Given-vector-a-i-j-k-c-j-k-a-b-c-and-a-b-3-then-b-

Question Number 132856 by bramlexs22 last updated on 17/Feb/21

Given vector \vec{a} = \hat{i} + \hat{j} + \hat{k}, \vec{c} = \hat{j} - \hat{k};

\vec{a} \times \vec{b} = \vec{c} and \vec{a} . \vec{b} = 3 then ∣ \vec{b} ∣ = ?

Answered by EDWIN88 last updated on 17/Feb/21

from \vec{a} \times \vec{b} = \vec{c} we get \vec{a} \times (\vec{a} \times \vec{b}) = \vec{a} \times \vec{c}

\vec{a} \times \vec{c} = | \begin{matrix} 1 1 1 \\ 0 1 - 1 \end{matrix} | = - 2 \hat{i} + \hat{j} + \hat{k}

and \vec{a} \times (\vec{a} \times \vec{b}) = (\vec{a} . \vec{b}) \vec{a} - (\vec{a} . \vec{a}) \vec{b} = - 2 \hat{i} + \hat{j} + \hat{k}

\Rightarrow 3 \vec{a} - 3 \vec{b} = - 2 \hat{i} + \hat{j} + \hat{k}

\vec{b} = \frac{(3, 3, 3) - (- 2, 1, 1)}{3} = \frac{5 \hat{i} + 2 \hat{j} + 2 \hat{k}}{3}

Therefore ∣ \vec{b} ∣ = \frac{\sqrt{25 + 4 + 4}}{3} = \frac{\sqrt{33}}{3} = \sqrt{\frac{11}{3}}

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