Question

If IAxBI = IA.B|, then the angle between A and B is supposed to be 9O°, but how ?
If I do ab sin ☆ = ab cos ☆, then I get ☆ as 45°
What am I doing wrong?

Answer 1

You probably mean IAxBI = IAI.IBI

Answer 2

\[|\vec A \times \vec B| = |\vec A|||\vec B|\sin \theta| \]
\[|\vec A \bullet \vec B| = |\vec A|||\vec B|\cos \theta| \]
if \[|\vec A \times \vec B| =|\vec A \bullet \vec B| \ then\]
\[\theta = \frac{\pi}{4}\]

ie the original proposition is wrong.

if

Answer 3

if \[\theta = \frac{\pi}{2} \ then\]
\[| \vec A \times \vec B | = |\vec A||\vec B|\]

Answer 4

but if \[\theta = \frac{\pi}{2} \ then\]
\[\vec A \bullet \vec B = 0, \ so \ |\vec A \bullet \vec B| = 0\]

Answer 5

If you take theta as 90 then you get ab sin 90 = ab cos 90 , ie ab(1) = ab (0), which are not equal. How do I prove for theta equals 90?

Answer 6

"then you get ab sin 90 = ab cos 90"

no, you don't, if you apply the formulae *correctly*. read above

in short, if theta = 90, then the dot prod A.B is zero and the cross prod A x B is |A||B|.

if this is not helping, you might be better posting a source question or an extract from your notes or the basis of the original incorrect proposition.

Answer 7

Just a rephrase of what IrishBoy123 said.

\[
\begin{align*}
\|\vec{A}\|\|\vec{B}\|\sin(\theta)&=\|\vec{A}\|\|\vec{B}\|\cos(\theta)\\
\sin(\theta)&=\cos(\theta)\\
\tan(\theta)&=1\\
\theta&=\frac{\pi}{4} \text{ or }\frac{5\pi}{4},\,0\leq\theta<2\pi
\end{align*}
\]
If the angle between \(\vec{A}\) and \(\vec{B}\) is 90°, then dot product is zero and cross product is \(\|\vec{A}\|\|\vec{B}\|\), so the statement "If \(\|\vec{A}\times\vec{B}\|=\|\vec{A}\cdot\vec{B}\|\), then the angle between A and B is supposed to be 90° " is wrong.

Answer 8

Ok I've got it! Thank you @Irishboy123 , @thomas5267 and @Vincent-Lyon.Fr :)