If A→=2i→−3j→+7k→, B→=i→+2k→ and C→=j→−k→ find magnitude of A→ . (B→×C→)

B→×C→=(i→+2k→)×(j→−k→) =i→×(j→−k→)+2k→×(j→−k→) =i→×j→−i→×k→+2k→×j→−2k→×k→ =k→+2i→−0=−2i→+j→+k→ A→ . (B→×C→)=(2i→−3j→+7k→) . (−2i→+j→+k→) =(2) (−2)+(−3) (1)+(7) (1) =0

If A→=2i→−3j→+7k→, B→=i→+2k→ and C→=j→−k→ find magnitude of A→ . (B→×C→)

B→×C→=(i→+2k→)×(j→−k→) =i→×(j→−k→)+2k→×(j→−k→) =i→×j→−i→×k→+2k→×j→−2k→×k→ =k→+2i→−0=−2i→+j→+k→ A→ . (B→×C→)=(2i→−3j→+7k→) . (−2i→+j→+k→) =(2) (−2)+(−3) (1)+(7) (1) =0

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If $\vec{A} = 2 \vec{i} - 3 \vec{j} + 7 \vec{k}, \vec{B} = \vec{i} + 2 \vec{k} a n d \vec{C} = \vec{j} - \vec{k}$ find magnitude of $\vec{A} . (\vec{B} \times \vec{C})$

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$\frac{\sqrt{17}}{13}$

$\frac{14}{3}$

$\frac{13}{\sqrt{7}}$

answer is D.

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Detailed Solution

$\vec{B} \times \vec{C} = (\vec{i} + 2 \vec{k}) \times (\vec{j} - \vec{k})$
$= \vec{i} \times (\vec{j} - \vec{k}) + 2 \vec{k} \times (\vec{j} - \vec{k})$
$= \vec{i} \times \vec{j} - \vec{i} \times \vec{k} + 2 \vec{k} \times \vec{j} - 2 \vec{k} \times \vec{k}$
$= \vec{k} + 2 \vec{i} - 0 = - 2 \vec{i} + \vec{j} + \vec{k}$
$\vec{A} . (\vec{B} \times \vec{C}) = (2 \vec{i} - 3 \vec{j} + 7 \vec{k}) . (- 2 \vec{i} + \vec{j} + \vec{k})$
$= (2) (- 2) + (- 3) (1) + (7) (1)$
$= 0$