First slide

Cross product of vectors or vector product of vectors

Question

The area of the triangle formed by the points $A (1, 2, 3) B (- 2, 1, 4), C (3, 4, - 2)$

Moderate

A

$\frac{1}{4} \sqrt{194}$
B

$\frac{1}{2} \sqrt{194}$
C

$\sqrt{194}$
D

$2 \sqrt{194}$

Solution

The given vertices are $A (1, 2, 3) B (- 2, 1, 4), C (3, 4, - 2)$
The area of the triangle formed by the points $A, B, C$ is $\frac{1}{2} | \bar{A B} \times \bar{A C} |$
$\bar{A B} = - 3 i - j + k, \bar{A C} = 2 i + 2 j - 5 k$
the area of the triangle is
$\begin{matrix} \frac{1}{2} | A B \times A C | = \frac{1}{2} |\begin{matrix} i & j & k \\ - 3 & - 1 & 1 \\ 2 & 2 & - 5 \end{matrix}| \\ = \frac{1}{2} | (3 i - 13 j - 4 k) | \\ = \frac{1}{2} \sqrt{9 + 169 + 16} \\ = \frac{\sqrt{194}}{2} \end{matrix}$
Therefore, the area of the triangle is $\frac{\sqrt{194}}{2} square units$

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