First slide

Introduction to 3-D Geometry

Question

lf there are three points A(2,3, 4), B(- 1,2, - 3) and C(-4,1, - 10) in a space, then they are

Easy

A

collinear
B

non-collinear
C

Do not say anything
D

None of the above

Solution

Given, points are A(2,3,4), B(-1,2, -3) and C(-4,1, -10).
Now, determine area of triangle
Area of triangle $Δ = \sqrt{Δ_{xy}^{2} + Δ_{γz}^{2} + Δ_{zx}^{2}}$
Where, $Δ_{xy} = \frac{1}{2} ||\begin{array}{l} x_{1} y_{1} 1 \\ x_{2} y_{2} 1 \\ x_{3} y_{3} 1 \end{array}||$ , $Δ_{yz} = \frac{1}{2} ||\begin{array}{l} y_{1} z_{1} 1 \\ y_{2} z_{2} 1 \\ y_{3} z_{3} 1 \end{array}||$ and $Δ_{zx} = \frac{1}{2} ||\begin{array}{l} z_{1} x_{1} 1 \\ z_{2} x_{2} 1 \\ z_{3} x_{3} 1 \end{array}||$
now,
$\begin{matrix} Δ_{xy} = \frac{1}{2} ||\begin{array}{ccc} 2 & 3 & 1 \\ - 1 & 2 & 1 \\ - 4 & 1 & 1 \end{array}|| \\ = \frac{1}{2} |[2 (2 - 1) - 3 (- 1 + 4)] + 1 (- 1 + 8)]| \\ = \frac{1}{2} | [2 - 9 + 7] | = 0 \end{matrix}$
$\begin{matrix} Δ_{yz} = \frac{1}{2} ||\begin{array}{ccc} 3 & 4 & 1 \\ 2 & - 3 & 1 \\ 1 & - 10 & 1 \end{array}|| \\ = \frac{1}{2} |[3 (- 3 + 10) - 4 (2 - 1) + 1 (- 20 + 3)]| \\ = \frac{1}{2} | 21 - 4 - 17 | = 0 \end{matrix}$
$\begin{matrix} Δ_{zx} = \frac{1}{2} ||\begin{array}{ccc} 4 & 2 & 1 \\ - 3 & - 1 & 1 \\ - 10 & - 4 & 1 \end{array}|| \\ = \frac{1}{2} |[4 (- 1 + 4) - 2 (- 3 + 10) + 1 (12 - 10)]| \\ = \frac{1}{2} | [12 - 14 + 2] | = 0 \\ Δ = \sqrt{0^{2} + 0^{2} + 0^{2}} = 0 \end{matrix}$
Hence, given points are collinear

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