Browse Questions

Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ are $\frac{\large 1}{\large \sqrt 3}, \: \frac{\large 1}{\large \sqrt 3}, \: \frac{\large 1}{\large \sqrt 3}$

$\begin{array}{1 1} (A) \large(\frac{1}{\sqrt3},\frac{1}{\sqrt3},-\frac{1}{\sqrt3}) \\ (B) \large(\frac{1}{\sqrt3},-\frac{1}{\sqrt3},\frac{1}{\sqrt3}) \\ (C) \large(-\frac{1}{\sqrt3},\frac{1}{\sqrt3},\frac{1}{\sqrt3}) \\ (D) \large(\frac{1}{\sqrt3},\frac{1}{\sqrt3},\frac{1}{\sqrt3}) \end{array}$

Toolbox:
• The cosine of the angle inclined by the vector to the axes is the direction cosines.
Step 1:
Let $a_1,a_2$ and $a_3$ be the direction ratios of the vector.
Since the vector is equally inclined to the axis.
Let $a_1=a_2=a_3=p$
$\Rightarrow \overrightarrow a=p\hat i+p\hat j+p\hat k$
Therefore $\sqrt{p^2+p^2+p^2}=p\sqrt 3$
Step 2:
Unit vector $\overrightarrow a=\large\frac{\overrightarrow a}{|\overrightarrow a|}$
$\qquad\qquad\;\;\;\;=\large\frac{p\hat i+p\hat j+p\hat k}{p\sqrt{3}}$
$\qquad\qquad\;\;\;\;=\large\frac{1}{\sqrt 3}$$\hat i+\large\frac{1}{\sqrt 3}$$\hat j+\large\frac{1}{\sqrt 3}$$\hat k Its direction cosines are \big(\large\frac{1}{\sqrt 3}$$\hat i+\large\frac{1}{\sqrt 3}$$\hat j+\large\frac{1}{\sqrt 3}$$\hat k\big)$
Step 3:
Hence the direction cosines are equally inclined to the axis.