Let L_1 be the line of intersection of the pl | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The direction vector of $L_1$ is $\vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & 1 \ 1 & 2 & 1 \end{vmatrix} = \hat{i}(1) - \ha

Vedclass · Accepted Answer

$A,C$

Vedclass · Answer

(C) The direction vector of $L_1$ is $\vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & 1 \ 1 & 2 & 1 \end{vmatrix} = \hat{i}(1) - \hat{j}(1) + \hat{k}(1) = \langle 1, -1, 1 angle$.Since $L_2$ is parallel to $L_1$ and passes through $P(2, -1, 3)$,its equation is $\frac{x-2}{1} = \frac{y+1}{-1} = \frac{z-3}{1} = \lambda$. Thus,any point on $L_2$ is $(\lambda+2, -\lambda-1, \lambda+3)$.For point $Q$,this point lies on plane $M: 2x+y-2z=6$. Substituting: $2(\lambda+2) + (-\lambda-1) - 2(\lambda+3) = 6 \Rightarrow 2\lambda+4-\lambda-1-2\lambda-6=6 \Rightarrow -\lambda-3=6 \Rightarrow \lambda=-9$.So,$Q = (-7, 8, -6)$.$PQ = \sqrt{(-7-2)^2 + (8-(-1))^2 + (-6-3)^2} = \sqrt{(-9)^2 + 9^2 + (-9)^2} = \sqrt{81+81+81} = 9\sqrt{3}$. Thus,$(A)$ is True.For $R$,the foot of the perpendicular from $P(2,-1,3)$ to $M: 2x+y-2z-6=0$,the line $PR$ has direction $\langle 2, 1, -2 angle$. So $R = (2\mu+2, \mu-1, -2\mu+3)$.Substituting into $M$: $2(2\mu+2) + (\mu-1) - 2(-2\mu+3) = 6 \Rightarrow 4\mu+4+\mu-1+4\mu-6=6 \Rightarrow 9\mu-3=6 \Rightarrow 9\mu=9 \Rightarrow \mu=1$.So,$R = (4, 0, 1)$.$QR = \sqrt{(4-(-7))^2 + (0-8)^2 + (1-(-6))^2} = \sqrt{11^2 + (-8)^2 + 7^2} = \sqrt{121+64+49} = \sqrt{234}$. Thus,$(B)$ is False.Area of $	riangle PQR = \frac{1}{2} |\vec{QP} 	imes \vec{QR}|$. $\vec{QP} = \langle 9, -9, 9 angle$,$\vec{QR} = \langle 11, -8, 7 angle$.$\vec{QP} 	imes \vec{QR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 9 & -9 & 9 \ 11 & -8 & 7 \end{vmatrix} = \hat{i}(-63+72) - \hat{j}(63-99) + \hat{k}(-72+99) = 9\hat{i} + 36\hat{j} + 27\hat{k} = 9(\hat{i} + 4\hat{j} + 3\hat{k})$.Magnitude $= 9\sqrt{1^2+4^2+3^2} = 9\sqrt{26}$. Area $= \frac{9}{2}\sqrt{26} = \frac{9}{2}\sqrt{\frac{234}{9}} = \frac{3}{2}\sqrt{234}$. Thus,$(C)$ is True.$\vec{PQ} = \langle -9, 9, -9 angle$,$\vec{PR} = \langle 2, 1, -2 angle$. $\cos 	heta = \frac{|\vec{PQ} \cdot \vec{PR}|}{|PQ||PR|} = \frac{|-18+9+18|}{9\sqrt{3} \cdot \sqrt{4+1+4}} = \frac{9}{9\sqrt{3} \cdot 3} = \frac{1}{3\sqrt{3}}$. Thus,$(D)$ is False.

Similar Questions

The plane passing through the line $L: \ell x-y+3(1-\ell)z=1, x+2y-z=2$ and perpendicular to the plane $3x+2y+z=6$ is $3x-8y+7z=4$. If $\theta$ is the acute angle between the line $L$ and the $y$-axis,then $415 \cos^{2} \theta$ is equal to...

If the line $\frac{x - 3}{2} = \frac{y + 2}{-1} = \frac{z + 4}{3}$ lies in the plane $lx + my - z = 9$,then $l^2 + m^2 = \dots$

If the lines $x = 1 + s, y = -3 - \lambda s, z = 1 + \lambda s, s \in R$ and $x = \frac{t}{2}, y = 1 + t, z = 2 - t, t \in R$ are coplanar,then $\lambda = $

The square of the distance of the point of intersection of the line $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+1}{6}$ and the plane $2x-y+z=6$ from the point $(-1,-1,2)$ is .... .

Vedclass Test Series

Exam Paper Generator

Online Exam Module