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 :)

  • $\begingroup$ 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, 2020 at 3:13

2 Answers 2


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

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


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

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


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

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


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .