Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Three-dimensional Geometry

Question:

If the equation of a line PQ is $\frac{x+1}{2}=\frac{2-y}{5}=\frac{z+6}{7}$, then the direction cosines of a line parallel to PQ are:

Options:

$\frac{2}{\sqrt{78}},\frac{5}{\sqrt{78}},\frac{7}{\sqrt{78}}$

$\frac{-5}{\sqrt{78}},\frac{-2}{\sqrt{78}},\frac{7}{\sqrt{78}}$

$\frac{5}{\sqrt{78}},\frac{-2}{\sqrt{78}},\frac{7}{\sqrt{78}}$

$\frac{2}{\sqrt{78}},\frac{-5}{\sqrt{78}},\frac{7}{\sqrt{78}}$

Correct Answer:

$\frac{2}{\sqrt{78}},\frac{-5}{\sqrt{78}},\frac{7}{\sqrt{78}}$

Explanation:

Line : $\frac{x+1}{2}=\frac{2-y}{5}=\frac{z+6}{7}$

⇒ line : $\frac{x+1}{2}=\frac{y-2}{-5}=\frac{z+6}{7}$

vector parallel to this is $2\hat i-5\hat j+7\hat k$

unit vector = $\frac{2\hat i-5\hat j+7\hat k}{\sqrt{2^2+(-5)^2+7^2}}=\frac{2\hat i-5\hat j+7\hat k}{\sqrt{78}}$

Direction cosines $\frac{2}{\sqrt{78}},\frac{-5}{\sqrt{78}},\frac{7}{\sqrt{78}}$