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Motion in a Plane — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsMotion in a Plane2 Marks
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.
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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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