Question

If a unit vector lies in yz–plane and makes angles of $30^{o}$ and $60^{o}$ with the positive y-axis and z-axis respectively, then its components along the co-ordinate axes will be

• A. $\frac{\sqrt{3}}{2},\frac{1}{2},0$
• B. $0,\frac{\sqrt{3}}{2},\frac{1}{2}$
• C. $\frac{\sqrt{3}}{2},0,\frac{1}{2}$
• D. $0,\frac{1}{2},\frac{\sqrt{3}}{2}$

Answer 1

Answer: (B)

$0,\frac{\sqrt{3}}{2},\frac{1}{2}$

Answer 2

Let unit vector be $y\mathbf{i} + z\mathbf{k},$ then $\sqrt{y^{2} + z^{2}} = 1$ …..(i)

Since given that $\cos 30{^\circ} = \frac{(y\mathbf{j} + z\mathbf{k}).(y\mathbf{j})}{|y\mathbf{j} + z\mathbf{k}||y\mathbf{j}|}$

$\Rightarrow \frac{y^{2}}{\left( \sqrt{y^{2} + z^{2}} \right)y} = \frac{\sqrt{3}}{2} \Rightarrow y = \frac{\sqrt{3}}{2}$,

$(\because\sqrt{y^{2} + z^{2}} = 1$by (i))

Similarly, $\cos 60{^\circ} = \frac{(y\mathbf{j} + z\mathbf{k}).z\mathbf{k}}{|y\mathbf{j} + z\mathbf{k}||z\mathbf{k}|} \Rightarrow z = \frac{1}{2}$

Hence the components of unit vector are $0,\frac{\sqrt{3}}{2},\frac{1}{2}.$

Trick : Since the vector lies in $yz -$plane, so it will be either $0\mathbf{i} + \frac{\sqrt{3}}{2}\mathbf{j} + \frac{1}{2}\mathbf{k}$ or $0\mathbf{i} + \frac{1}{2}\mathbf{j} + \frac{\sqrt{3}}{2}\mathbf{k}.$ But the vector $\frac{\sqrt{3}}{2}\mathbf{j} + \frac{1}{2}\mathbf{k}$ makes angle $30{^\circ}$with $y -$axis and that of $60{^\circ}$ with z-axis.