Question #63619

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Question #63619

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Answer 1 · 2016-06-06T12:22:45

$θ_{z} = 90^{\circ}$ $theta_z = 90^@$

Explanation:

Let $\to v = {v_{x}, v_{y}, v_{z}}$ $vec v = {v_x,v_y,v_z}$ and ${ˆ e}_{x}, {ˆ e}_{y}, {ˆ e}_{z}$ $hat e_x,hat e_y,hat e_z$ the axis unity vectors.

We have

$⟨ \to v, {ˆ e}_{x} ⟩ = ∥ ∥ \to v ∥ ∥ cos (θ_{x}) = v_{x}$ $<< vec v, hat e_x>> = norm(vec v)cos(theta_x) = v_x$
$⟨ \to v, {ˆ e}_{y} ⟩ = ∥ ∥ \to v ∥ ∥ cos (θ_{y}) = v_{y}$ $<< vec v, hat e_y>> = norm(vec v)cos(theta_y) = v_y$
$⟨ \to v, {ˆ e}_{z} ⟩ = ∥ ∥ \to v ∥ ∥ cos (θ_{z}) = v_{z}$ $<< vec v, hat e_z>> = norm(vec v)cos(theta_z) = v_z$

also

$v_{x}^{2} + v_{y}^{2} + v_{z}^{2} = {∥ ∥ \to v ∥ ∥}^{2} = {∥ ∥ \to v ∥ ∥}^{2} {cos}^{2} (θ_{x}) + {∥ ∥ \to v ∥ ∥}^{2} {cos}^{2} (θ_{y}) + {∥ ∥ \to v ∥ ∥}^{2} {cos}^{2} (θ_{z})$ $v_x^2+v_y^2+v_z^2=norm(vec v)^2 = norm(vec v)^2cos^2(theta_x)+norm(vec v)^2cos^2(theta_y)+norm(vec v)^2cos^2(theta_z)$

simplifying we have

${cos}^{2} (θ_{x}) + {cos}^{2} (θ_{y}) + {cos}^{2} (θ_{z}) = 1$ $cos^2(theta_x)+cos^2(theta_y)+cos^2(theta_z) = 1$

so

$cos (θ_{z}) = \pm \sqrt{1 - {cos}^{2} (θ_{x}) - {cos}^{2} (θ_{y})}$ $cos(theta_z) = pm sqrt(1-cos^2(theta_x)-cos^2(theta_y))$

having now $θ_{x} = 150^{\circ}$ $theta_x = 150^@$ and $θ_{y} = 60^{\circ}$ $theta_y = 60^@$

we have $cos (θ_{z}) = 0$ $cos(theta_z)=0$ so $θ_{z} = \pm \frac{π}{2} = \pm 90^{\circ}$ $theta_z = pm pi/2 = pm 90^@$

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Question #63619

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