Let $\vec{a}=2\hat{i}+2\hat{j}+2\hat{k}$ and $\vec{b}=2\hat{i}+\hat{j}-2\hat{k}$. Then a vector of magnitude 24, which is perpendicular to the vectors $\vec{a} +\vec{b}$ and $\vec{a}-\vec{b}$, is :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $\frac{36\sqrt{26}}{13}\hat i-\frac{48\sqrt{26}}{13}+\frac{12\sqrt{26}}{13}\hat k$
$\vec{a}=2\hat{i}+2\hat{j}+2\hat{k}$
$\vec{b}=2\hat{i}+\hat{j}-2\hat{k}$
$\vec{a}+\vec{b}=4\hat i+3\hat j$
$\vec{a}-\vec{b}=\hat j+4\hat k$
$\vec v=(\vec{a}+\vec{b})×(\vec{a}-\vec{b})=\begin{vmatrix}\hat i&\hat j&\hat k\\4&3&0\\0&1&4\end{vmatrix}$
$\vec v=12\hat i-16\hat j+4\hat k$
$\hat v=\frac{\vec v}{|\vec v|}=\frac{3\sqrt{26}}{26},-\frac{48\sqrt{26}}{13},\frac{\sqrt{26}}{26}$
$24\hat v=\frac{36\sqrt{26}}{13}\hat i-\frac{48\sqrt{26}}{13}+\frac{12\sqrt{26}}{13}\hat k$