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VECTOR ALGEBRA — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsVECTOR ALGEBRA4 Marks
Question
$\text{Let}\ \vec{\text{a}}=4\hat{\text{i}}+5\hat{\text{j}}-\hat{\text{k}},\vec{\text{b}}=\hat{\text{i}}-4\hat{\text{j}}+5\hat{\text{k}}$ and $\vec{\text{c}}=3\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}$ Find a vector $\vec{\text{d}}$ which is perpendicular to both $\vec{\text{c}}\ \text{and }\vec{\text{b}}\ \text{and}\ \vec{\text{d}}\cdot\vec{\text{a}}=21.$
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Answer

$\vec{\text{a}}=4\hat{\text{i}}+5\hat{\text{j}}-\hat{\text{k}}$
$\vec{\text{b}}=-\hat{\text{i}}-4\hat{\text{j}}+5\hat{\text{k}}$
$\vec{\text{c}}=3\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}$
To find vector $\vec{\text{d}}$ such that
$\vec{\text{d}}\cdot\vec{\text{c}}=0$
$\vec{\text{d}}\cdot\vec{\text{b}}=0$
$\vec{\text{d}}\cdot\vec{\text{a}}=21$
$\text{Let}\ \vec{\text{d}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}$
$\Big(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}\Big)\cdot\Big(3\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\Big)=0$
$3\text{x}+\text{y}-\text{z}=0\ \ \ ....(1)$
$\Big(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}\Big)\cdot\Big(\hat{\text{i}}+5\hat{\text{j}}-\hat{\text{k}}\Big)=0$
$\text{x}-4\text{y}+5\text{z}=0\ \ \ ....(2)$
$\Big(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}\Big)\cdot\Big(4\hat{\text{i}}+5\hat{\text{j}}-\hat{\text{k}}\Big)=21$
$4\text{x}+5\text{y}-\text{z}=21\ \ \ ....(3)$
$\text{eq}^\text{n}(1)\times4+\text{eq}^\text{n}(2)$

$\text{eq}^\text{n}(2)\times5+\text{eq}^\text{n}(3)\times4$

$21(\text{x}+\text{z})=84$
$\text{x}+\text{z}=4\ \ \ ....(5)$
$\text{eq}.(4)-(5)$
$12\text{x}=-4$
$\text{x}=\frac{-4}{12}=\frac{-1}{3}$
$\text{z}=4-\text{x}$
$\text{z}=4+\frac{1}{3}=\frac{13}{3}$
$\text{Put x}\ \&\ \text{z in }(1)$
$3\text{x}+\text{y}-\text{z}=0$
$3\times\Big(\frac{-1}{3}\Big)+\text{y}-\frac{13}{3}=0$
$\text{y}=\frac{13}{3}+1$
$\text{y}=\frac{16}{3}$
$\vec{\text{d}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}$
$\vec{\text{d}}=\frac{-1}{3}\hat{\text{i}}+\frac{16}{3}\hat{\text{j}}+\frac{13}{3}\hat{\text{k}}$

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