If $\vec a=-\hat i+\hat j+\hat k,\vec b=2\hat i+0\hat j+\hat k$, then a vector $\vec X$ satisfying the conditions:
(i) that it is coplanar with $\vec a$ and $\vec b$
(ii) that is perpendicular to $\vec b$,
(iii) that $\vec a.\vec X=7$, is
Answer & explanation
Correct answer: option 2
Since $\vec X$ is in the plane of $\vec a$ and $\vec b$ perpendicular to $\vec b$.
$∴\vec X=λ\{\vec b×(\vec a×\vec b)\}$
$⇒\vec X=λ[(\vec b.\vec b)\vec a-(\vec b.\vec a)\vec b]$
$⇒\vec X=λ[5(-\hat i+\hat j+\hat k)-(-1)(2\hat i+0\hat j+\hat k)]$
$⇒\vec X=λ[-5\hat i +5\hat j +5\hat k+2\hat i+0\hat j +\hat k]$
$⇒\vec X=λ[-3\hat i+5\hat j+6\hat k]$
Now,
$\vec a.\vec X=7$
$⇒(-\hat i+\hat j+\hat k).λ(-3\hat i+5\hat j+6\hat k)=7$
$⇒λ(3+5+6)=7⇒λ=\frac{1}{2}$
Hence, $\vec X=\frac{1}{2}(-3\hat i+5\hat j+6\hat k)$