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Mathematics Question on Vector Algebra

Let a⃗=j^−k^\vec{a} = \hat{j} - \hat{k} and c⃗=i^−j^−k^\vec{c} = \hat{i} - \hat{j} - \hat{k}. Then vector b⃗\vec{b} satisfying a⃗×b⃗+c⃗=0⃗\vec{a} \times\vec{b}+\vec{c} = \vec{0} and a⃗⋅b⃗=3\vec{a} \cdot\vec{b} = 3 is

A

2i^−j^+2k^2 \hat{i} - \hat{j} +2 \hat{k}

B

i^−j^−2k^ \hat{i} - \hat{j} -2 \hat{k}

C

i^+j^−2k^ \hat{i} + \hat{j} -2 \hat{k}

D

−i^+j^−2k^- \hat{i} + \hat{j} -2 \hat{k}

Answer

−i^+j^−2k^- \hat{i} + \hat{j} -2 \hat{k}

Explanation

Solution

c⃗=b⃗×a⃗\vec{c} =\vec{b}\times\vec{a} ⇒b⃗⋅c⃗=0\Rightarrow \vec{b}\cdot \vec{c} = 0 ⇒(b1i^+b2j^+b3k^)⋅(i^−j^−k^)=0\Rightarrow \left(b_{1}\hat{i} + b_{2} \hat{j} +b_{3} \hat{k}\right)\cdot \left(\hat{i} - \hat{j} - \hat{k}\right) = 0 b1−b2−b3=0b_{1} - b_{2} - b_{3} = 0 and a⃗⋅b⃗=3\vec{a}\cdot \vec{b} = 3 ⇒b2−b3=3b1=b2+b3=3+2b3\Rightarrow b_{2} - b_{3} = 3 b_{1} = b_{2} + b_{3} = 3 + 2b_{3} b⃗=(3+2b3)i^+(3+b3)j^+b3k^.\vec{b} = \left(3+2b_{3}\right)\hat{i} +\left(3+b_{3}\right)\hat{j} +b_{3}\hat{k}.