Question

The direction cosines of the line passing through $P(2,3,-1)$ and the origin are

• A. $\frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}},\frac{1}{\sqrt{14}}$
• B. $\frac{2}{\sqrt{14}},\frac{-3}{\sqrt{14}},\frac{1}{\sqrt{14}}$
• C. $\frac{-2}{\sqrt{14}},\frac{-3}{\sqrt{14}},\frac{1}{\sqrt{14}}$
• D. $\frac{2}{\sqrt{14}},\frac{-3}{\sqrt{14}},\frac{-1}{\sqrt{14}}$

Answer 1

The correct option is C

$\frac{-2}{\sqrt{14}},\frac{-3}{\sqrt{14}},\frac{1}{\sqrt{14}}$

Given points are $P(2,3,-1)=(x_1,y_1,z_1)$ and $(0,0,0)=(x_2,y_2,z_2)$

Direction cosines of a line passing through these two points is given by

$\cos \alpha =\frac{x_2-x_1}{\sqrt{x_1^2+y_1^2+z_1^2}}$

$\cos \beta =\frac{y_2-y_1}{\sqrt{x_1^2+y_1^2+z_1^2}}$

$\cos \gamma =\frac{z_2-z_1}{\sqrt{x_1^2+y_1^2+z_1^2}}$

$⟹$ $\cos \alpha =\frac{0-2}{\sqrt{2^2+3^2+(-1{)}^2}}=\frac{-2}{\sqrt{14}}$

$\cos \beta =\frac{0-3}{\sqrt{2^2+3^2+(-1{)}^2}}=\frac{-3}{\sqrt{14}}$

$\cos \gamma =\frac{0-(-1)}{\sqrt{2^2+3^2+(-1{)}^2}}=\frac{1}{\sqrt{14}}$

Hence, the direction cosines are $\frac{-2}{\sqrt{14}},\frac{-3}{\sqrt{14}},\frac{1}{\sqrt{14}}$.