# Write down a unit vector in XY plane, making an angle of 30 degress with the positive direction of $$x$$ axis.

$\begin{array}{1 1} (A) \large-\frac{\sqrt3}{2}\hat i+\large\frac{1}{2}\hat j\\ (B) \large\frac{\sqrt3}{2}\hat i-\large\frac{1}{2}\hat j \\ (C) \large\frac{\sqrt3}{2}\hat i+\large\frac{1}{2}\hat j\\ (D) \large-\frac{\sqrt3}{2}\hat i-\large\frac{1}{2}\hat j \end{array}$

Toolbox:
• The angles made by the vector with the respective axes is the direction angles.
• The cosine of these angles is the direction cosines.
• $|\overrightarrow a|=\sqrt{a_1^2+a_2^2+a_3^2}$
• Magnitude of a unit vector is 1.
Step 1:
Let $OX,OY,OZ$ be the co-ordinate axes.The line $OP$ lies on the $XY$ plane.
Hence $\angle XOP=30^{\large\circ}$
$\qquad\angle POY=60^{\large\circ}$
$\qquad\angle POZ=90^{\large\circ}$
This is clearly shown in the figure.
Therefore the direction angles of $OP$ are $30^{\large\circ}$,$60^{\large\circ}$ and $90^{\large\circ}$
Step 2:
Hence its direction cosines are $\cos 30^{\large\circ}$,$\cos 60^{\large \circ}$,$\cos 90^{\large \circ}$ which are equal to $\large\frac{\sqrt 3}{2},\frac{1}{2}$,$0$ respectively.
$\Rightarrow \overrightarrow {OP}=\large\frac{\sqrt 3}{2}$$\hat i+\large\frac{1}{2}$$\hat j$
$|\overrightarrow {OP}|=\sqrt{\big(\large\frac{\sqrt 3}{2}\big)^2+\big(\large\frac{1}{2}^2\big)}$
$\qquad=\sqrt{\large\frac{3}{4}+\frac{1}{4}}$
$\qquad=1.$
Step 3:
Hence it is a unit vector.
Therefore the required vector is $\large\frac{\sqrt3}{2}$$\hat i+\frac{1}{2}$$\hat j$
edited May 23, 2013