If vertices of a triangle are A(1,- 1, Z), B(2,0,- 1) and C(0,2,1), then the area of a triangle is

A
$\sqrt{6}$
B
$2 \sqrt{6}$
C
$3 \sqrt{6}$
D
$4 \sqrt{6}$

Solution:

Now,

$\begin{array}{l} Δ_{xy} = \frac{1}{2} ||\begin{array}{ccc} 1 & - 1 & 1 \\ 2 & 0 & 1 \\ 0 & 2 & 1 \end{array}|| \\ = \frac{1}{2} | 1 (0 - 2) + 1 (2 - 0) + 1 (4 - 0) | \\ = \frac{1}{2} | - 2 + 2 + 4 | = 2 \end{array}$

$\begin{array}{l} Δ_{yz} = \frac{1}{2} ||\begin{array}{ccc} - 1 & 2 & 1 \\ 0 & - 1 & 1 \\ 2 & 1 & 1 \end{array}|| \\ = \frac{1}{2} | [- 1 (- 1 - 1) - 2 (0 - 2) + 1 (0 + 2)] | \\ = \frac{1}{2} | [2 + 4 + 2] | = 4 \end{array}$

$\begin{array}{l} Δ_{zx} = \frac{1}{2} ||\begin{array}{ccc} 2 & 1 & 1 \\ - 1 & 2 & 1 \\ 1 & 0 & 1 \end{array}|| \\ = \frac{1}{2} | [2 (2 - 0) - 1 (- 1 - 1) + 1 (0 - 2)] | \\ = \frac{1}{2} | [4 + 2 - 2] | = 2 \end{array}$

$∴$ Area of triangle,

$\begin{array}{l} Δ = \sqrt{Δ_{xy}^{2} + Δ_{yz}^{2} + Δ_{zx}^{2}} \\ = \sqrt{2^{2} + 4^{2} + 2^{2}} = \sqrt{24} = 2 \sqrt{6} \end{array}$

Solution:

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