The point A(1, - 1, 3), B(2, - 4,8) and C(5,- 18, 11) are

A
collinear
B
non-collinear
C
Do not say anything
D
None of the above

Solution:

Now,

$\begin{matrix} Δ_{xy} = \frac{1}{2} ||\begin{array}{ccc} 1 & - 1 & 1 \\ 2 & - 4 & 1 \\ 5 & - 13 & 1 \end{array}|| \\ = \frac{1}{2} | [1 (- 4 + 13) + 1 (2 - 5) + 1 (- 26 + 20)] | \\ = \frac{1}{2} | [9 - 3 - 6] | = 0 \\ Δ_{yz} = \frac{1}{2} ||\begin{array}{ccc} - 1 & 3 & 1 \\ - 4 & 5 & 1 \\ - 13 & 11 & 1 \end{array}|| \\ = \frac{1}{2} | [- 1 (5 - 11) - 3 (- 4 + 13) + 1 (- 44 + 65)] | \\ = \frac{1}{2} | [6 - 27 + 21] | = 0 \\ Δ_{zx} = \frac{1}{2} ||\begin{array}{ccc} 3 & 1 & 1 \\ 5 & 2 & 1 \\ 11 & 5 & 1 \end{array}|| \\ = \frac{1}{2} | [3 (2 - 5) - 1 (5 - 11) + 1 (25 - 22)] | \\ = \frac{1}{2} | [- 9 + 6 + 3] | = 0 \end{matrix}$

$∴$ Area of triangle

$\begin{matrix} = \sqrt{Δ_{xy}^{2} + Δ_{yz}^{2} + Δ_{zx}^{2}} \\ = \sqrt{0 + 0 + 0} = 0 \end{matrix}$

Hence, given points are collinear.

Solution:

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