Given $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}, \vec{b} = 3\hat{i} - \hat{k}$ and $\vec{c} = 2\hat{i} + \hat{j} - 2\hat{k}$. Find a vector $\vec{d}$ which is perpendicular to both $\vec{a}$ and $\vec{b}$ and $\vec{c} \cdot \vec{d} = 3$.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $3\hat{i} + 15\hat{j} + 9\hat{k}$ ##
$\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$
$\vec{b} = 3\hat{i} - \hat{k}$
$\vec{c} = 2\hat{i} + \hat{j} - 2\hat{k}$
$\vec{d} = (x\hat{i} + y\hat{j} + z\hat{k})$
Since, $\vec{d}$ is perpendicular to $\vec{a}$ and $\vec{b}$
$\vec{d} \cdot \vec{a} = (x\hat{i} + y\hat{j} + z\hat{k}) \cdot (2\hat{i} - \hat{j} + \hat{k})$
$= 2x - y + z = 0$
$\vec{d} \cdot \vec{b} = (x\hat{i} + y\hat{j} + z\hat{k}) \cdot (3\hat{i} - \hat{k})$
$= 3x - z = 0$
$\vec{c} \cdot \vec{d} = (2\hat{i} + \hat{j} - 2\hat{k}) \cdot (x\hat{i} + y\hat{j} + z\hat{k})$
$= 2x + y - 2z = 3$
On solving $x = 3, y = 15, z = 9$
vector $\vec{d} = x\hat{i} + y\hat{j} + z\hat{k}$
$= 3\hat{i} + 15\hat{j} + 9\hat{k}$