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Question

Show that area of the prallelogram whose diagonals are given by $\to a a n d \to b i s \frac{\to a \times \to b}{2}$ . Also, find the area of the parallelogram, whose diagonals are $2^i -^j + k a n d^i + 3^j -^k .$

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Solution

Let ABCD be a parallelogram such that

$- - \to A B = \to p . - - \to A D = \to q \Rightarrow - - \to B C = \to q$

By triangle law of addition, we get

$- - \to A C = \to p + \to q = \to a [s a y] . . . . . (i)$

Similarly, $- - \to B D = - \to p + \to q = \to b [s a y] . . . . (i i)$

On adding Eqs. (i) and (ii), we get

$\to a + \to b = 2 \to q \Rightarrow \to q = \frac{1}{2} (\to a + \to b)$

On subtracting Eq. (ii) from Eq. (i), we get

$\to a - \to b = 2 \to p \Rightarrow \to p = \frac{1}{2} (\to a - \to b) N o w, \to p \times \to q = \frac{1}{4} (\to a - \to b) \times (\to a + \to b) = \frac{1}{4} (\to a \times \to a + \to a \times \to b - \to b \times \to a - \to b \times \to b) = \frac{1}{4} [\to a \times \to b + \to a \times \to b] = \frac{1}{2} (\to a \times \to b) So, area of a parallelogram ABCD = | \to p \times \to q | = \frac{1}{2} | \to a \times \to b |$

Now, area of a parallelogram, whose diagonals are $2^i -^j +^k a n d^i + 3^j -^k .$

$= \frac{1}{2} | (2^i -^j +^k) \times (^i + 3^j -^k) | = \frac{1}{2} ∥ ∥ ∥ ∥ ∥ \begin{matrix} ^i & ^j & ^k 2 & - 1 & 1 1 & 3 & - 1 \end{matrix} ∥ ∥ ∥ ∥ ∥ = \frac{1}{2} | [^i (1 - 3) -^j (- 2 - 1) +^k (6 + 1)] | = \frac{1}{2} | - 2^i + 3^j + 7^k | = \frac{1}{2} \sqrt{4 + 9 + 49} = \frac{1}{2} \sqrt{62} s q u n i t s$

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