# Is there a vector that forms angles of 30 and 45 degrees with the x and y axes respectively?

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 :)

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

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.

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

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

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