# Finding two angles between vectors (analytic-geometry, vectors)

The vector $$\vec a$$ makes an angle of $$π/3$$ with the $$x$$−axis, an angle of $$π/4$$ with the $$y$$−axis, and some sharp angle with the $$z$$−axis. The vector $$\vec b$$ forms an angle of $$π/6$$ with the $$x$$−axis. What are the angles that vector $$\vec b$$ makes with $$y$$ and $$z$$−axis if it is known that vectors $$\vec a$$ and $$\vec b$$ are perpendicular?

Here is what I know:

$$\cos^2(\alpha) + \cos^2(\beta)+\cos^2(\gamma)=1 \\ \rightarrow\cos^2(π/3)+ \cos^2(π/4) +\cos^2(\gamma)=1 \\ \implies\cos^2(\gamma)=\pm1/2$$

Vectors $$\vec a$$ and $$\vec b$$ are perpendicular that means angle between $$\vec a$$ and $$\vec b$$ is $$π/2$$ and $$\vec a\cdot\vec b=0$$. I don't know what to do next and how to find angles between vector $$b$$ and $$y$$ and $$z$$ axis. I would be thankful for any help.

Wlog $$\|\vec a\|=\|\vec b\|=1.$$ If so then $$\vec a=\left(\frac12,\frac1{\sqrt2},\frac12\right),\quad \vec b=\left(\frac{\sqrt3}2,u,v\right)$$ with $$u^2+v^2=\frac14\quad\text{and}\quad\frac{\sqrt3}4+\frac u{\sqrt2}+\frac v2=0,$$ i.e. $$u=-\frac1{\sqrt6},\quad v=-\frac1{2\sqrt3}.$$ $$u,v$$ are the cosines of the angles that $$b$$ makes with the $$y$$- and $$z$$-axis.

• Thank you so much <3 Commented Feb 24, 2023 at 9:11

$$\def\uv#1{\hat{\mathbf#1}}$$ WLOG, consider the normalised vectors of $$\vec a$$ and $$\vec b$$, respectively $$\uv a$$ and $$\uv b$$.

Let $$\uv i$$, $$\uv j$$ and $$\uv k$$ be the standard unit vectors along the $$x$$-, $$y$$- and $$z$$-axes respectively. Then decompose $$\uv a$$ into its scalar components:

\begin{align*} \uv a &= a_x \uv i + a_y \uv j + a_z \uv z\\ &= \cos\frac\pi3\cdot \uv i + \cos \frac\pi4\cdot \uv j + a_z \uv k\\ &= \frac12 \uv i + \frac1{\sqrt 2} \uv j + a_z \uv k \end{align*}

Like what OP already did in the question, $$\uv a$$ has length $$1$$:

\begin{align*} \left(\frac12\right)^2 + \left(\frac1{\sqrt2}\right)^2 + a_z^2 &= 1\\ a_z^2 &= 1-\frac14 - \frac12\\ &= \frac14\\ a_z &= \frac12 \end{align*}

Assuming the "sharp" angle that $$\vec a$$ makes with the $$z$$-axis means an "acute" angle, $$a_z$$ is taken to be positive.

Similarly, decompose $$\uv b$$ into its scalar components:

\begin{align*} \uv b &= \cos\frac\pi6 \cdot \uv i + b_y \uv j + b_z \uv k\\ &= \frac{\sqrt3}2 \uv i + b_y \uv j + b_z \uv k \end{align*}

$$\vec a$$ and $$\vec b$$ are perpendicular, so $$\uv a \cdot \uv b = 0$$:

\begin{align*} \frac12 \cdot \frac{\sqrt3}2 + \frac1{\sqrt2} b_y + \frac12 b_z&= 0\\ b_z &= -\frac{\sqrt3}2 - \sqrt2 b_y \end{align*}

$$\uv b$$ has length $$1$$:

\begin{align*} \left(\frac{\sqrt3}2\right)^2 + b_y ^2 + b_z^2 &= 1\\ \left(\frac{\sqrt3}2\right)^2 + b_y ^2 + \left(-\frac{\sqrt3}2 - \sqrt2 b_y\right)^2 &= 1\\ \frac{3}4 + b_y ^2 + \frac{3}4 + \sqrt 6 b_y +2 b_y^2 &= 1\\ 3b_y^2 + \sqrt6 b_y +\frac12 &= 0\\ b_y &= \frac{-\sqrt6 \pm\sqrt{6-4\cdot 3/2}}{2\cdot 3}\\ &= -\frac{1}{\sqrt 6}\\ b_z &= -\frac{\sqrt3}{2} + \sqrt2 \cdot \frac{1}{\sqrt6}\\ &= -\frac{1}{2\sqrt3}\\ \uv b &= \frac{\sqrt3}2 \uv i - \frac1{\sqrt6} \uv j - \frac{1}{2\sqrt3}\uv k \end{align*}

So $$\vec b$$ makes the angle $$\arccos b_y = \arccos\left(- \frac1{\sqrt6}\right)$$ with the $$y$$-axis, and makes the angle $$\arccos b_z = \arccos\left(\frac{1}{2\sqrt3}\right)$$ with the $$z$$-axis.

• Thank you so much <3 Commented Feb 24, 2023 at 9:09