value of $|\vec a\times \vec b - \vec a \times \vec c |$

Question

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.

Answer 1

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.

Answer 2

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 $$|z|^2 = |a|^2|b|^2 + |a|^2|c|^2 - 2 (\vec a \times \vec b). (\vec a \times \vec c)=1+1-2\frac{1}{2}=1$$

Answer 3

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