1
$\begingroup$

Given two position vectors $(x_1, y_1)$ and $(x_2, y_2)$, is there an intuitive way to see why the angle between them is < 90 when $x_1x_2 + y_1y_2$ is positive ?

I am trying to figure this without using trig ratios as this is linear algebra. Any suggestions on getting hold of the concept ?

$\endgroup$
2
  • $\begingroup$ It seems a little silly to insist on using no trigonometry. You're talking about angles, and trigonometry is the study of angles. Is there an intuitive way to see why the angle between them is $90^{\circ}$ when that quantity is zero? A basic understanding of simple trig is essential if you want to understand these things, I think. In the end, the answer is that this quantity is proportional to the cosine of the angle, and that cosine is positive precisely when the angle is less than $90^{\circ}$. $\endgroup$
    – MPW
    Oct 3 '14 at 15:55
  • $\begingroup$ Thank you :) so it seems there is no way to see $a\cdot b \gt 0$ means the angle is $\lt 90$ without appealing to cosine ? $\endgroup$
    – AgentS
    Oct 3 '14 at 16:00
2
$\begingroup$

with complex numbers $z_i = x_i+y_i$ you are asking that the real part of $\frac{z_1}{z_2}$ is positive. $$ \frac{z_1}{z_2} = (x_2^2+y_2^2)^{-\frac12}\left( (x_1x_2 + y_1y_2) + i(x_2y_1-x_1y_2)\right) $$

$\endgroup$
2
  • $\begingroup$ thank you :) I see the complex number z1/z2 has an argument equal to angle between z1 and z2. but I don't see how that forces the real part to be positive yet sorry im slow $\endgroup$
    – AgentS
    Oct 3 '14 at 16:42
  • $\begingroup$ think of points on the unit circle, since length is irrelevant here. then you can think of division by$z_2$ as aligning $z_2$ along the positive real axis, and rotating $z_1$ by the same amount. $\endgroup$ Oct 3 '14 at 17:14
2
$\begingroup$

You need to impose something. For example, you can say that if the angle is less than 90, then the resulting length of the remaining side of the triangle $ABO$ is less than those of a right triangle, intuitively. Call $A$ the point $(x_1,y_1)$ and $B$ the point $(x_2,y_2)$

Than you have:

$OA^2 =x_1^2 + y_1^2 $

$OB^2 =x_2^2 + y_2^2 $

$AB^2 =(x_2-x_1)^2 + (y_2-y_1)^2 $

If you say $AB^2 < OA^2 + OB^2$ then

$x_1^2 + y_1^2 + x_2^2 + y_2^2 - 2x_1x_2 - 2y_1y_2 < x_1^2 + y_1^2 + x_2^2 + y_2^2 $

$ - 2x_1x_2 - 2y_1y_2 < 0 $

which, if you reverse it and divide by 2, gives what you look for.

$\endgroup$
2
  • $\begingroup$ thanks! exactly what I have been looking for xD based on Pythagoras theorem $$a^T.a + b^T.a \lt (a+b)^T.(a+b) $$ $\endgroup$
    – AgentS
    Oct 3 '14 at 16:39
  • $\begingroup$ ** $$a^T.a + b^T.b \lt (a-b)^T.(a-b) $$ $\endgroup$
    – AgentS
    Oct 3 '14 at 16:53

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.