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I need your expertise in understanding the following problem:

Given three vectors $v1,v2,v3 \in \mathbb{R}^n$ such that $v1$ is orthogonal to $v2$. Let $\theta \in [-180, 180]$ be the angle between $v1$ and $v3$.

Can we know what will be the angle between $v2$ and $v3$? (maybe as a function of $\theta$?)

Please advise and thanks in advance.

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  • $\begingroup$ So there is no way to bound the angle? $\endgroup$ – user3492773 Jul 25 '17 at 17:29
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    $\begingroup$ I would recommend getting out of the habit of using degrees for angles and using radians instead, or at the very least label them as degrees as in 180^\circ $180^\circ$. In any case, we wouldn't be able to pinpoint exactly the value of the angle between $v_2$ and $v_3$, but we could at least find a range of values for it. Try it yourself, take your hand and form an $L$ shape with your thumb and forefinger at a right angle. Call your thumb $v_1$. With your other hand, take a pencil and form an angle with your thumb, spinning it around the thumb maintains the angle with $v_1$ but not $v_2$ $\endgroup$ – JMoravitz Jul 25 '17 at 17:30
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First think in the plane: We can pick $v_1$ and $v_2$ to be the $x$ and $y$ axis directions. Then $$ v_3 = (\cos \theta, \sin \theta) $$ or $$ v_3 = (\cos \theta, -\sin \theta) $$ (assuming we're working with unit vectors). So the answer is that the angle between $v_1$ and $v_3$ is either $\frac{\pi}{2} - \theta$ or $\frac{\pi}{2} + \theta$. (Draw a picture to confirm).

In 3-space, a similar analysis turns out to show that the angle can be anything between these two values. Why? Because with $$ v_1 = (1, 0, 0)\\ v_2 = (0, 1, 0) $$ the vector $v$ (I'm using $v$ instead of $v_3$) must be $$ v = (\cos \theta, \sin u \sin \theta, \cos u \sin \theta) $$ (Why?? See below) and the angle between this and $v_2$ must have cosine equal to $$ v_2 \cdot v = \sin u \sin \theta $$ which ranges from $-\sin \theta$ to $\sin \theta$.

Why must $v$ be that vector? Because it has to lie on a cone around $v_1$ with center angle $\theta$, and be a unit vector, and therefore must lie on a circle on that cone -- a circle of constant $x$-value (namely $\cos \theta$), so it's just $P = (\cos \theta, 0, 0)$ plus a circle in the $yz$ plane. Since the point $(\cos \theta, \sin \theta, 0) = P + (0, \sin \theta, 0)$ lies on the circle, the radius of the circle must be $\sin \theta$. So each point of the circle looks like $P + Q$, where $$ Q = \sin \theta (0, \cos u, \sin u) $$ i.e., where $Q$ is a point of a unit-circle, scaled by $\sin \theta$. When you write that out, you get the formula that I gave.

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Here's an image that might help you visualize the problem.

The image shows two (fixed) vectors $v_1$ and $v_2$ that are perpendicular. The third vector is $v_3$. Notice that specifying the angle between $v_1$ and $v_3$ constrains $v_3$ to a cone, but not to a specific position in space. As $v_3$ sweeps around the cone, the angle between $v_3$ and $v_1$ remains constant, but the angle between $v_3$ and $v_1$ does not.

enter image description here

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