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

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.

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.

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

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

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}