Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Three-dimensional Geometry

Question:

The direction cosines of the line which is perpendicular to the lines with direction cosines proportional to (1, – 2, – 2), (0, 2, 1) is

Options:

$\left(\frac{2}{3}, \frac{1}{3}, \frac{2}{3}\right)$

$\left(\frac{2}{3}, \frac{-1}{3}, \frac{-2}{3}\right)$

$\left(\frac{2}{3}, \frac{-1}{3}, \frac{2}{3}\right)$

$\left(\frac{-2}{3}, \frac{-1}{3}, \frac{-2}{3}\right)$

Correct Answer:

$\left(\frac{2}{3}, \frac{-1}{3}, \frac{2}{3}\right)$

Explanation:

Let direction ratios of the required line be <a, b, c>

Therefore a − 2 b − 2 c = 0

And 2 b + c = 0

⇒ c = − 2 b

a − 2 b + 4b = 0 ⇒ a = − 2 b

Therefore direction ratios of the required line are <− 2b, b, − 2b> = <2, − 1, 2> direction cosines of the required line

$\left(\frac{2}{\sqrt{2^2+1^2+2^2}}, \frac{-1}{\sqrt{2^2+1^2+2^2}}, \frac{2}{\sqrt{2^2+1^2+2^2}}\right)=\left(\frac{2}{3}, \frac{-1}{3}, \frac{2}{3}\right)$