0
$\begingroup$

Everyone!

I've a problem that asks

Is there any 3D-vector that forms angles of 30 and 45 degrees with x-axis and y-axis, respectively?

My attempt was using the directional cosines, and I got that the square of the angle between the vector and the z-axis is negative, so it doesn't exist! However, is this a valid way to prove that this vector does not exist?

Thanks in advance :)

$\endgroup$
1
  • $\begingroup$ Given two vectors, you can find the angle between them using dot product: $\vec{u}\cdot\vec{v}=|\vec{u}||\vec{v}|\cos\theta$ $\endgroup$ – Ray Oct 9 '20 at 3:13
1
$\begingroup$

Assume such a vector exists, call it $\vec{v} = (a,b,c)$ and WLOG let it be a unit vector. Then:

$$ \begin{align} \vec{v} \cdot \vec{i} &= \cos 30^{\circ} \\ \vphantom{} \\ (a,b,c) \cdot (1,0,0) &= \frac{\sqrt{3}}{2} \\ \vphantom{} \\ a &= \frac{\sqrt{3}}{2} \end{align} $$

Similarly,

$$ \begin{align} \vec{v} \cdot \vec{j} &= \cos 45^{\circ} \\ \vphantom{} \\ (a,b,c) \cdot (0,1,0) &= \frac{\sqrt{2}}{2} \\ \vphantom{} \\ b &= \frac{\sqrt{2}}{2} \end{align} $$

But now since $a^2 + b^2 > 1$, $\vec{v}$ cannot be a unit vector so we have a contradiction.

$\endgroup$
1
$\begingroup$

Yes, this is a valid way of proving no such vector exists. You proved by contradiction. You found that $n^2 \le 0$ ($n$ is direction cosine wrt $z$-axis) which is not possible.

To visualize, $\vec v$ should lie on the surface of a cone whose axis is positive x-axis and apex angle is $60^{\circ}$. And simultaneously on the surface of cone whose axis is positive y-axis and apex angle is $90^{\circ}$.

To find the constraint, use the direction cosines,

$$ l^2+m^2+n^2=1$$

Since $0 \le n^2 \le 1$,

$$ 0 \le l^2+m^2 \le 1$$

which reads

$$ \{ \text{cosine (angle with + x-axis)} \}^2 + \{ \text{cosine (angle with + y-axis)} \}^2 \in [0,1]$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.