The equation of the line passing through (1, | vedclass

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

Question

(A) Let the direction ratios of the required line be $(a, b, c)$.The direction ratios of the first line are $(1, 2, 4)$ and the second line are $(2, 2

Vedclass · Accepted Answer

$\frac{x-1}{6} = \frac{2-y}{7} = \frac{z-3}{2}$

Vedclass · Answer

(A) Let the direction ratios of the required line be $(a, b, c)$.The direction ratios of the first line are $(1, 2, 4)$ and the second line are $(2, 2, 1)$.Since the required line is perpendicular to both,its direction vector is the cross product of the direction vectors of the given lines:$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 4 \ 2 & 2 & 1 \end{vmatrix} = \hat{i}(2-8) - \hat{j}(1-8) + \hat{k}(2-4) = -6\hat{i} + 7\hat{j} - 2\hat{k}$.Thus,the direction ratios are proportional to $(-6, 7, -2)$ or $(6, -7, 2)$.The line passes through $(1, 2, 3)$,so its equation is $\frac{x-1}{6} = \frac{y-2}{-7} = \frac{z-3}{2}$.This can be rewritten as $\frac{x-1}{6} = \frac{2-y}{7} = \frac{z-3}{2}$.

The equation of the line passing through $(1, 2, 3)$ and perpendicular to the lines $x-1 = \frac{y+2}{2} = \frac{z+4}{4}$ and $\frac{x-1}{2} = \frac{y-2}{2} = z+3$ is

Similar Questions

Find the angle between the lines whose direction cosines are given by the equations $3l+m+5n=0$ and $6mn-2nl+5lm=0$.

The point collinear with $(1, -2, -3)$ and $(2, 0, 0)$ among the following is

Line $L$ passes through two points $(2, -3, 1)$ and $(3, -4, -5)$. If point $(0, a, b)$ lies on the line $L$,then $a+b =$ . . . . . . .

Let $L_1: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $L_2: \frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$ be two lines. Then which of the following points lies on the line of the shortest distance between $L_1$ and $L_2$?

The vector equation of the line whose Cartesian equations are $y=2$ and $4x-3z+5=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module