Question

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 :

• A. \(12 \sqrt{29}\)
• B. \(24 \sqrt{29}\)
• C. \(24 \sqrt7\)
• D. \(48 \sqrt7\)

Answer 1

The Correct Option is A

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:

\[ \text{Area} = 12 \sqrt{29} \]