First slide

Introduction to 3-D Geometry

Question

The area of the triangle, whose vertices are at the points (2, 1, 1), (3, 1,2) and (- 4,0, 1) is

Moderate

A

$\sqrt{19}$
B

$\frac{1}{2} \sqrt{19}$
C

$\frac{1}{2} \sqrt{38}$
D

$\frac{1}{2} \sqrt{57}$

Solution

$\begin{array}{l} ∵ Δ_{yz} = \frac{1}{2} ||\begin{array}{lll} y_{1} & z_{1} & 1 \\ y_{2} & z_{2} & 1 \\ y_{3} & z_{3} & 1 \end{array}||, Δ_{zx} = \frac{1}{2} ||\begin{array}{lll} z_{1} & x_{1} & 1 \\ z_{2} & x_{2} & 1 \\ z_{3} & x_{3} & 1 \end{array}||, Δ_{xy} = \frac{1}{2} ||\begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array}|| \\ ∴ Δ_{yz} = \frac{1}{2} ||\begin{array}{l} 1 1 1 \\ 1 2 1 \\ 0 1 1 \end{array}|| \end{array}$
$\begin{array}{l} = \frac{1}{2} | [1 (2 - 1) - 1 (1 - 0) + 1 (1 - 0)] | \\ = \frac{1}{2} | [1 - 1 + 1] | = \frac{1}{2} \end{array}$
$Δ_{zx} = \frac{1}{2} ||\begin{array}{ccc} 1 & 2 & 1 \\ 2 & 3 & 1 \\ 1 & - 4 & 1 \end{array}||$
$\begin{array}{l} = \frac{1}{2} | [1 (3 + 4) - 2 (2 - 1) + 1 (- 8 - 3)] | \\ = \frac{1}{2} | [7 - 2 - 11] | = 3 \end{array}$
$Δ_{xy} = \frac{1}{2} ||\begin{array}{ccc} 2 & 1 & 1 \\ 3 & 1 & 1 \\ - 4 & 0 & 1 \end{array}||$
$= \frac{1}{2} | [2 (1 - 0) - 1 (3 + 4) + 1 (0 + 4)] | = \frac{1}{2}$
Area of triangle,
$\begin{matrix} Δ = \sqrt{Δ_{yz}^{2} + Δ_{zx}^{2} + Δ_{xy}^{2}} \\ = \sqrt{{(\frac{1}{2})}^{2} + (3)^{2} + {(\frac{1}{2})}^{2}} \\ = \sqrt{\frac{1}{4} + 9 + \frac{1}{4}} = \frac{\sqrt{38}}{2} sq unit \end{matrix}$

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