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Vector or Cross Product — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsVector or Cross Product3 Marks
Question
Find a vector of magnitude 49, which is perpendicular to both the vectors $2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}$ and $3\hat{\text{i}}-6\hat{\text{j}}+2\hat{\text{k}}.$
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Answer

Let, $\vec{\text{a}}=2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}},\vec{\text{b}}=3\hat{\text{i}}-6\hat{\text{j}}+2\hat{\text{k}}$
If $\vec{\text{a}}=\text{a}_1\hat{\text{i}}+\text{b}_1\hat{\text{j}}+\text{c}_1\hat{\text{k}}$ and
$\vec{\text{b}}=\text{a}_2\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{c}_2\hat{\text{k}},$ then,
$\vec{\text{a}}\times\vec{\text{b}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\\text{a}_1&\text{b}_1&\text{c}_1\\\text{a}_2&\text{b}_2&\text{c}_2 \end{vmatrix}$
$=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&3&6\\3&-6&2 \end{vmatrix}$
$=\hat{\text{i}}(6+36)-\hat{\text{j}}(4-18)+\hat{\text{k}}(-12-9)$
$=42\hat{\text{i}}+14\hat{\text{j}}-21\hat{\text{k}}$
$=7\big(6\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\big)$
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=7\sqrt{(6)^2+(2)^2(-3)^2}$
$=7\sqrt{36+4+9}$
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=7\sqrt{49}$
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=7\times7$
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=49$

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