The equation of the line passing through the | vedclass

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

Question

(B) The given planes are $P_1: x - 2y - z + 5 = 0$ and $P_2: x + y + 3z = 6$.Their normal vectors are $\vec{N}_1 = \hat{i} - 2\hat{j} - \hat{k}$ and $

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The given planes are $P_1: x - 2y - z + 5 = 0$ and $P_2: x + y + 3z = 6$.Their normal vectors are $\vec{N}_1 = \hat{i} - 2\hat{j} - \hat{k}$ and $\vec{N}_2 = \hat{i} + \hat{j} + 3\hat{k}$.The direction vector $\vec{b}$ of the line of intersection is given by the cross product of the normals:$\vec{b} = \vec{N}_1 	imes \vec{N}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -2 & -1 \ 1 & 1 & 3 \end{vmatrix} = \hat{i}(-6 - (-1)) - \hat{j}(3 - (-1)) + \hat{k}(1 - (-2)) = -5\hat{i} - 4\hat{j} + 3\hat{k}$.The line passes through $(2, 3, 1)$ and is parallel to $\vec{b} = -5\hat{i} - 4\hat{j} + 3\hat{k}$.Thus,the equation of the line is $\frac{x-2}{-5} = \frac{y-3}{-4} = \frac{z-1}{3}$.

The equation of the line passing through the point $(2, 3, 1)$ and parallel to the line of intersection of the planes $x - 2y - z + 5 = 0$ and $x + y + 3z = 6$ is:

Similar Questions

The equation of the plane passing through the intersection of the lines $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-5}{-3}$ and $\frac{x+5}{3}=\frac{y-4}{-1}=\frac{z+3}{4}$ and parallel to the $xy$-plane is

The equation of the plane $\pi$ through the line of intersection of the planes $\pi_1 \equiv x+3y-6=0$ and $\pi_2 \equiv 3x-y+4z=0$ is $\pi_1+\lambda \pi_2=0$. If the plane $\pi$ is at unit distance from the origin,then an equation of the plane $\pi$ is

The distance of the point $(1, -5, 9)$ from the plane $x - y + z = 5$ measured along the line $x = y = z$ is . . . . . . units.

The vector equation of the plane passing through the point $(2, 1, -1)$ and the line of intersection of the planes $r \cdot (i + 3j - k) = 0$ and $r \cdot (j + 2k) = 0$ is:

The direction cosines of the line formed by the intersection of the planes $x - y + 2z = 5$ and $3x + y + z = 6$ are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module