Question

Given two vectors are $\hat{i}-\hat{j}$ and $\hat{i}+2\hat{j},$ the unit vector is coplanar with the two vectors and perpendicular to the first. Find the vector?

• A. $+\frac{1}{\sqrt{2}}(i+k)$
• B. $+\frac{1}{\sqrt{5}}(2i+j)$
• C. $+\frac{1}{\sqrt{2}}(i+j)$
• D. $+\frac{1}{\sqrt{2}}(2i+3j)$

Answer 1

Answer: (C)

Let given two vectors are $a=i-j$ and $b=i+2j$
Again let third unit vector is c.
$\because$ c is coplanar with a, b
$\therefore$ $c=xa+yb$
$=x(i-j)+y(i+2j)$
$\Rightarrow$ $c=(a+y)i+(-x+2y)j$ ..(i)
Also, c is perpendicular to a $\therefore$
$a.c=0$
$\Rightarrow$ $(i-j).\{(x+y)i+(-x+2y)j\}=0$
$\Rightarrow$ $(x+y)-(-x+2y)=0$
$\Rightarrow$ $x+y+x-2y=0$
$\Rightarrow$ $2x-y=0\Rightarrow y=2x$
On putting this value of y in Eq. (i), we get
$c=(x+2x)i+(-x+4x)j$
$\Rightarrow$ $c=3xi+3xj$ ..(ii)
But c is a unit vector. So, $|c|=1$
$\Rightarrow$ ${(3x)}^2+{(3x)}^2=1\Rightarrow 9x^2+9x^2=1$
$\Rightarrow$ $x^2=\frac{1}{18}\Rightarrow x=\frac{1}{3\sqrt{2}}$
On putting this value of x in Eq. (ii), we get $c=3.\frac{1}{3\sqrt{2}}i+3.\frac{1}{3\sqrt{2}}j\Rightarrow c=\frac{1}{\sqrt{2}}(i+j)$