If a line L is common to the planes x-y+z+2=0 | vedclass

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

Question

(A) The line $L$ is the intersection of the two planes $x-y+z+2=0$ and $2x+y-2z+5=0$.The normal vectors to these planes are $\vec{n_1} = \hat{i} - \ha

Vedclass · Accepted Answer

$\left(\frac{1}{\sqrt{26}}, \frac{4}{\sqrt{26}}, \frac{3}{\sqrt{26}}\right)$

Vedclass · Answer

(A) The line $L$ is the intersection of the two planes $x-y+z+2=0$ and $2x+y-2z+5=0$.The normal vectors to these planes are $\vec{n_1} = \hat{i} - \hat{j} + \hat{k}$ and $\vec{n_2} = 2\hat{i} + \hat{j} - 2\hat{k}$.The direction vector $\vec{v}$ of the line $L$ is perpendicular to both normals,so $\vec{v} = \vec{n_1} 	imes \vec{n_2}$.$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & 1 \ 2 & 1 & -2 \end{vmatrix} = \hat{i}(2-1) - \hat{j}(-2-2) + \hat{k}(1+2) = \hat{i} + 4\hat{j} + 3\hat{k}$.The direction ratios of the line are $(1, 4, 3)$.The magnitude of the direction vector is $\sqrt{1^2 + 4^2 + 3^2} = \sqrt{1 + 16 + 9} = \sqrt{26}$.The direction cosines are given by $\left(\frac{1}{\sqrt{26}}, \frac{4}{\sqrt{26}}, \frac{3}{\sqrt{26}}ight)$.

If a line $L$ is common to the planes $x-y+z+2=0$ and $2x+y-2z+5=0$,then the direction cosines of the line $L$ are

Similar Questions

In what ratio does the $xy$-plane divide the line segment joining the points $(1, 2, 3)$ and $(4, 2, 1)$?

Let the plane containing the line of intersection of the planes $P_1: x+(\lambda+4)y+z=1$ and $P_2: 2x+y+z=2$ pass through the points $(0,1,0)$ and $(1,0,1)$. Then the distance of the point $(2\lambda, \lambda, -\lambda)$ from the plane $P_2$ is (in $\sqrt{6}$)

The equation of the plane which is parallel to the line $\frac{x - 4}{1} = \frac{y + 3}{-4} = \frac{z + 1}{7}$ and passes through the points $(0, 0, 0)$ and $(3, -1, 2)$ is

$A$ line with positive direction cosines passes through the point $P(2,-1,2)$ and makes equal angles with the coordinate axes. The line meets the plane $2x+y+z=9$ at point $Q$. The length of the line segment $PQ$ is equal to:

Let the line $L: \frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-3}{1}$ intersect the plane $2x+y+3z=16$ at the point $P$. Let the point $Q$ be the foot of the perpendicular from the point $R(1, -1, -3)$ on the line $L$. If $\alpha$ is the area of triangle $PQR$,then $\alpha^2$ is equal to $...........$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module