Question

If the two directional cosines of a vectors are 1/√2 and 1/√3 then the value of third
directional cosine is
Answer - 1/√6

Answer 1

Answer:

1/✓6

Explanation:

i am guessing that the vector is unit vector. assuming this the sum of components of vector should be equal to squre of mod value.

1*1= 1/2 + 1/3 + c^2

hence the square value is c^2= 1-(1/2+1/3) = 1/6

hence value of the third component is c=1/✓6

Answer 2

$z^{2}$Answer:

cosγ=1/$\sqrt{6$

Explanation:

cos α=x/a = 1/$\sqrt{2$ = 1x$\sqrt{3$/$\sqrt{2$x$\sqrt{3$ =$\sqrt{3$/$\sqrt{6$ [ therefore a will be same]

cosβ=y/a = 1/$\sqrt{3$ = $\sqrt{2$/$\sqrt{6$

cosγ=z/a

a vector = $\sqrt{x^2+y^2+z^2$

$\sqrt{6$ = $\sqrt{x^2+y^2+z^2$ [ both roots will be canceled]

6 = $\sqrt{3}^2$ + $\sqrt{2}^2$ + $z^{2}$

6 = 3+2+$z^{2}$

6 = 5+$z^{2}$

$z^{2}$ = 1

z = 1

Therefore, cosγ=z/a=1/$\sqrt{6$

orelse, u can use cos^2α+cos^2β+cos^2γ=1 formua to solve this problem,...