0,a-0,a-0)$

$\therefore (a,a,a) i.e. (1,1,1)$

The vector equation of the plane through the point

$\vec i+2\vec j-\vec k$ and

$\bot$ to the line of intersection of the plane

$\overrightarrow { r } .\left( 3\vec i-\vec j+\vec k \right) =1$ and

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

0%
$\overrightarrow { r } .\left( 2\vec i+7\vec j-13\vec k \right) =1$
0%
$\overrightarrow { r } .\left( 2\vec i-7\vec j-13k \right) =1$
0%
$\overrightarrow { r } .\left( 2\vec i+7\vec j+13\vec k \right) =0$
0%
None of these

Explanation

The line of inetrsection of the planes

$\overrightarrow { r } .\left( 3\vec i-\vec j+\vec k \right) =1$ and

$\overrightarrow { r } .\left( \vec i+4\vec j-2\vec k \right) =2$ is common to both the planes.

Therefore, it is

$\bot$ to normals to the two planes i.e.

$\overrightarrow { { n }_{ 1 } } =3\vec i-\vec j+\vec k$ and

$\overrightarrow { { n }_{ 2 } } =\vec i+4\vec j-2\vec k$ .

Hence, it is parallel to the vector

$\overrightarrow { { n }_{ 1 } } \times \overrightarrow { { n }_{ 2 } } =-2\vec i+7\vec j+13\vec k$ .

Thus, we have to find the equation of the plane passing through

$\overrightarrow { a } =\vec i+2\vec j-\vec k$ and normal to the vector

$\overrightarrow { n } =\overrightarrow { { n }_{ 1 } } \times \overrightarrow { { n }_{ 2 } }$ .

The equation of the required plane is

$\left( \overrightarrow { r } -\overrightarrow { a } \right) .\overrightarrow { n } \Rightarrow \overrightarrow { r } .\overrightarrow { n } =\overrightarrow { a } .\overrightarrow { n } \\ \Rightarrow \overrightarrow { r } .\left( -2\vec i+7\vec j+13\vec k \right) =\left( \vec i+2\vec j+\vec k \right) .\left( -2\vec i+7\vec j+13\vec k \right) \\ \Rightarrow \overrightarrow { r } .\left( 2\vec i-7\vec j-13\vec k \right) =1$

Equation of the line which passes through the point with p.v. (2, 1, 0) and perpendicular to the plane containing the vectors

$\widehat{i}+\widehat{j}\:and\: \widehat{j}+\widehat{k}$ is

0%
$\vec{r}=(2, 1, 0)+ t(1,-1, 1)$
0%
$\vec{r}=(2, 1, 0)+ t(-1,1, 1)$
0%
$\vec{r}=(2, 1, 0)+ t(1,1, -1)$
0%
$\vec{r}=(2, 1, 0)+ t(1,1, 1)$

Explanation

if line perpendicular to plane then eq

$\vec{r}=position\ vector\ \vec{a}+t(normal\ vector\ \vec{n})$

normal vector of plane

Given

$\vec{a}=\hat{i}+\hat{j}$

$\vec{b}=\hat{j}+\hat{k}$

$\vec{n}=\vec{a}\times\vec{b}$

$\vec{a}\times\vec{b}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&1&0\\0&1&1\end{vmatrix}$

$\vec{a}\times\vec{b}=\hat{i}(1-0)-\hat{j}(1-0)+\hat{k}(1-0)$

$\vec{a}\times\vec{b}=\hat{i}-\hat{j}+\hat{k}$

eq become

$\vec{r}=(2,1,0)+t(1,-1,1)$

CBSE Questions for Class 12 Commerce Maths Three Dimensional Geometry Quiz 9 - 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 9 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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