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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.

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2 Answers 2

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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.

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  • $\begingroup$ Thank you so much <3 $\endgroup$ Commented Feb 24, 2023 at 9:11
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$\def\uv#1{\hat{\mathbf#1}}$ 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.

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

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