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Question
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
Let A(3, 5,−4), B(−1, 1, 2) and C(−5, −5, −2).
Direction ratio of AB = (−1 − 3), (1 − 5), (2 − (−4))
= (−4, −4, 6)
|AB| = `sqrt((-4)^2 + (-4)^2 + (6)^2)`
= `sqrt(16 + 16 + 36)`
= `sqrt68`
= `2sqrt17`
Direction ratio of BC = (−5 − (−1), −5 − 1, −2 − 2)
= (−4, −6, −4)
|BC| = `sqrt((-4)^2 + (-6)^2 + (-4)^2)`
= `sqrt(16 + 36 + 16)`
= `sqrt68`
= `2sqrt17`
Direction ratio of CA = (−5 − 3, −5 − 5, −2 − (−4))
= (−8, −10, 2)
|CA| = `sqrt((-8)^2 + (-10)^2 + (2)^2)`
= `sqrt(64 + 100 + 4)`
= `sqrt168`
= `2sqrt42`
∴ AB are `< (-1 - 3)/|AB|, (1 - 5)/|AB|, (2 + 4)/|AB| >`
i.e., `< (-2)/sqrt17, (-2)/sqrt17, 3/sqrt17 >`
∴ d.c. of BC are `< (-5 + 1)/|BC|, (- 5 - 1)/|BC|, (- 2 -2)/|BC|>`
i.e., `< (-2)/sqrt17, (-3)/sqrt17, (-2)/sqrt17 >`
∴ d.c of CA are `< (3 + 5)/|CA|, (5 + 5)/|CA|, (- 4 + 2)/|CA|`
i.e., `< 4/sqrt42, 5/sqrt42, (-1)/sqrt42 >`
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