Find the direction cosines of the vector u given below. $$\mathbf { u } = ( - 2 , - 2 , - 5 )$$
A) $\cos ( \alpha ) = \frac { 2 } { \sqrt { 33 } } , \cos ( \beta ) = \frac { 2 } { \sqrt { 33 } } , \cos ( y ) = \frac { 5 } { \sqrt { 33 } }$
B) $\cos ( \alpha ) = \frac { 4 } { \sqrt { 33 } } , \cos ( \beta ) = \frac { 4 } { \sqrt { 33 } } , \cos ( \gamma ) = \frac { 10 } { \sqrt { 33 } }$
C) $\cos ( \alpha ) = \frac { 4 } { \sqrt { 33 } } , \cos ( \beta ) = \frac { 4 } { \sqrt { 33 } } , \cos ( \gamma ) = \frac { 25 } { \sqrt { 33 } }$
D) $\cos ( \alpha ) = \frac { - 2 } { \sqrt { 33 } } , \cos ( \beta ) = \frac { - 2 } { \sqrt { 33 } } , \cos ( \gamma ) = \frac { - 5 } { \sqrt { 33 } }$
E) $\cos ( \alpha ) = \frac { - 4 } { \sqrt { 33 } } , \cos ( \beta ) = \frac { - 4 } { \sqrt { 33 } } , \cos ( \gamma ) = \frac { - 10 } { \sqrt { 33 } }$

Question

Quizplus · Accepted Answer

The direction cosines of a vector $\mathbf{v} = (a,b,c)$ are given by $\frac{a}{\sqrt{a^2+b^2+c^2}}, \frac{b}{\sqrt{a^2+b^2+c^2}}, \frac{c}{\sqrt{a^2+b^2+c^2}}$. Using this formula, we get $\cos(\alpha) = \frac{-2}{\sqrt{33}}, \cos(\beta) = \frac{-2}{\sqrt{33}}, \cos(\gamma) = \frac{-5}{\sqrt{33}}$. Therefore, the answer is $\boxed{	ext{(D)}}$.

Find the Direction Cosines of the Vector U Given Below u=(−2,−2,−5)\mathbf { u } = ( - 2 , - 2 , - 5 )u=(−2,−2,−5)

Find the Direction Cosines of the Vector U Given Below $\mathbf { u } = ( - 2 , - 2 , - 5 )$