Is a vector making equal angles with three non-zero coplanar vectors perpendicular to that plane of three vectors?

Question

Question:

Is a vector making equal angles with three non-zero coplanar vectors perpendicular to that plane of three vectors?

It is easy to see the converse, but try as I may, I am not able to prove it in the general case.

I have tried to equate dot products , but all I got is two equations of plane intersecting at origin. I know there is some path here, but can we "see" this more elegantly??

Answer 1

If $a,b,c$ are coplanar and $a$ and $b$ are not parallel then $c=\lambda_1a+\lambda_2b$. Suppose $d$ makes the same angle $\theta$ with $a,b$ and $c$. Then

$|d||c|\cos\theta = d \cdot c = \lambda_1 d\cdot a + \lambda_2 d\cdot b = \lambda_1 |a||d|\cos \theta + \lambda_2 |b||d|\cos \theta$

We know $|d|$ is non-zero. If $\cos\theta$ is also not $0$ then we can cancel $|d|\cos\theta$ from both side to get

$|c| = \lambda_1|a| + \lambda_2|b| \\ \Rightarrow |c|^2 = \lambda_1^2|a|^2 + 2\lambda_1\lambda_2|a||b| + \lambda_2^2|b|^2$

But

$|c|^2 = c \cdot c = \lambda_1^2a \cdot a + 2\lambda_1\lambda_2a \cdot b + \lambda_2^2 b \cdot b = \lambda_1^2|a|^2 + 2\lambda_1\lambda_2|a||b|\cos \phi + \lambda_2^2|b|^2$

where $\phi$ is the angle between $a$ and $b$. These two expressions for $|c|^2$ can only be equal if:

(1) $\lambda_1=0$ in which case $c$ is parallel to $b$

(2) $\lambda_2=0$ in which case $c$ is parallel to $a$

(3) $\cos \phi=1$ - but this is not the case since $a$ and $b$ are assumed not parallel.

Therefore either some pair out of $a, b$ and $c$ are parallel or $\cos \theta =0$, in which case $d$ is perpendicular to $a,b$ and $c$.

Answer 2

Without loss of generality, normalise all vectors and set the three coplanar vectors $a,b,c$ to lie in the $xy$-plane. Consider the fourth vector $d$, which is such that $$d\cdot a=d\cdot b=d\cdot c$$ Subtracting each of the three equations from the next cyclically gives $$d\cdot(b-a)=d\cdot(c-b)=d\cdot(a-c)=0$$ and since $a,b,c$ lie in the plane, so do $b-a,c-b,a-c$, so each relation to 0 constrains $d$ to a plane containing the $z$-axis. Since $a,b,c$ are distinct (an implicit assumption of the question), there will be two non-parallel constraining planes, which means $d$ must lie on the $z$-axis and so is perpendicular to $a,b,c$.