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Algebra of Vectors — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAlgebra of Vectors2 Marks
Question
If $\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}},\ \vec{\text{b}}=4\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$and $\vec{\text{c}}=\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$, find a vecctor of magnitude 6 units which is parallel to the vector $2\vec{\text{a}}-\vec{\text{b}}+3\vec{\text{c}}$.
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Answer

We have, $\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}},\ \vec{\text{b}}=4\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$and $\vec{\text{c}}=\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$Then,
$2\vec{\text{a}}-\vec{\text{b}}+3\vec{\text{c}}=2\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)\\-\big(4\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}\big)+3\big(\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}\big)$ $=\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}$ $\therefore$ A unit vector parallel to $2\vec{\text{a}}-\vec{\text{b}}+3\vec{\text{c}}$ is $\frac{2\vec{\text{a}}-\vec{\text{b}}+3\vec{\text{c}}}{\big|2\vec{\text{a}}-\vec{\text{b}}+3\vec{\text{c}}\big|}=\frac{(\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}})}{\sqrt{1^2+(-2)^2+2^2}}$ $=\frac{(\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}})}{\sqrt9}$ $=\frac{(\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}})}3$ Hence, Required vector $=\frac{6}3\big(\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}\big)$ $=2\hat{\text{i}}-4\hat{\text{j}}+4\hat{\text{k}}$

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