Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Vector Algebra
0 votes

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} $

Can you answer this question?

1 Answer

0 votes
  • 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$
$|\overrightarrow {OP}|=\sqrt{\big(\large\frac{\sqrt 3}{2}\big)^2+\big(\large\frac{1}{2}^2\big)}$
Step 3:
Hence it is a unit vector.
Therefore the required vector is $\large\frac{\sqrt3}{2}$$\hat i+\frac{1}{2}$$\hat j$
answered May 23, 2013 by sreemathi.v
edited May 23, 2013 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App