Find the equation of the line passing through | vedclass

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

Question

(A) The equations of the two planes are $x+2y=0$ and $3y-z=0$.Let $\vec{n}_{1}$ and $\vec{n}_{2}$ be the normals to the two planes,respectively.$\vec{

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The equations of the two planes are $x+2y=0$ and $3y-z=0$.Let $\vec{n}_{1}$ and $\vec{n}_{2}$ be the normals to the two planes,respectively.$\vec{n}_{1} = \hat{i} + 2\hat{j} + 0\hat{k}$ and $\vec{n}_{2} = 0\hat{i} + 3\hat{j} - \hat{k}$.Since the required line is parallel to both planes,it must be parallel to the cross product of their normals,$\vec{b} = \vec{n}_{1} 	imes \vec{n}_{2}$.$\vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 0 \ 0 & 3 & -1 \end{vmatrix} = \hat{i}(-2 - 0) - \hat{j}(-1 - 0) + \hat{k}(3 - 0) = -2\hat{i} + \hat{j} + 3\hat{k}$.The line passes through the point $(3, 0, 1)$.The equation of the line in symmetric form is $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.Substituting the point $(3, 0, 1)$ and the direction vector $(-2, 1, 3)$,we get $\frac{x-3}{-2} = \frac{y-0}{1} = \frac{z-1}{3}$.

Find the equation of the line passing through the point $(3, 0, 1)$ and parallel to the planes $x+2y=0$ and $3y-z=0$.

Similar Questions

What is the reflection of the point $P(1, 3, 4)$ in the plane $2x - y + z + 3 = 0$?

The plane containing the line $\frac{x - 3}{2} = \frac{y + 2}{-1} = \frac{z - 1}{3}$ and also containing its projection on the plane $2x + 3y - z = 5$ contains which one of the following points?

The ratio in which the line segment joining the points $(2, -4, 3)$ and $(-4, 5, -6)$ is divided by the plane $3x + 2y + z - 4 = 0$ is:

Find the vector equation of the line passing through $(1, 2, 3)$ and parallel to the planes $\vec{r} \cdot (\hat{i} - \hat{j} + 2\hat{k}) = 5$ and $\vec{r} \cdot (3\hat{i} + \hat{j} + \hat{k}) = 6$.

Find the position vector of the point where the line $r = (i - j + k) + t(i + j + k)$ meets the plane $r \cdot (i + j - k) = 5$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module