Question

The directional cosines of a vector i + j + √2k.

Answer 1

Hey There,

The Direction Cosines of a vector represent the cosines of the angles that the vector makes with the Co-ordinate Axes. 
They are represented as $\cos\alpha \, , \,\cos \beta \, \, and\, \, \cos \gamma$ , or they can be simply called l, m and n respectively.


To find the Direction Cosines, we first find the Unit-Vector in the Direction of the given vector. Then, each component of that unit-vector simply represents the Direction Cosines.


[To find unit-vector of any given vector, we simply divide the vector by its magnitude]



Here, Let

$\vec{a} = \hat{\imath}+\hat{\jmath}+\sqrt{2}\hat{k} \\ \\ \\ \implies \hat{a} = \frac{\vec{a}}{\mid \vec{a} \mid} \\ \\ \\ \implies \hat{a} = \frac{\hat{\imath}+\hat{\jmath}+\sqrt{2}\hat{k}}{\sqrt{1^2+1^2+(\sqrt{2})^2}} \\ \\ \\ \implies \hat{a} = \frac{1}{2} (\hat{\imath}+\hat{\jmath}+\sqrt{2}\hat{k}) \\ \\ \\ \implies \hat{a} = \frac{1}{2}\hat{\imath}+\frac{1}{2}\hat{\jmath}+\frac{1}{\sqrt{2}}\hat{k}$


Thus, we have found the unit vector. Now, each component of this unit-vector represents the direction cosines. 
This means that the direction cosines are 

$\frac{1}{2} \, , \, \, \frac{1}{2} \, \, and \, \, \frac{1}{\sqrt{2}}$


Hope it helps
Purva
Brainly Community

Answer 2

Explanation:

answer will be 1 by 2, 1 by 2 1 by root 2 pocket helps you mark it s brainliest answer