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VECTOR ALGEBRA — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsVECTOR ALGEBRA3 Marks
Question
If $\vec{\text{a}}=2\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}},\vec{\text{b}}=-\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}},$and $\vec{\text{c}}=3\hat{\text{i}}+\hat{\text{j}}$ are such that $\vec{\text{a}}+\lambda\vec{\text{b}}$ is perpendicular to $\vec{\text{c}},$ then find the value of $\lambda.$
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Answer

The given vectors are $\vec{\text{a}}=2\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}},\vec{\text{b}}=-\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}},$ and $\vec{\text{c}}=3\hat{\text{i}}+\hat{\text{j}}$.
Now,
$\vec{\text{a}}+\lambda\vec{\text{b}}=\big(2\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)+\lambda\big(-\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}\big)\\=(2-\lambda)\hat{\text{i}}+(2+2\lambda)\hat{\text{j}}+(3+\lambda)\hat{\text{k}}$
If $\big(\vec{\text{a}}+\lambda\vec{\text{b}}\big)$ is perpendicular to $\vec{\text{c}},$ then
$\big(\vec{\text{a}}+\lambda\vec{\text{b}}\big).\vec{\text{c}}=0.$
$\Rightarrow\Big[\big(2-\lambda\big)\hat{\text{i}}+(2+2\lambda)\hat{\text{j}}+(3+\lambda)\hat{\text{k}}\Big].\big(3\hat{\text{i}}+\hat{\text{j}}\big)=0$
$\Rightarrow(2-\lambda)3+(2+2\lambda)1+(3+\lambda)0=0$
$\Rightarrow6-3\lambda+2+2\lambda=0$
$\Rightarrow-\lambda+8=0$
$\Rightarrow\lambda=8$
Hence, the required value of $\lambda$ is 8.

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