Find the direction angles of the vector. Round to the nearest degree, if necessary.
-$$\mathbf { v } = 3 \mathbf { i } + 2 \mathbf { j } - 6 \mathbf { k }$$
A) $\alpha = 86 ^ { \circ } , \beta = 88 ^ { \circ } , \gamma = 97 ^ { \circ }$
B) $\alpha = 62 ^ { \circ } , \beta = 72 ^ { \circ } , \gamma = 203 ^ { \circ }$
C) $\alpha = 68 ^ { \circ } , \beta = 76 ^ { \circ } , \gamma = 139 ^ { \circ }$
D) $\alpha = 65 ^ { \circ } , \beta = 73 ^ { \circ } , \gamma = 149 ^ { \circ }$

Question

Quizplus · Accepted Answer

The direction angles $\alpha$, $\beta$, and $\gamma$ of a vector $\mathbf{v} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k}$ can be found using the formulas $\cos(\alpha) = \frac{a}{\|\mathbf{v}\|}$, $\cos(\beta) = \frac{b}{\|\mathbf{v}\|}$, and $\cos(\gamma) = \frac{c}{\|\mathbf{v}\|}$, where $\|\mathbf{v}\|$ is the magnitude of the vector. For $\mathbf{v} = 3\mathbf{i} + 2\mathbf{j} - 6\mathbf{k}$, the magnitude is $\sqrt{3^2 + 2^2 + (-6)^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$. Thus, $\cos(\alpha) = \frac{3}{7}$, $\cos(\beta) = \frac{2}{7}$, and $\cos(\gamma) = \frac{-6}{7}$. Calculating the inverse cosines gives $\alpha \approx 65^\circ$, $\beta \approx 73^\circ$, and $\gamma \approx 149^\circ$.

Find the Direction Angles of the Vector v=3i+2j−6k\mathbf { v } = 3 \mathbf { i } + 2 \mathbf { j } - 6 \mathbf { k }v=3i+2j−6k

Find the Direction Angles of the Vector $\mathbf { v } = 3 \mathbf { i } + 2 \mathbf { j } - 6 \mathbf { k }$