Question
Let $$a=2i+j+k,\,b=i+2j-k$$ and a unit vector $$c$$ be coplanar. If $$c$$ is perpendicular to $$a,$$ then $$c=?$$
A.
$$\frac{1}{{\sqrt 2 }}\left( { - j + k} \right)$$
B.
$$\frac{1}{{\sqrt 3 }}\left( {- i - j - k} \right)$$
C.
$$\frac{1}{{\sqrt 5 }}\left( { i - 2j} \right)$$
D.
$$\frac{1}{{\sqrt 3 }}\left( { i - j - k} \right)$$
Answer :
$$\frac{1}{{\sqrt 2 }}\left( { - j + k} \right)$$
Solution :
As $$c$$ is coplanar with $$a$$ and $$b,$$ we take,
$$c = \alpha a + \beta b.....(1)$$
where $$\alpha ,\beta $$ are scalars.
As $$c$$ is perpendicular to $$a,\,c.a=0$$
$$ \therefore $$ From (1) we get, $$0 = \alpha \,a.a + \beta \,b.a$$
$$\eqalign{
& \Rightarrow 0 = \alpha \left( 6 \right) + \beta \left( {2 + 2 - 1} \right) = 3\left( {2\alpha + \beta } \right) \Rightarrow \beta = - 2\alpha \cr
& {\text{Thus, }}c = \alpha \left( {a - 2b} \right) = \alpha \left( { - 3j + 3k} \right) = 3\alpha \left( { - j + k} \right) \cr
& \Rightarrow {\left| {\vec c} \right|^2} = 9{\alpha ^2}\left( {1 + 1} \right) = 18{\alpha ^2}\,\, \Rightarrow 1 = 18{\alpha ^2} \cr
& \Rightarrow \alpha = \pm \frac{1}{{3\sqrt 2 }} \cr
& \therefore c = \pm \frac{1}{{\sqrt 2 }}\left( { - j + k} \right) \cr} $$
Thus, we may take $$c = \frac{1}{{\sqrt 2 }}\left( { - j + k} \right).$$