Find the direction cosines of the sides of the triangle whose vertices are $(3, 5, - 4), (- 1, 1, 2)$ and $(- 5, - 5, - 2)$ .

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Solution

$A B$

$A (3, 5, - 4)$

$B (- 1, 1, 2)$

Direction ratios

$= - 1 - 3, 1 - 5, 2 - (- 4) = - 4, - 4, 6$

$A B = \sqrt{(- 1 - 3)^{2} + (1 - 5)^{2} + (2 + 4)^{2}} = \sqrt{68} = 2 \sqrt{17}$

Direction cosins

$\frac{- 4}{2 \sqrt{17}}, \frac{- 4}{2 \sqrt{17}}, \frac{6}{2 \sqrt{17}}$

$= \frac{- 2}{\sqrt{17}}, \frac{- 2}{\sqrt{17}}, \frac{3}{\sqrt{17}}$

$B C$

$B (- 1, 1, 2)$
$C (- 5, - 5, - 2)$

Direction ratios
$= - 5 - (- 1), - 5 - 1, - 2 - 2 = - 4, - 6, - 4$

$C A = \sqrt{(- 5 + 1)^{2} + (- 5 - 1)^{2} + (- 2 - 2)^{2}} = \sqrt{68} = 2 \sqrt{17}$

Direction cosins

$\frac{- 4}{2 \sqrt{17}}, \frac{- 6}{2 \sqrt{17}}, \frac{- 4}{2 \sqrt{17}}$

$= \frac{- 2}{\sqrt{17}}, \frac{- 3}{\sqrt{17}}, \frac{- 2}{\sqrt{17}}$

$C A$

$C (- 5, - 5, - 2)$
$A (3, 5, - 4)$

Direction ratios
$= 3 - (- 5), 5 - (- 5), - 4 - (- 2) = 8, 10, - 2$

$C A = \sqrt{(3 + 5)^{2} + (5 + 5)^{2} + (- 4 + 2)^{2}} = \sqrt{168} = 2 \sqrt{42}$

Direction cosins

$\frac{8}{2 \sqrt{42}}, \frac{10}{2 \sqrt{42}}, \frac{- 2}{2 \sqrt{42}}$

$= \frac{4}{\sqrt{42}}, \frac{5}{\sqrt{42}}, \frac{- 1}{\sqrt{42}}$

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Similar questions

(3, 5, - 4)

(- 1, 1, 2)

(- 5, - 5, - 2)

Which of the following are not the direction ratios of the sides of the triangle whose vertices are (3, 5, -4), (-1, 1, 2) and (-5, -5, -2)

Find the direction cosines of the side AB of the triangle whose vertices are A(3, 5, -4), B(-1, 1, 2) and C(-5, -5, -2)

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