Let a = i +4 j +2 k , b =3 i -2 j +7 k and c =2 i - j +4 k . Find… - Vidyadip

Question

$\text{Let}\ \vec{\text{a}}=\hat{\text{i}}+4\hat{\text{j}}+2\hat{\text{k}},\ \vec{\text{b}}=3\hat{\text{i}}-2\hat{\text{j}}+7\hat{\text{k}}$ and $ \vec{\text{c}}=2\hat{\text{i}}-\hat{\text{j}}+4\hat{\text{k}}.$ Find a vector $\vec{\text{d}}$ which is perpendicular to both $\vec{\text{a}}\ \text{and}\ \vec{\text{b}},\ \text{and}\ \vec{\text{c}}\cdot\vec{\text{d}}=15.$

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Answer

$\text{Let}\ \vec{\text{d}}=\text{d}_{1}\hat{\text{i}}+\text{d}_{2}\hat{\text{j}}+\text{d}_{3}\hat{\text{k}}$ Since $\vec{\text{d}}$ is perpendicular to both $\vec{\text{a}}\ \text{and}\ \vec{\text{b}},$ we have: $\vec{\text{d}}\cdot\vec{\text{a}}=0$ $\Rightarrow \text{d}_{1}+4\text{d}_{2}+2\text{d}_{3}=0\ \ \ ...\text{(i)}$ And, $\vec{\text{d}}\cdot\vec{\text{b}}=0$ $\Rightarrow 3\text{d}_{1}-2\text{d}_{2}+7\text{d}_{3}=0\ \ \ ...\text{(ii)}$ Also, it is given that: $\vec{\text{c}}\cdot\vec{\text{d}}=15$ $\Rightarrow 2\text{d}_{1}-\text{d}_{2}+4\text{d}_{3}=15\ \ \ ...\text{(iii)}$ On solving (i), (ii), and (iii), we get: $\text{d}_{1}=\frac{160}{3},\text{d}_{2}=-\frac{5}{3}\ \text{and}\ \text{d}_{3}=-\frac{70}{3}$ $\therefore\vec{\text{d}}=\frac{160}{3}\hat{\text{i}}-\frac{5}{3}\hat{\text{j}}-\frac{70}{3}\hat{\text{k}}=\frac{1}{3}\Big(160\hat{\text{i}}-5\hat{\text{j}}-70\hat{\text{k}}\Big)$Hence, the required vector is $\frac{1}{3}\Big(160\hat{\text{i}}-5\hat{\text{j}}-70\hat{\text{k}}\Big)$

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