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Find the direction cosines of the sides ...

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

A

`(-4)/sqrt(17),(-4)/sqrt(17),(6)/sqrt(17)`

B

`(-2)/sqrt(17),(-2)/sqrt(17),(3)/sqrt(17)`

C

`(-2)/sqrt(17),(2)/sqrt(17),(-3)/sqrt(17)`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
A, B
Let the vertices of the triangle be A(3,5,-4),B(-1,1,2). The direction ratios of side AB are [-1,-3,1-5,2-(-4)] i.e., (-4,-4,6)
Now `|AB|=sqrt((-4)^2+(-4)^2+(6)^2)`
`=sqrt(16+16+36)=sqrt(68)=2sqrt(17)`
Therefore, the direction cosines of AB are `((-4)/(2sqrt(17)),(-4)/(2sqrt(17)),6/(2sqrt(17))) or (-2)/sqrt(17),(-2)/sqrt(17),3/sqrt(17)`
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