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VECTOR ALGEBRA — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsVECTOR ALGEBRA3 Marks
Question
The two adjacent sides of a parallelogram are $2\hat{\text{i}}-4\hat{\text{j}}+5\hat{\text{k}}\ \text{and}\ \hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}.$ Find the unit vector parallel to its diagonal. Also, find its area.
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Answer

Adjacent sides of a parallelogram are given as: $\vec{\text{a}}=2\hat{\text{i}}-4\hat{\text{j}}+5\hat{\text{k}}\ \text{and}\ \vec{\text{b}}=\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}$
Then, the diagonal of a parallelogram is given by $\vec{\text{a}}+\vec{\text{b}}.$
$\vec{\text{a}}+\vec{\text{b}}=(2+1)\hat{\text{i}}+(-4-2)\hat{\text{j}}+(5-3)\hat{\text{k}}$ $=3\hat{\text{i}}-6\hat{\text{j}}+2\hat{\text{k}}$
Thus, the unit vector parallel to the diagonal is
$\frac{\vec{\text{a}}+\vec{\text{b}}}{\Big|\vec{\text{a}}+\vec{\text{b}}\Big|}=\frac{3\hat{\text{i}}-6\hat{\text{j}}+2\hat{\text{k}}}{\sqrt{3^2+(-6)^2+2^2}}=\frac{3\hat{\text{i}}-6\hat{\text{j}}+2\hat{\text{k}}}{\sqrt{9+36+4}}$ $=\frac{3\hat{\text{i}}-6\hat{\text{j}}+2\hat{\text{k}}}{7}=\frac{3}{7}\hat{\text{i}}-\frac{6}{7}\hat{\text{j}}+\frac{2}{7}\hat{\text{k}}.$
$\therefore$ Area of parallelogram ABCD $=\big|\vec{\text{a}}\times\vec{\text{b}}\big|$
$\vec{\text{a}}\times\vec{\text{b}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&-4&5\\1&-2&-3\end{vmatrix}$
$=\hat{\text{i}}(12+10)-\hat{\text{j}}(-6-5)+\hat{\text{k}}(-4+4)$
$=22\hat{\text{i}}+11\hat{\text{j}}$
$=11\big(2\hat{\text{i}}+\hat{\text{j}}\big)$
$\therefore\big|\vec{\text{a}}\times\vec{\text{b}}\big|=11\sqrt{2^2+1^2}=11\sqrt{5}$
Hence, the area of the parallelogram is $11\sqrt{5}$ square units.

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