# Show that the area of the parallelogram whose diagonals are given by $\overrightarrow{a}$ and $\overrightarrow{b}$ is $\frac{\mid \overrightarrow{a} \times \overrightarrow{b}\mid}{2}$.Also find the area of the parallelogram whose diagonals are $2\hat i+\hat j+\hat k$ and $\hat i+3\hat j+\hat k$.

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• The area of a parallelogram whose diagonals are $\overrightarrow a$ and $\overrightarrow b$ is $\large\frac{1}{2}$$|\overrightarrow a \times \overrightarrow b| Step 1: Let ABCD be a parallelogram with A as the origin, let the position vectors of B and D be \overrightarrow p and \overrightarrow q respectively.Then, \overrightarrow {AB}= \overrightarrow p and \overrightarrow {AD}= \overrightarrow q But we know \overrightarrow {BC}= \overrightarrow {AD} Therefore \overrightarrow {BC}= \overrightarrow q By triangle law of vectors, we have \overrightarrow {AB}+ \overrightarrow {BC}=\overrightarrow {AC} \overrightarrow {AB}+ \overrightarrow {AD}=\overrightarrow {AC} Therefore \overrightarrow p+ \overrightarrow q=\overrightarrow {AC}-----(1) Now let \overrightarrow {AC}= \overrightarrow a and \overrightarrow {BD}=\overrightarrow b, be the diagonals of the parallelogram ABCD Then from equ(1) we have \overrightarrow p+ \overrightarrow q=\overrightarrow a------(2) and \overrightarrow {BD}= \overrightarrow {OD}-\overrightarrow {OB} =>\overrightarrow b= \overrightarrow q-\overrightarrow p-----(3) Add equ (1) and equ (2) \overrightarrow {2q}= \overrightarrow {a}+\overrightarrow {b} =>\overrightarrow {q}=\large\frac{1}{2}$$(\overrightarrow {a}+\overrightarrow {b})$
Step 2:
Subtract equ(2) and equ(3) we get,
$2\overrightarrow p= \overrightarrow a-\overrightarrow b$
$=>\overrightarrow p= \large\frac{1}{2}$$(\overrightarrow {a}-\overrightarrow {b}) Therefore \overrightarrow p \times \overrightarrow q=\large\frac{1}{2}$$(\overrightarrow a-\overrightarrow b) \times \large\frac{1}{2}(\overrightarrow a+\overrightarrow b)$
$\qquad\qquad\qquad=\large\frac{1}{4} $$[(\overrightarrow a-\overrightarrow b) \times (\overrightarrow a+\overrightarrow b)] On expanding we get, \overrightarrow p \times \overrightarrow q=\large\frac{1}{4}$$\bigg[\overrightarrow a \times \overrightarrow a+ \overrightarrow a \times \overrightarrow b-\overrightarrow b \times \overrightarrow a-\overrightarrow b \times \overrightarrow b\bigg]$
But $\overrightarrow a \times \overrightarrow a=\overrightarrow b \times \overrightarrow b=0$ and $-\overrightarrow b \times \overrightarrow a=\overrightarrow a \times \overrightarrow b$
Therefore $\overrightarrow p \times \overrightarrow q=\large\frac{1}{4} $$\bigg[\overrightarrow a \times \overrightarrow b+ \overrightarrow a \times \overrightarrow b\bigg] \quad\qquad\qquad\qquad=\large\frac{1}{2}$$\bigg[\overrightarrow a \times \overrightarrow b\bigg]$
So area of the paralleogram $ABCD$
$=|\overrightarrow p \times \overrightarrow q|=\large\frac{1}{2} $$|\overrightarrow a \times \overrightarrow b| Step 3: Let \overrightarrow a=2 \hat i+\hat j+\hat k and \overrightarrow b=\hat i+3 \hat j+\hat k Hence the area of the parallelogram is \large\frac{1}{2}$$|\overrightarrow a \times \overrightarrow b|$
$=>\large\frac{1}{2} $$|\overrightarrow a \times \overrightarrow b|=\large\frac{1}{2}$$|(2 \hat i+\hat j+\hat k) \times (\hat i+3 \hat j+\hat k)|$
Let us determine $\overrightarrow a \times \overrightarrow b$
$\overrightarrow a \times \overrightarrow b=\begin {vmatrix} \hat i & \hat j & \hat k \\ 2 & 1 & 1 \\ 1 & 3 & 1 \end {vmatrix}$
$\quad\qquad=\hat i(1-3)-\hat j(2-1)+\hat k (6-1)$
$\quad\qquad=-2 \hat i-\hat j+5 \hat k$
$|\overrightarrow a \times \overrightarrow b|=\sqrt {(-2)^2+(-1)^2+(5)^2}$
$\qquad\qquad=\sqrt {4+1+25}$
$\qquad\qquad=\sqrt {30}$
Hence the area of the parallogram is $\large\frac{1}{2}$$\sqrt {30} sq.units$
edited May 31, 2013 by meena.p