Let b⃗=i^+j^+λk^\vec b=\hat i+\hat j+\lambda\hat kb=i^+j^+λk^, λ∈R\lambda\in\mathbb Rλ∈R. If a⃗\vec aa is a vector such that a⃗×b⃗=13i^−j^−4k^\vec a\times\vec b=13\hat i-\hat j-4\hat ka×b=13i^−j^−4k^ and a⃗⋅b⃗+21=0\vec a\cdot\vec b+21=0a⋅b+21=0, then (b⃗−a⃗)⋅(k^−j^)+(b⃗+a⃗)⋅(i^−k^)\left(\vec b-\vec a\right)\cdot\left(\hat k-\hat j\right)+\left(\vec b+\vec a\right)\cdot\left(\hat i-\hat k\right)(b−a)⋅(k^−j^)+(b+a)⋅(i^−k^) is equal to ______.
Official answer: 14
Find λ\lambdaλ. a⃗×b⃗⊥b⃗\vec a\times\vec b\perp\vec ba×b⊥b: 13−1−4λ=013-1-4\lambda=013−1−4λ=0, λ=3\lambda=3λ=3, b⃗=(1,1,3)\vec b=(1,1,3)b=(1,1,3), ∣b⃗∣2=11|\vec b|^2=11∣b∣2=11.
Find a⃗\vec aa. b⃗×(a⃗×b⃗)=∣b⃗∣2a⃗−(a⃗⋅b⃗)b⃗\vec b\times(\vec a\times\vec b)=|\vec b|^2\vec a-(\vec a\cdot\vec b)\vec bb×(a×b)=∣b∣2a−(a⋅b)b; with b⃗×(13,−1,−4)=(−1,43,−14)\vec b\times(13,-1,-4)=(-1,43,-14)b×(13,−1,−4)=(−1,43,−14): a⃗=(−1,43,−14)−21(1,1,3)11=(−2,2,−7)\vec a=\dfrac{(-1,43,-14)-21(1,1,3)}{11}=(-2,2,-7)a=11(−1,43,−14)−21(1,1,3)=(−2,2,−7).
Value. (b⃗−a⃗)=(3,−1,10)(\vec b-\vec a)=(3,-1,10)(b−a)=(3,−1,10) gives 1+10=111+10=111+10=11; (b⃗+a⃗)=(−1,3,−4)(\vec b+\vec a)=(-1,3,-4)(b+a)=(−1,3,−4) gives −1+4=3-1+4=3−1+4=3. Total 14.
Answer: 14