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

(C) The required line passes through the point $(3, 1, 2)$ and is perpendicular to the lines with direction vectors $\vec{v_1} = (1, 2, 3)$ and $\vec{

Vedclass · Accepted Answer

$\frac{x-3}{2}=\frac{y-1}{-7}=\frac{z-2}{4}$

Vedclass · Answer

(C) The required line passes through the point $(3, 1, 2)$ and is perpendicular to the lines with direction vectors $\vec{v_1} = (1, 2, 3)$ and $\vec{v_2} = (-3, 2, 5)$.The direction vector $\vec{b}$ of the required line is given by the cross product of $\vec{v_1}$ and $\vec{v_2}$:$\vec{b} = \vec{v_1} 	imes \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 3 \ -3 & 2 & 5 \end{vmatrix}$$\vec{b} = \hat{i}(2 	imes 5 - 3 	imes 2) - \hat{j}(1 	imes 5 - 3 	imes (-3)) + \hat{k}(1 	imes 2 - 2 	imes (-3))$$\vec{b} = \hat{i}(10 - 6) - \hat{j}(5 + 9) + \hat{k}(2 + 6) = 4\hat{i} - 14\hat{j} + 8\hat{k}$We can simplify the direction vector by dividing by $2$: $\vec{b'} = 2\hat{i} - 7\hat{j} + 4\hat{k}$.The equation of the line passing through $(x_1, y_1, z_1)$ with direction vector $(a, b, c)$ is $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.Substituting the point $(3, 1, 2)$ and direction vector $(2, -7, 4)$,we get:$\frac{x-3}{2} = \frac{y-1}{-7} = \frac{z-2}{4}$.

The equation of the line passing through the point $(3,1,2)$ and perpendicular to the lines $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}$ and $\frac{x}{-3}=\frac{y}{2}=\frac{z}{5}$ is

Similar Questions

$A$ line $L$ passes through points $A(1, 3, 2)$ and $B(2, 2, 1)$. If the mirror image of point $P(1, 1, -1)$ in the line $L$ is $(x, y, z)$,then $x+y+z=$

Let a line having direction ratios $1, -4, 2$ intersect the lines $\frac{x-7}{3}=\frac{y-1}{-1}=\frac{z+2}{1}$ and $\frac{x}{2}=\frac{y-7}{3}=\frac{z}{1}$ at the points $A$ and $B$ respectively. Then $( AB )^{2}$ is equal to

The Cartesian equation of a line is $2x - 3 = 3y + 1 = 5 - 6z$. The vector equation of the line passing through the point $(7, -5, 0)$ and parallel to the given line is

The shortest distance between the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x - 2}{3} = \frac{y - 4}{4} = \frac{z - 5}{5}$ is:

If the foot of the perpendicular from point $(4,3,8)$ on the line $L_{1}: \frac{x-a}{l}=\frac{y-2}{3}=\frac{z-b}{4},$ $l \neq 0$ is $(3,5,7),$ then the shortest distance between the line $L_{1}$ and line $L_{2}: \frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module