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Vector or Cross Product — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsVector or Cross Product5 Marks
Question
The two adjacent sides of a parallelogram are $2\hat{\text{i}}-4\hat{\text{j}}-5\hat{\text{k}}$ and $2\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}.$ Find the two unit vectors parallel to its diagonals. Using the diagonal vectors, find the area of the parallelogram.
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Answer

The two adjacent sides of a parallelogram are $2\hat{\text{i}}-4\hat{\text{j}}-5\hat{\text{k}}$ and $2\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}.$ suppose $\vec{\text{a}}=2\hat{\text{i}}-4\hat{\text{j}}-5\hat{\text{k}}$ and $\vec{\text{b}}=2\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$

Then any one diagonal of a parallelogram is $\vec{\text{P}}=\vec{\text{a}}+\vec{\text{b}}.$
$\vec{\text{P}}=\vec{\text{a}}+\vec{\text{b}}$
$=2\hat{\text{i}}-4\hat{\text{j}}-5\hat{\text{k}}+2\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$
$=4\hat{\text{i}}-2\hat{\text{j}}-2\hat{\text{k}}$
Therefore, unit vector along the diagonal is $\frac{\vec{\text{P}}}{\big|\vec{\text{P}}\big|}=\frac{4\hat{\text{i}}-2\hat{\text{j}}-2\hat{\text{k}}}{\sqrt{16+4+4}}=\frac{2\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}}{\sqrt{6}}.$
Another diagonal of a parallelogram is $\vec{\text{P}}'=\vec{\text{b}}-\vec{\text{a}}.$
$\vec{\text{P}}'=\vec{\text{b}}-\vec{\text{a}}$
$=2\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}-2\hat{\text{i}}+4\hat{\text{j}}+5\hat{\text{k}}$
$=6\hat{\text{j}}+8\hat{\text{k}}$
Therefore, unit vector along the diagonal is $\frac{\vec{\text{P}}}{\big|\vec{\text{P}}\big|}=\frac{6\hat{\text{j}}+8\hat{\text{k}}}{\sqrt{36+64}}=\frac{6\hat{\text{j}}+8\hat{\text{k}}}{10}=\frac{3\hat{\text{j}}+4\hat{\text{k}}}{5}.$
Now,
$\vec{\text{P}}\times\vec{\text{P}}'=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\4&-2&-2\\0&6&8\end{vmatrix}$
$=\hat{\text{i}}(-16+12)-\hat{\text{j}}(32-0)+\hat{\text{k}}(24-0)$
$=-4\hat{\text{i}}-32\hat{\text{j}}+24\hat{\text{k}}$
Area of parallelogram $\frac{\big|\vec{\text{p}}\times\vec{\text{p}'}\big|}{2}=\frac{\sqrt{16+1024+576}}{2}=\frac{\sqrt{1616}}{2}$
$=\frac{4\sqrt{101}}{2}=2\sqrt{101}\text{ square units}$

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