Let a= i- j, b= i+ j+ k, and c satisfy a c+ b= 0, a c=4.

Question

Let $\vec a=\hat i-\hat j$, $\vec b=\hat i+\hat j+\hat k$, and $\vec c$ satisfy $\vec a	imes\vec c+\vec b=\vec 0$, $\vec a\cdot\vec c=4$. Then $|\vec c|^2$ is:

Accepted Answer

$\vec a\times\vec c=-\vec b \implies |\vec a\times\vec c|^2=|\vec b|^2=3$.

$$|\vec a\times\vec c|^2 = |\vec a|^2|\vec c|^2-(\vec a\cdot\vec c)^2 \implies 3 = 2|\vec c|^2-16 \implies |\vec c|^2=\frac{19}2$$

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