If the lines frac{x-1}{-3}=frac{y-2}{2k}=frac | vedclass

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

Question

(A) The direction ratios of the lines $\frac{x-1}{-3}=\frac{y-2}{2k}=\frac{z-3}{2}$ and $\frac{x-1}{3k}=\frac{y-1}{1}=\frac{z-6}{-5}$ are $(-3, 2k, 2)

Vedclass · Accepted Answer

$-\frac{10}{7}$

Vedclass · Answer

(A) The direction ratios of the lines $\frac{x-1}{-3}=\frac{y-2}{2k}=\frac{z-3}{2}$ and $\frac{x-1}{3k}=\frac{y-1}{1}=\frac{z-6}{-5}$ are $(-3, 2k, 2)$ and $(3k, 1, -5)$ respectively.Two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ are perpendicular if $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$.Substituting the given values,we get:$-3(3k) + (2k)(1) + (2)(-5) = 0$$-9k + 2k - 10 = 0$$-7k - 10 = 0$$-7k = 10$$k = -\frac{10}{7}$Thus,the value of $k$ is $-\frac{10}{7}$.

If the lines $\frac{x-1}{-3}=\frac{y-2}{2k}=\frac{z-3}{2}$ and $\frac{x-1}{3k}=\frac{y-1}{1}=\frac{z-6}{-5}$ are perpendicular,find the value of $k$.

Similar Questions

If the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$ intersect,then the value of $k$ is

Show that the points $(2,3,4), (-1,-2,1),$ and $(5,8,7)$ are collinear.

The equation of a line passing through the point $(2, -1, 1)$ and parallel to the line joining the points $\hat{i} + 2\hat{j} + 2\hat{k}$ and $-\hat{i} + 4\hat{j} + \hat{k}$ is

Find the equation of the line in vector and in Cartesian form that passes through the point with position vector $2\hat{i}-\hat{j}+4\hat{k}$ and is in the direction $\hat{i}+2\hat{j}-\hat{k}$.

The coordinates of the point where the line passing through $A(3, 4, 1)$ and $B(5, 1, 6)$ crosses the $xy$-plane are

Vedclass Test Series

Exam Paper Generator

Online Exam Module