Let a, b, c satisfy a b= c, b c= a, | a|=2.

Question

Let $\vec a,\vec b,\vec c$ satisfy $\vec a	imes\vec b=\vec c$, $\vec b	imes\vec c=\vec a$, $|\vec a|=2$. Which is not true?

Accepted Answer

Since $\vec a\times\vec b=\vec c$ and $\vec b\times\vec c=\vec a$, $\vec a,\vec b,\vec c$ are mutually perpendicular; $|\vec c|=|\vec a\times\vec b|=|\vec a||\vec b|\Rightarrow$ with $|\vec a|=2$: from $\vec b\times\vec c=\vec a$, $|\vec b||\vec c|=2$, and $|\vec c|=2|\vec b|$ (from first relation), giving $|\vec b|=1,\ |\vec c|=2$.

Checking (A),(B),(C) each verify true using standard triple-product manipulations. For (D):
$$|3\vec a+\vec b-2\vec c|^2 = 9|\vec a|^2+|\vec b|^2+4|\vec c|^2 = 9(4)+1+4(4) = 36+1+16=53 \ne 51$$

So (D) is not true.

Solutions · 1