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Scalar Or Dot Product — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsScalar Or Dot Product3 Marks
Question
If $\vec{\text{a}}=2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}},\vec{\text{b}}=\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}$ and $\vec{\text{c}}=\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}}.$ find $\lambda$ such that $\vec{\text{a}}$ is perpendicular to $\lambda\vec{\text{b}}+\vec{\text{c}}.$
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Answer

The given vectors are $\vec{\text{a}}=2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}},\vec{\text{b}}=\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}$ and $\vec{\text{c}}=\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}}.$
Now,
$\lambda\vec{\text{b}}+\vec{\text{c}}=\lambda(\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}})+(\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}})\\=(\lambda+1)\hat{\text{i}}+(\lambda+3)\hat{\text{j}}-(2\lambda+1)\hat{\text{k}}$
It is given
$\vec{\text{a}}\perp\big(\lambda\vec{\text{b}}+\vec{\text{c}}\big)$
$\Rightarrow\vec{\text{a}}.\big(\lambda\vec{\text{b}}+\vec{\text{c}}\big)=0$
$\Rightarrow\big(2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big).\Big[(\lambda+1)\hat{\text{i}}+(\lambda+3)\hat{\text{j}}-(2\lambda+1)\hat{\text{k}}\Big]=0$
$\Rightarrow2(\lambda+1)-(\lambda+3)-(2\lambda+1)=0$
$\Rightarrow2\lambda+2-\lambda-3-2\lambda-1=0$
Thus, the value of $\lambda$ is -2

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