Single Choice

If the two directional cosines of a vectors are $$\dfrac { 1 }{ \sqrt { 2 }  } $$ and $$\dfrac { 1 }{ \sqrt { 3}  } $$ then the value of third directional  cosines is

A$$\dfrac { 1 }{ \sqrt { 6 } } $$
Correct Answer
B$$\dfrac { 1 }{ \sqrt { 5 } } $$
C$$\dfrac { 1 }{ \sqrt { 7 } } $$
D$$\dfrac { 1 }{ \sqrt { 10 } } $$

Solution

we have, $$\cos ^2\alpha +\cos ^2\beta +\cos ^2\gamma =1$$ $$\left ( \dfrac{1}{\sqrt{2}} \right )^2+\left ( \dfrac{1}{\sqrt{3}} \right )^2+\cos ^2\gamma =1$$ $$\cos ^2\gamma =1-\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{1}{6}$$ $$\therefore \cos \gamma =\dfrac{1}{\sqrt{6}}$$


SIMILAR QUESTIONS

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