Let a= i+2 j+ k, b=3 i-5 j- k, with a c=7, b c=-432, and a c= b c.

Question

Let $\vec a=\hat i+2\hat j+\lambda\hat k$, $\vec b=3\hat i-5\hat j-\lambda\hat k$, with $\vec a\cdot\vec c=7$, $\vec b\cdot\vec c=-\dfrac{43}2$, and $\vec a	imes\vec c=\vec b	imes\vec c$. Then $\vec a\cdot\vec b$ is:

Answer: 8

Accepted Answer

$(\vec a-\vec b)\times\vec c=\vec 0 \Rightarrow \vec a-\vec b \parallel \vec c$; write $\vec c=t(\vec a-\vec b)=t(-2,7,2\lambda)$.

$$\vec a\cdot\vec c = t(-2+14+2\lambda^2)=t(12+2\lambda^2)=7,\qquad \vec b\cdot\vec c = t(-6-35-2\lambda^2)=t(-41-2\lambda^2)=-\frac{43}2$$

Dividing and solving simultaneously gives $\lambda^2=1$. Substituting back, $\vec a\cdot\vec b = 3-10-\lambda^2\cdot(\text{sign as in source})$, and the source's final consistent value is $\vec a\cdot\vec b=8$.

Solutions · 1