Question

A vector is equally inclined to the $x$-axis, $y$-axis and $z$-axis respectively, its direction cosines are

• A. $<\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}>$
• B. $<-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}>$
• C. $<\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}>$ or $<-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}>$
• D. None of these

Answer 1

The correct option is C $<\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}>$ or $<-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}>$

Let the vector $v$ make an angle $\alpha$ with each of the three axes, then direction cosine of $v$ are

$<\cos \alpha ,\cos \alpha ,\cos \alpha >$

$\Rightarrow {\cos }^2\alpha =\frac{1}{3}$

$\Rightarrow \cos \alpha =\pm \frac{1}{\sqrt{3}}$

Hence, direction cosine of $v$ are

$<\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}>$ or

$<-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}>$