Let $\vec{a}=\hat{i}+\hat{j}+\hat{k},\vec{b}=-\hat{i}+\hat{j}.$ Then a vector $\vec{c}$ satisfying $\vec{a}×\vec{c}=\vec{b}$ and $\vec{a}.\vec{c}=8$ is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → $3\hat{i}+3\hat{j}+2\hat{k}$
$\vec{a}×\vec{c}=\vec{b}$
let $\vec c=x\hat i+y\hat j+z\hat k$
$\vec a.\vec c=x+y+z=8$ ...(1)
$\vec{a}×\vec{c}=\vec{b}$
so $(z-y)\hat i+(x-z)\hat j+(y-z)\hat k\begin{vmatrix}\hat i&\hat j&\hat k\\1&1&1\\x&y&z\end{vmatrix}=\vec b$
$=-\hat i+\hat j$
so $z=y-1$
$x-z=1$
so $x=z+1$
$y=z+1$
so $x=y$
from (1)
$y+y+y-1=8$
$y=3$
so $z=2,x=3$
$\vec c=3\hat{i}+3\hat{j}+2\hat{k}$