Question

A vector makes an angle with $X,Y$ and $Z$ axes that are in ratio $1:2:3$ respectively. The angle made by the vector with Y- axis is :
A. $\dfrac{\pi }{3}$
B. $\dfrac{\pi }{6}$
C. $\dfrac{\pi }{4}$
D. $\dfrac{\pi }{2}$

Answer 1

We know that x, y and z axis are perpendicular to each other. Hence we will start by assuming angles made with x, y and z axes respectively. Later on we will apply the formula based on angles . Then after finding the angles we will check what angle it makes with just y axis.

Formula used:
The formula that we will use here is the formula of direction of cosines:
${\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\chi = 1 - - - - - - - (1)$
$\Rightarrow {\cos ^2}\alpha = \dfrac{{1 + \cos 2\alpha }}{2} - - - - - - - (2)$
$\Rightarrow \cos 3\theta = 4{\cos ^3}\theta - 3\cos \theta - - - - - - - (3)$

Complete step by step answer:
Let us start by assuming that the angles made by the x, y and z axes are A, B and C respectively. If we look at the question it is given that A:B:C =1:2:3; so as mentioned in the hint , let us start by assuming that
$A = \theta$;
$\Rightarrow B = 2\theta$;
And $C = 3\theta$
Hence by using formula for direction cosines , we get:
${\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\chi = 1 \\\$
$\Rightarrow {\cos ^2}\theta + {\cos ^2}2\theta + {\cos ^2}3\theta = 1 \\\$
WE also know that : ${\cos ^2}\alpha = \dfrac{{1 + \cos 2\alpha }}{2}$,

Therefore by substituting ;
$\dfrac{{1 + \cos 2\theta }}{2} + \dfrac{{1 + \cos 4\theta }}{2} + \dfrac{{1 + \cos 6\theta }}{2} = 1$
$\Rightarrow 1 + \cos 2\theta + 1 + \cos 4\theta + 1 + \cos 6\theta = 2 \\\\$
\Rightarrow \cos 2\theta + \cos 4\theta + \cos 6\theta = 2 - 3 \\\
\Rightarrow \cos 2\theta + \cos 4\theta + \cos 6\theta = - 1 \\\
Again by using equation 3 that is: $\cos 3\theta = 4{\cos ^3}\theta - 3\cos \theta \\\$
And assuming $2\theta = p$we get;
\cos p + \cos 2p + \cos 3p = - 1 \\\
\Rightarrow \cos p + (2{\cos ^2}p - 1) + (4{\cos ^3}p - 3\cos p) = - 1 \\\
\Rightarrow \cos p + 2{\cos ^2}p + 4{\cos ^3}p - 3\cos p = 0 \\\
\Rightarrow 2{\cos ^2}p + 4{\cos ^3}p - 2\cos p = 0 \\\
\Rightarrow 2\cos p(\cos p + 2{\cos ^2}p - 1) = 0 \\\
\Rightarrow (2\cos p - 1)(\cos p + 1)\cos p = 0 \\\
\Rightarrow \cos p = 0, - 1,\dfrac{1}{2} \\\
As we before assumed that $2\theta = p$;
Hence, $\theta = \dfrac{\pi }{4},\dfrac{\pi }{2},\dfrac{\pi }{6}$
Now when ,
Case1: $\theta = \dfrac{\pi }{4},B = 2\theta = \dfrac{\pi }{2}$
Case2: $\theta = \dfrac{\pi }{2},B = 2\theta = \pi$
Case3: $\theta = \dfrac{\pi }{6},B = 2\theta = \dfrac{\pi }{3}$

Hence the correct answer can be option A or D.

Note: Here a common mistake that can be made is by not substituting the value of $2\theta = p$. if you solve the question considering only ‘p’ the values of the corresponding angles that is $\theta$ will be wrong. Also to check if your angles are correct you can divide the above mentioned angles with each other and check their ratios. In the above solved question we see that after dividing the values of $\theta$, they are in the ratio 1:2:3. Hence it is correct.