If A(l, –1, 2), B(5, 7, –6), C(3, 4, –10) and D(–l, –4, –2)... | Mathematics | SolveeIt

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

$12 \sqrt{29}$

$24 \sqrt{29}$

$24 \sqrt7$

$48 \sqrt7$

Answer

$12 \sqrt{29}$

Explanation

Given points:

$A(1, -1, 2), \, B(5, 7, -6), \, C(3, 4, -10), \, D(-1, -4, -2)$

The area is given by:

$\text{Area} = \frac{1}{2} | \vec{AC} \times \vec{BD} | = \frac{1}{2} | (2i + 5j - 12k) \times (6i + 11j - 4k) |$

Calculating the cross product:

$= \frac{1}{2} | 12i - 64j - 8k |$

Taking the magnitude:

$= \frac{1}{2} \sqrt{(12)^2 + (-64)^2 + (-8)^2}$

$= \frac{1}{2} \sqrt{144 + 4096 + 64}$

$= \frac{1}{2} \sqrt{4304}$

$= \frac{1}{2} \times 2 \sqrt{1076}$

$= \sqrt{1076}$

Therefore:

$\text{Area} = 12 \sqrt{29}$

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