If $\vec{a} = 2\hat{i} + 2\hat{j} + 3\hat{k}$, $\vec{b} = -\hat{i} + 2\hat{j} + \hat{k}$ and $\vec{c} = 3\hat{i} + \hat{j}$ are such that the vector $(\vec{a} + \lambda \vec{b})$ is perpendicular to vector $\vec{c}$, then find the value of $\lambda$. |
$8$ $-8$ $2$ $-2$ |
$8$ |
The correct answer is Option (1) → $8$ ## If $\vec{a} + \lambda\vec{b}$ is perpendicular to $\vec{c}$, then $(\vec{a} + \lambda\vec{b}) \cdot \vec{c} = 0$ $\Rightarrow \{(2-\lambda)\hat{i} + (2+2\lambda)\hat{j} + (3+\lambda)\hat{k}\} \cdot (3\hat{i} + \hat{j})= 0$ $\Rightarrow (2-\lambda) \cdot 3 + (2+2\lambda) \cdot 1 + (3+\lambda) \cdot 0 = 0$ $\Rightarrow 6 - 3\lambda + 2 + 2\lambda = 0$ $\Rightarrow -\lambda + 8 = 0$ $\Rightarrow \lambda = 8$ |