Practicing Success
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 : |
$16\hat{i}+16\hat{j}+8\hat{k}$ $16\hat{i}+16\hat{j}-8\hat{k}$ $\frac{36\sqrt{26}}{13}\hat i-\frac{48\sqrt{26}}{13}+\frac{12\sqrt{26}}{13}\hat k$ $16\hat{i}-16\hat{j}-8\hat{k}$ |
$\frac{36\sqrt{26}}{13}\hat i-\frac{48\sqrt{26}}{13}+\frac{12\sqrt{26}}{13}\hat k$ |
The correct answer is Option (3) → $\frac{36\sqrt{26}}{13}\hat i-\frac{48\sqrt{26}}{13}+\frac{12\sqrt{26}}{13}\hat k$ $\vec{a}=2\hat{i}+2\hat{j}+2\hat{k}$ $\vec{b}=2\hat{i}+\hat{j}-2\hat{k}$ $\vec{a}+\vec{b}=4\hat i+3\hat j$ $\vec{a}-\vec{b}=\hat j+4\hat k$ $\vec v=(\vec{a}+\vec{b})×(\vec{a}-\vec{b})=\begin{vmatrix}\hat i&\hat j&\hat k\\4&3&0\\0&1&4\end{vmatrix}$ $\vec v=12\hat i-16\hat j+4\hat k$ $\hat v=\frac{\vec v}{|\vec v|}=\frac{3\sqrt{26}}{26},-\frac{48\sqrt{26}}{13},\frac{\sqrt{26}}{26}$ $24\hat v=\frac{36\sqrt{26}}{13}\hat i-\frac{48\sqrt{26}}{13}+\frac{12\sqrt{26}}{13}\hat k$ |