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Mathematics Question on Vector Algebra

Let a⃗=i^−2 j^+k^\vec{a}=\hat{i}-2\,\hat{j}+\hat{k} and b⃗=i^−j^+k^\vec{b}=\hat{i}-\hat{j}+\hat{k} be two vectors. If c⃗\vec{c} is a vector such that b⃗×c⃗=b⃗×a⃗\vec{b}\times\vec{c}=\vec{b}\times\vec{a} and c⃗⋅a⃗=0,\vec{c}\cdot\vec{a}=0, then c⃗⋅b⃗\vec{c} \cdot \vec{b} is equal to :

A

12\frac{1}{2}

B

−32- \frac{3}{2}

C

−12- \frac{1}{2}

D

−1-1

Answer

−12- \frac{1}{2}

Explanation

Solution

b⃗×c⃗−b⃗×a⃗=0⃗\vec{b}\times\vec{c}-\vec{b}\times\vec{a} = \vec{0}
b⃗×(c⃗−a⃗)=0⃗\vec{b}\times \left(\vec{c}- \vec{a}\right) = \vec{0}
b⃗=λ(c⃗−a⃗)...(i)\vec{b} = \lambda\left(\vec{c}- \vec{a}\right) \quad...\left(i\right)
a⃗⋅b⃗=λ(a⃗⋅c⃗−a⃗2)\vec{a}\cdot \vec{b} = \lambda\left( \vec{a}\cdot \vec{c}- \vec{a}^{2}\right)
4=λ(0−6)⇒λ=−46=−234 = \lambda\left(0 - 6\right) \Rightarrow \lambda = \frac{-4}{6} = \frac{-2}{3}
from (i)b⃗=−23(c⃗−a⃗)\left(i\right) \vec{b} = \frac{-2}{3}\left(\vec{c}- \vec{a}\right)
c⃗=−32b⃗+a⃗=−12(i^+j^+k^)\vec{c} = \frac{-3}{2}\vec{b}+ \vec{a} = \frac{-1}{2}\left(\hat{i}+\hat{j}+\hat{k}\right)

b⃗⋅c⃗=−12\vec{b}\cdot \vec{c} = -\frac{1}{2}