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Home  >>  CBSE XII  >>  Math  >>  Vector Algebra
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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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Toolbox:
  • The diagonal of a parallelogram is the vector addition of its adjacent sides.
  • Area of a parallelogram is the magnitude of the cross product of its adjacent sides.
Step 1:
Let $\overrightarrow a=2\hat i-4\hat j+5\hat k$ and $\overrightarrow b=\hat i-2\hat j-3\hat k$
According to law of triangles,its diagonal $\overrightarrow {AC}=\overrightarrow a+\overrightarrow b$
$\overrightarrow{AC}=(2\hat i-4\hat j+5\hat k)+(\hat i-2\hat j-3\hat k)$
$\quad\quad=3\hat i-6\hat j+2\hat k$
Step 2:
Next let us find its magnitude
$|\overrightarrow{AC}|=\sqrt{(3)^2+(-6)^2+(2)^2}$
$\quad\quad=\sqrt{9+36+4}$
$\quad\quad=\sqrt{49}$
$\quad\quad=7$
Unit vector parallel to $\overrightarrow {a}=\large\frac{\overrightarrow a}{|\overrightarrow a|}$
Step 3:
Unit vector parallel to diagonal of the parallelogram =$\large\frac{1}{7}$$(3\hat i-6\hat j+2\hat k)$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\;=\large\frac{3}{7}$$\hat i-\large\frac{6}{7}$$\hat j+\large\frac{2}{7}$$\hat k$
Step 4:
Now let us find its area
Area of the parallelogram is the cross product of its adjacent sides.
(i.e)Area of the parallelogram =$|\overrightarrow a\times\overrightarrow b|$
First we have to find $\overrightarrow a\times\overrightarrow b$
$\overrightarrow a\times\overrightarrow b=\begin{vmatrix}\hat i &\hat j&\hat k\\2 & -4 & 5\\1 & -2 & -3\end{vmatrix}$
$\qquad=\hat i(12+10)-\hat j(5+6)+\hat k(-4+4)$
$\qquad=22\hat i-11\hat j$
Step 5:
Now we have to find $|\overrightarrow a\times\overrightarrow b|$
$|22\hat i-11\hat j|=\sqrt{(22)^2+(-11)^2}$
$\qquad\qquad\;\;=11\sqrt{4+1}$
$\qquad\qquad\;\;=11\sqrt{5}$sq.units.
Hence the area of the parallelogram is $11\sqrt{5}$ sq. units.
answered May 24, 2013 by sreemathi.v
edited May 24, 2013 by sreemathi.v
 

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