Question
If vectors $\vec{a}=2 \hat{ i }+2 \hat{\jmath}+3 \hat{ k }, \vec{b}=-\hat{ i }+2 \hat{\jmath}+\hat{ k }$ and $\vec{c}=3 \hat{ i }+\hat{ j }$ are such that $\vec{b}+\lambda \vec{c}$ is perpendicular to $\vec{a}$, then find the value of $\lambda$.

Answer

We have $\overrightarrow{ b }+\lambda \overrightarrow{ c }=(-1+3 \lambda) \hat{ i }+(2+\lambda) \hat{\jmath}+\hat{ k }$ $(\vec{b}+\lambda \vec{c}) \cdot \vec{a}=0 \Rightarrow 2(-1+3 \lambda)+2(2+\lambda)+3=0$
$\lambda=-\frac{5}{8}$

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