The two adjacent sides of a parallelogram are and <img alt="2\hat{i}+2\hat{j}+3\hat{k}\cdot" src="https://entrance

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

Answers (1)

Let ABCD be a parallelogram such that
       $\overrightarrow{AB}= \vec{a}= 2\hat{i}-4\hat{j}-5\hat{k}$
and $\overrightarrow{BC}= \vec{b}= 2\hat{i}+2\hat{j}+3\hat{k}$
Then $\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$
$\overrightarrow{AC}= \vec{a}+\vec{b}= 4\hat{i}-2\hat{j}-2\hat{k}$

and $\overrightarrow{AB}+\overrightarrow{BD}= \overrightarrow{AD}$

$\Rightarrow \overrightarrow{BD}= \overrightarrow{AD}- \overrightarrow{AB}$
$= \overrightarrow{BD}= \vec{b}-\vec{a}= 0\hat{i}+6\hat{j}+8\hat{k}$
Now, $= \overrightarrow{AC}= 4\hat{i}-2\hat{j}-2\hat{k}$
$\left | AC \right |= \sqrt{\left ( 4 \right )^{2}+\left ( -2 \right )^{2}+\left ( -2 \right )^{2}}= \sqrt{24}= 2\sqrt{6}$
and
     $\left | \overrightarrow{BD} \right |=\sqrt{\left ( 0 \right )^{2}+\left ( 6 \right )^{2}+\left ( 8 \right )^{2}}= \sqrt{100}= 10$
unit vector along AC $= \frac{\overrightarrow{AC}}{\left | \overrightarrow{AC} \right |}= \frac{1}{2\sqrt{6}}\left ( 4\hat{i} -2\hat{j}-2\hat{k}\right )$
                              $= \frac{1}{\sqrt{6}}\left ( 2\hat{i}-\hat{j}-\hat{k} \right )$
unit vector along $\overrightarrow{BD}=\frac{\overrightarrow{BD}}{\left | \overrightarrow{BD} \right |}= \frac{1}{10}\left ( 6\hat{j}+8\hat{k} \right )$
                                                        $= \frac{1}{5}\left ( 3\hat{j}+4\hat{k} \right )$
Now, area of parallelogram $= \frac{1}{2}\left | \overrightarrow{AC}\times\overrightarrow{BD} \right |$
$\Rightarrow \overrightarrow{AC}\times\overrightarrow{BD}= \begin{vmatrix} \hat{i} & \hat{j} &\hat{k} \\ 4& -2 &-2 \\ 0 &6 & 8 \end{vmatrix}$
$=-4 \hat{i} -32 \hat{j} +24\hat{k}$
Area of parallelogram $=\frac{1}{2}\left | \overrightarrow{AC}\times \overrightarrow{BD} \right |$
                                  $=\sqrt{404}= 2\sqrt{101}\, sq.units$