Question

Let $\vec{a}=\hat{i}-2\,\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+\hat{k}$ be two vectors. If $\vec{c}$ is a vector such that $\vec{b}\times\vec{c}=\vec{b}\times\vec{a}$ and $\vec{c}\cdot\vec{a}=0,$ then $\vec{c} \cdot \vec{b}$ is equal to :

• A. $\frac{1}{2}$
• B. $- \frac{3}{2}$
• C. $- \frac{1}{2}$
• D. $-1$

Answer 1

The Correct Option is C

\(\vec{b}\times\vec{c}-\vec{b}\times\vec{a} = \vec{0}\)
\(\vec{b}\times \left(\vec{c}- \vec{a}\right) = \vec{0}\)
\(\vec{b} = \lambda\left(\vec{c}- \vec{a}\right) \quad...\left(i\right)\)
\(\vec{a}\cdot \vec{b} = \lambda\left( \vec{a}\cdot \vec{c}- \vec{a}^{2}\right)\)
\(4 = \lambda\left(0 - 6\right) \Rightarrow \lambda = \frac{-4}{6} = \frac{-2}{3}\) 
from \(\left(i\right) \vec{b} = \frac{-2}{3}\left(\vec{c}- \vec{a}\right)\)
\(\vec{c} = \frac{-3}{2}\vec{b}+ \vec{a} = \frac{-1}{2}\left(\hat{i}+\hat{j}+\hat{k}\right)\)

\(\vec{b}\cdot \vec{c} = -\frac{1}{2}\)