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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

Posted by

Ravindra Pindel

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