If a= i+2 k, b= i+ j+ k, c=7 i-3 j+4 k, r b= c b, and r a=0, then r c is equal to:

Question

If $\vec a=\hat i+2\hat k$, $\vec b=\hat i+\hat j+\hat k$, $\vec c=7\hat i-3\hat j+4\hat k$, $\vec r	imes\vec b=\vec c	imes\vec b$, and $\vec r\cdot\vec a=0$, then $\vec r\cdot\vec c$ is equal to:

Accepted Answer

$\vec r=\vec c+\lambda\vec b$. From $\vec r\cdot\vec a=0$: $\vec c\cdot\vec a+\lambda(\vec b\cdot\vec a)=0$.

$\vec c\cdot\vec a=7+8=15$, $\vec b\cdot\vec a=1+2=3$: $\lambda=-\dfrac{15}3=-5$.

$$\vec r\cdot\vec c = |\vec c|^2+\lambda(\vec b\cdot\vec c) = 74+(-5)(8) = 74-40=34$$

(using $|\vec c|^2=49+9+16=74$ and $\vec b\cdot\vec c=7-3+4=8$)

Solutions · 1