Let $\vec a =\hat i + 2\hat j+3\hat k, \vec b = −\hat i+ 2\hat j +\hat k,\vec c = 3\hat i+\hat j$ be three vectors. If $(\vec a+\vec λ\vec b)$ is perpendicular to $\vec c$, then the value of $λ$ is

5

The correct answer is Option (2) → 5

$\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$

$\vec{b} = -\hat{i} + 2\hat{j} + \hat{k}$

$\vec{c} = 3\hat{i} + \hat{j}$

Condition: $(\vec{a} + \lambda \vec{b}) \cdot \vec{c} = 0$

Compute $\vec{a} + \lambda \vec{b}$:

$\vec{a} + \lambda \vec{b} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(-\hat{i} + 2\hat{j} + \hat{k})$

$= (1 - \lambda)\hat{i} + (2 + 2\lambda)\hat{j} + (3 + \lambda)\hat{k}$

Now dot with $\vec{c} = 3\hat{i} + \hat{j}$:

$(\vec{a} + \lambda \vec{b}) \cdot \vec{c} = (1 - \lambda)\cdot3 + (2 + 2\lambda)\cdot1 + (3 + \lambda)\cdot0$

$= 3 - 3\lambda + 2 + 2\lambda = 5 - \lambda$

For perpendicularity: $5 - \lambda = 0 \Rightarrow \lambda = 5$

$\lambda = 5$