Target Exam

CUET

Subject

Section B1

Chapter

Three-dimensional Geometry

Question:

Find the direction cosines of the line passing through the two points $(-2, 4, -5)$ and $(1, 2, 3)$.

Options:

$(\frac{1}{\sqrt{77}}, \frac{2}{\sqrt{77}}, \frac{3}{\sqrt{77}})$

$(\frac{3}{\sqrt{77}}, -\frac{2}{\sqrt{77}}, \frac{8}{\sqrt{77}})$

$(\frac{3}{77}, -\frac{2}{77}, \frac{8}{77})$

$(-3, 2, -8)$

Correct Answer:

$(\frac{3}{\sqrt{77}}, -\frac{2}{\sqrt{77}}, \frac{8}{\sqrt{77}})$

Explanation:

The correct answer is Option (2) → $(\frac{3}{\sqrt{77}}, -\frac{2}{\sqrt{77}}, \frac{8}{\sqrt{77}})$ ##

We know the direction cosines of the line passing through two points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ are given by

$\frac{x_2 - x_1}{PQ}, \quad \frac{y_2 - y_1}{PQ}, \quad \frac{z_2 - z_1}{PQ}$

where $PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$.

Here $P \equiv (-2, 4, -5)$ and $Q \equiv (1, 2, 3)$.

So $PQ = \sqrt{(1 - (-2))^2 + (2 - 4)^2 + (3 - (-5))^2} = \sqrt{3^2 + (-2)^2 + 8^2} = \sqrt{9 + 4 + 64} = \sqrt{77}$.

Thus, the direction cosines of the line joining two points is $\frac{3}{\sqrt{77}}, \frac{-2}{\sqrt{77}}, \frac{8}{\sqrt{77}}$.