Let b= i+ j+ k, R.

Question

Let $\vec b=\hat i+\hat j+\lambda\hat k$, $\lambda\in\mathbb R$. If $\vec a	imes\vec b=13\hat i-\hat j-4\hat k$ and $\vec a\cdot\vec b+21=0$, then $\big[(\vec b-\vec a)\cdot(\hat k-\hat j)\big] + \big[(\vec b+\vec a)\cdot(\hat i-\hat k)\big]$ is equal to\_\_\_\_

Answer: 14

Accepted Answer

Let $\vec a=(x,y,z)$. $\vec a\times\vec b = (y\lambda-z,\ z-x\lambda,\ x-y) = (13,-1,-4)$:
$$x-y=-4,\qquad y\lambda-z=13,\qquad z-x\lambda=-1$$

Solving together with $\vec a\cdot\vec b = x+y+\lambda z=-21$ gives $\lambda=3,\ x=-2,\ y=2,\ z=-7$, so $\vec a=(-2,2,-7)$, $\vec b=(1,1,3)$.

$\vec b-\vec a=(3,-1,10)$, $\vec b+\vec a=(-1,3,-4)$.

$$(\vec b-\vec a)\cdot(\hat k-\hat j) = 1(-1)\cdot(-1)+10(1) = 1+10=11,\qquad (\vec b+\vec a)\cdot(\hat i-\hat k) = -1+4=3$$

$11+3=14$

Solutions · 1