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 9{{x}^{2}}+9{{x}^{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) $