Question
Determine $\lambda$ such that:
$\vec{\text{A}}=2\hat{\text{i}}+\lambda\hat{\text{j}}+\hat{\text{k}};\ \vec{\text{B}}=4\hat{\text{i}}-2\hat{\text{j}}-2\hat{\text{k}}$ are perpendicular to each other.

Answer

$\vec{\text{A}}=2\hat{\text{i}}+\lambda\hat{\text{j}}+\hat{\text{k}}$
$\vec{\text{B}}=4\hat{\text{i}}-2\hat{\text{j}}-2\hat{\text{k}}$
$\vec{\text{A}}.\vec{\text{B}}=(2\hat{\text{i}}+\lambda\hat{\text{j}}+\hat{\text{k}}).(4\hat{\text{i}}-2\hat{\text{j}}-2\hat{\text{k}})$
$=8-2\lambda-2$
$0=6-2\lambda$ or $\lambda=3$

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