If | c|^2=60 and c( i+2 j+5 k)= 0, then a value of c(-7 i+2 j+3 k) is:

Question

If $|\vec c|^2=60$ and $\vec c	imes(\hat i+2\hat j+5\hat k)=\vec 0$, then a value of $\vec c\cdot(-7\hat i+2\hat j+3\hat k)$ is:

Accepted Answer

Since $\vec c\times(\hat i+2\hat j+5\hat k)=\vec0$, $\vec c$ is parallel to $\hat i+2\hat j+5\hat k$: $\vec c=t(\hat i+2\hat j+5\hat k)$.

$$|\vec c|^2=t^2(1+4+25)=30t^2=60 \implies t=\pm\sqrt2$$

$$\vec c\cdot(-7\hat i+2\hat j+3\hat k) = \pm\sqrt2(-7+4+15) = \pm12\sqrt2$$

Taking $t=\sqrt2$ gives $12\sqrt2$.

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