If A(1, –1, 2), B(5, 7, –6), C(3, 4, –10) and D(–1, –4, –2) are the vertices of a quadrilateral ABCD, then its area is : [2024]

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Solution

Q.
If A(1, –1, 2), B(5, 7, –6), C(3, 4, –10) and D(–1, –4, –2) are the vertices of a quadrilateral ABCD, then its area is : [2024]

1 $24 \sqrt{7}$

2 $12 \sqrt{29}$

3 $48 \sqrt{7}$

4 $24 \sqrt{29}$

Ans.
(2)

We have, A(1, –1, 2), B(5, 7, –6), C(3, 4, –10) and D(–1, –4, –2)

$\vec{A C}$ and $\vec{B D}$ are diagonals of ABCD.

Now, $\vec{A C} = 2 \hat{i} + 5 \hat{j} - 12 \hat{k}$

and $\vec{B D} = 6 \hat{i} + 11 \hat{j} - 4 \hat{k}$

Area, of ABCD = $\frac{1}{2}$ $| | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 5 & - 12 \\ 6 & 11 & - 4 \end{matrix} | |$

$= \frac{1}{2} | 112 \hat{i} - 64 \hat{j} - 8 \hat{k} | = \frac{1}{2} \times \sqrt{16704}$

$= \frac{1}{2} \times 24 \sqrt{29} = 12 \sqrt{29}$ .

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