Target Exam

CUET

Subject

Section B1

Chapter

Three-dimensional Geometry

Question:

If the lines $\frac{x - 2}{2k} = \frac{y - 3}{3} = \frac{z + 2}{-1}$ and $\frac{x - 2}{8} = \frac{y - 3}{6} = \frac{z + 2}{-2}$ are parallel, then find the value of $k$.

Options:

1

2

3

4

Correct Answer:

2

Explanation:

The correct answer is Option (2) → 2 ##

Given lines are:

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

and $\frac{x - 2}{8} = \frac{y - 3}{6} = \frac{z + 2}{-2}$

The direction ratios of the first line are $(2k, 3, -1)$ and the direction ratios of the second line are $(8, 6, -2)$.

Lines are parallel;

So, $\frac{2k}{8} = \frac{3}{6} = \frac{-1}{-2} \Rightarrow \frac{k}{4} = \frac{1}{2} = \frac{1}{2}$

$\Rightarrow k = 2$