Question

Let $a =\hat{ i }+2 \hat{ j }-\hat{ k }$ and $b =\hat{ i }+\hat{ j }+\hat{ k } .$ If $p$ is a
unit vector such that $[abp]$ is maximum. then $p =$

• A. $\frac{1}{\sqrt{6}}(\hat{ i }-2 \hat{ j }+\hat{ k })$
• B. $\frac{1}{\sqrt{3}}(\hat{ i }+\hat{ j }-\hat{ k })$
• C. $\frac{1}{\sqrt{14}}(3 \hat{ i }-2 \hat{ j }-\hat{ k })$
• D. $\frac{1}{\sqrt{14}}(\hat{ i }+2 \hat{ j }+3 \hat{ k })$

Answer 1

Answer: (C)

Given, $a =\hat{ i }+2 \hat{ j }-\hat{ k }, b =\hat{ i }+\hat{ j }+\hat{ k }$
$a \times b =\begin{vmatrix}\hat{ i } & \hat{ j } & \hat{ k } \\ 1 & 2 & -1 \\ 1 & 1 & 1\end{vmatrix}=3 \hat{ i }-2 \hat{ j }-\hat{ k }$
$[ a\,b \,p ]=p \cdot( a \times b )=p \cdot(3 \hat{ i }-2 \hat{ j }-\hat{ k })$
$[ a \,b \,p ]=|P|| a \times b | \cos \theta$
$[a \,b\, p ]$ is maximum
$\therefore p =\frac{ a \times b }{| a \times b |} $
$p =\frac{1}{\sqrt{14}}(3 \hat{ i }-2 \hat{ j }-\hat{ k })$