A vector overline{OP} makes angles of 45^{cir | vedclass

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(A) Let the direction cosines of the vector $\overline{OP}$ be $l, m, n$.Given that the angles with the axes are $\alpha = 45^{\circ}$,$\beta = 60^{\c

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(A) Let the direction cosines of the vector $\overline{OP}$ be $l, m, n$.Given that the angles with the axes are $\alpha = 45^{\circ}$,$\beta = 60^{\circ}$,and let the angle with $OZ$ be $\gamma$.Then,$l = \cos(45^{\circ}) = \frac{1}{\sqrt{2}}$,$m = \cos(60^{\circ}) = \frac{1}{2}$,and $n = \cos(\gamma)$.We know that $l^2 + m^2 + n^2 = 1$.Substituting the values: $(\frac{1}{\sqrt{2}})^2 + (\frac{1}{2})^2 + n^2 = 1$.$\frac{1}{2} + \frac{1}{4} + n^2 = 1$.$\frac{3}{4} + n^2 = 1 \Rightarrow n^2 = 1 - \frac{3}{4} = \frac{1}{4}$.Thus,$n = \pm \frac{1}{2}$.Since $n = \cos(\gamma)$,we have $\cos(\gamma) = \frac{1}{2}$ or $\cos(\gamma) = -\frac{1}{2}$.Therefore,$\gamma = 60^{\circ}$ or $\gamma = 120^{\circ}$.

$A$ vector $\overline{OP}$ makes angles of $45^{\circ}$ with $OX$ and $60^{\circ}$ with $OY$. Find the angle that $\overline{OP}$ makes with $OZ$ (in degrees).

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