Question

The two adjacent sides of a parallelogram are $2\hat{i}-4\hat{j}-5\hat{k}$ and $2\hat{i}+2\hat{j}+3\hat{k}$. Find the two unit vectors parallel to its diagonals. Using the diagonals vectors, find the area of the parallelogram.

Answer 1

Given:The two adjacent sides of a parallelogram are $2\hat{i}-4\hat{j}-5\hat{k}$ and $2\hat{i}+2\hat{j}+3\hat{k}$

Suppose $\overrightarrow{a}=2\hat{i}-4\hat{j}-5\hat{k}$ and $\overrightarrow{b}=2\hat{i}+2\hat{j}+3\hat{k}$

Then any one diagonal of a parallelogram is $\overrightarrow{P_1}=\overrightarrow{a}+\overrightarrow{b}$

$\overrightarrow{P_1}=\overrightarrow{a}+\overrightarrow{b}$

$=2\hat{i}-4\hat{j}-5\hat{k}+2\hat{i}+2\hat{j}+3\hat{k}=4\hat{i}-2\hat{j}-2\hat{k}$

$\therefore$unit vector along the diagonal is

$\frac{\overrightarrow{P_1}}{\left|\overrightarrow{P_1}\right|}=\frac{4\hat{i}-2\hat{j}-2\hat{k}}{\sqrt{16+4+4}}=\frac{4\hat{i}-2\hat{j}-2\hat{k}}{\sqrt{24}}=\frac{4\hat{i}-2\hat{j}-2\hat{k}}{2\sqrt{6}}=\frac{2\hat{i}-\hat{j}-\hat{k}}{\sqrt{6}}$

Another diagonal of a parallelogram is $\overrightarrow{P_2}=\overrightarrow{b}-\overrightarrow{a}$

$=2\hat{i}+2\hat{j}+3\hat{k}-2\hat{i}+4\hat{j}+\hat{k}=6\hat{j}+8\hat{k}$

$\therefore$ unit vector along the diagonal is $\frac{\overrightarrow{P_2}}{\left|\overrightarrow{P_2}\right|}=\frac{6\hat{j}+8\hat{k}}{\sqrt{36+64}}=\frac{6\hat{j}+8\hat{k}}{\sqrt{100}}=\frac{6\hat{j}+8\hat{k}}{10}=\frac{3\hat{i}+4\hat{j}}{5}$

Now,

$\overrightarrow{P_1}\times \overrightarrow{P_2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 4& -2& -2\\ 0& 6& 8\end{array}\right|$

$=\hat{i}(-16+12)-\hat{j}(32-0)+\hat{k}(24-0)$

$=-4\hat{i}-32\hat{j}+24\hat{k}$

Area of parallelogam$=\frac{1}{2}|\overrightarrow{P_1}\times \overrightarrow{P_2}|=\frac{1}{2}\times \sqrt{16+1024+576}=\frac{\sqrt{1616}}{2}=\frac{4\sqrt{101}}{2}=2\sqrt{101}$sq.units