0
$\begingroup$

If $\vec a,\vec b,\vec c$ are unit vectors such that $\vec a.\vec b = 0 = \vec a.\vec c$ , and the angle between $b$ and $c$ is $\pi/3$, then the value of $|\vec a\times \vec b - \vec a \times \vec c |$ is?

Attempt:

Let $\vec z = \vec a \times \vec b- \vec a \times \vec c$

(Avoiding \vec with modulus)

$\implies |z|^2 = |a|^2|b|^2 + |a|^2|c|^2 - 2 (\vec a \times \vec b). (\vec a \times \vec c) $ (since angle between a and b is pi/2, $\sin \theta = 1$)

$\implies |z|^2 = 1+1 - 2[\vec a\times \vec b ~~~~\vec a ~~~~\vec c]$

Where [] denotes box product (scalar triple product)

from property of scalar triple product we get:

$ \implies |z|^2 = 2 - 2[\vec c ~~~~ \vec a\times \vec b ~~~~\vec a] = 2-2 \vec c\times ((\vec a \times \vec b) . \vec a)$

Now $a $ is perpendicular to $a\times b$ so last term should be zero.

$\implies |z|^2 = 2 \implies |z| = \sqrt 2$

But answer given is $1$. Please let me know my mistake.

$\endgroup$
1
$\begingroup$

The error is that $[c\ \ a\times b\ \ a]$ does not equal $c\times((a\times b) \cdot a)$. Inside this last expression, $(a\times b)\cdot a$ is a scalar, and one cannot take the vector product of a vector with a scalar. In fact $$[c\ \ a\times b\ \ a]=c\cdot((a\times b)\times a).$$ Now one can use the vector triple product formula to simplify this $$(a\times b)\times a=(a\cdot a)b-(b\cdot a)a$$ etc.

But a simpler way to approach this problem is to note that $a\times b-a\times c =a\times (b-c)$. Since $a$ and $b-c$ are orthogonal, $|a\times (b-c)|=|a||b-c|$ etc.

$\endgroup$
  • $\begingroup$ How would we find |b-c| ? $\endgroup$ – Abcd Nov 28 '18 at 6:13
  • $\begingroup$ @Abcd Either by simple plane geometry, or by working out $|b-c|^2$. $\endgroup$ – Lord Shark the Unknown Nov 28 '18 at 6:15
1
$\begingroup$

After writing

$|z|^2 = |a|^2|b|^2 + |a|^2|c|^2 - 2 (\vec a \times \vec b). (\vec a \times \vec c)$

Notice that the vectors $\vec c, \vec b, (\vec a \times \vec c)$, and $(\vec a \times \vec b)$ are coplanar and that the angle between $(\vec a \times \vec b)$ and $(\vec a \times \vec c)$ is also $\frac{\pi}{3}$

So,$(\vec a \times \vec b). (\vec a \times \vec c)=1×1×\cos \frac{\pi}{3}= \frac{1}{2}$

Now substitue in the first equation and you'll get that $$|z|^2 = |a|^2|b|^2 + |a|^2|c|^2 - 2 (\vec a \times \vec b). (\vec a \times \vec c)=1+1-2\frac{1}{2}=1$$

$\endgroup$
0
$\begingroup$

Use the Gram-determinant: $$\begin{align} \|a\times(b-c)\|^2&=\|a\|^2\|b-c\|^2-\langle a,b-c\rangle^2\\ &=1\cdot(\|b\|^2-2\langle b,c\rangle+\|c\|^2)-(\langle a,b\rangle-\langle a,c\rangle)^2\\ &=1-2\cos(\pi/3)+1-(0-0)^2\\ &=1. \end{align}$$

$\endgroup$

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.