Angle between vectors by using cross product

cross productvectors

Given $A=(0,10,2)$ and $B=(0,-4,0.5)$, find the angle between $A$ and $B$ using the cross product.

My attempt by using dot product:

$$\cos (\theta)=\frac{A\cdot B}{|A||B|}=\frac{(0)(0)+10(-4)+2(0.5)}{\sqrt{0^2+10^2+2^2}\sqrt{0^2+(-4)^2+(0.5)^2}}=\frac{-39}{41.1}$$

Then $\theta=\arccos(-0.948)= 161.44$

Now using the cross product:

$$\sin(\theta)=\frac{|A\times B|}{|A||B|}=\frac{ \begin{vmatrix}
i & j & k \\
0 & 10 & 2 \\
0 & -4 & 0.5 \\
\end{vmatrix}}{\sqrt{0^2+10^2+2^2} \sqrt{0^2+(-4)^2+(0.5)^2}}=
\frac{|10(0.5)i+0(-4)k+0(2)j-0(10)k-2(-4)i-(0.5)(0)j|}{\sqrt{1690}}=\frac{|13i|}{41.1}=\frac{13}{41.1}=0.316$$

but then I get $\theta=\arcsin(0.316)=18.42$

What am I doing wrong?

Best Answer

Note that your two angles add to $180^\circ$. The $\arcsin$ function is defined to have a limited range so it is a function, but you need to select the correct angle of all the ones that have the sine you calculate. As the vectors are almost in opposite directions the correct answer is the one you got from the dot product.

Related Solutions

Geometry – Using Cross Product to Find Angle Between Vectors

Hint The cross product satisfies $$||{\bf a} \times {\bf b}|| = ||{\bf a}|| \, ||{\bf b}|| \sin \theta,$$ where $\theta \in [0, \pi]$ is the angle between $\bf a$ and $\bf b$.

(In fact this property is very nearly one of the common definitions of the cross product; see, e.g., Defn. 7.4 of Dennis G. Zill, Michael R. Cullen (2006). Advanced engineering mathematics (3rd ed.). Jones & Bartlett Learning.)

Calculus – Maximum Angle Between Vectors

Unless you're working under peculiar constraints, doing this algebraically is almost certainly a tactical error. Instead, draw a diagram:

      A
     / \
   u/   \v
  |/     \|
  B------>C
      v'

The angles are $\phi$ at C and $\theta$ at A.

$\phi$ will certainly be a maximum if we can make it $\pi$ which happens if $|u|>|v|$ and $\theta=0$.

Then consider the case $|u|<|v|$. If we have already decided on the positions of A and C, the possible positions of B is a circle with center A. Among those, the ones that maximize the angle at C are the ones where BC is tangent to the circle -- which are the ones where the angle is B is right!

Basic trigonometry then gives us $|u|=|v|\cos\theta$.

Best Answer

Related Solutions

Geometry – Using Cross Product to Find Angle Between Vectors

Calculus – Maximum Angle Between Vectors

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