Given a→=(1,1,1),c→=(0,1,-1),a→·b→=3 and a→×b→=c→, find b→

Let b→=b1i^+b2j^+b3k^,a→·b→=3⇒b1+b2+b3=3a→×b→=c→⇒i^j^k^111b1b2b3=j^-k^b2-b3=0,b1-b3=1,b2-b1=-1.Solving together with b1+b2+b3=3, we getb1=53,b2=b3=23 ∴ b→=53,23,23

Given a→=(1,1,1),c→=(0,1,-1),a→·b→=3 and a→×b→=c→, find b→

Let b→=b1i^+b2j^+b3k^,a→·b→=3⇒b1+b2+b3=3a→×b→=c→⇒i^j^k^111b1b2b3=j^-k^b2-b3=0,b1-b3=1,b2-b1=-1.Solving together with b1+b2+b3=3, we getb1=53,b2=b3=23 ∴ b→=53,23,23

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Given $\vec{a} = (1, 1, 1), \vec{c} = (0, 1, - 1), \vec{a} \cdot \vec{b} = 3$ and $\vec{a} \times \vec{b} = \vec{c}, find \vec{b}$

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$\frac{5}{3} i + \frac{2}{3} \hat{j} + \frac{2}{3} \hat{k}$

$\frac{1}{3} (5 \hat{i} + \hat{j} + \hat{k})$

$5 \hat{i} + 2 \hat{j} + 2 \hat{k}$

$\frac{1}{5} (\hat{i} + \hat{j} + \hat{k})$

answer is A.

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Detailed Solution

$Let \vec{b} = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}, \vec{a} \cdot \vec{b} = 3 \Rightarrow b_{1} + b_{2} + b_{3} = 3$

$\vec{a} \times \vec{b} = \vec{c} \Rightarrow |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ b_{1} & b_{2} & b_{3} \end{matrix}| = \hat{j} - \hat{k}$