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  • The two adjacent sides of a parallelogram are 2i − 4 j + 5k and i − 2 j − 3k . Find the unit vector parallel to its diagonal. Also, find its area.

Q10   The two adjacent sides of a parallelogram are 2 \hat i - 4 \hat j + 5 \hat k \: \:and \: \: \hat i - 2 \hat j - 3 \hat k . Find the unit vector parallel to its diagonal. Also, find its area.

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Given, two adjacent sides of the parallelogram 

2 \hat i - 4 \hat j + 5 \hat k \: \:and \: \: \hat i - 2 \hat j - 3 \hat k

The diagonal will be the resultant of these two vectors. so

resultant R:

\vec R=2 \hat i - 4 \hat j + 5 \hat k \: +\: \hat i - 2 \hat j - 3 \hat k=3\hat i-6\hat j+2\hat k

Now unit vector in direction of R 

\vec u=\frac{3\hat i-6\hat j+2\hat k}{\sqrt{3^2+(-6)^2+2^2}}=\frac{3\hat i-6\hat j+2\hat k}{\sqrt{49}}=\frac{3\hat i-6\hat j+2\hat k}{7}

Hence unit vector along the diagonal of the parallelogram 

\vec u={\frac{3}{7}\hat i-\frac{6}{7}\hat j+\frac{2}{7}\hat k}

Now,

Area of parallelogram 

A=(2 \hat i - 4 \hat j + 5 \hat k )\: \times \: \: (\hat i - 2 \hat j - 3 \hat k)

A=\begin{vmatrix} \hat i &\hat j &\hat k \\ 2& -4 &5 \\ 1&-2 &-3 \end{vmatrix}=|\hat i(12+10)-\hat j(-6-5)+\hat k(-4+4)|=|22\hat i +11\hat j|

A=\sqrt{22^2+11^2}=11\sqrt{5}

Hence the area of the parallelogram is 11\sqrt{5}.

Posted by

Pankaj Sanodiya

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