First slide

Vector triple product

Question

$Let \bar{A} = \bar{i} - 3 \vec{j} + 4 \bar{k}, \bar{B} = 6 \bar{i} + 4 \vec{j} - 8 \vec{k}, \bar{C} = 5 \bar{i} + 2 \bar{j} + 5 \bar{k} and a vector \bar{R} satisfies$
$\bar{R} \times \bar{B} = \bar{C} \times \bar{B}, \bar{R} \cdot \bar{A} = 0 then value of \frac{| \bar{B} |}{| \bar{R} - \bar{C} |} is$

Moderate

A

1
B

3
C

4
D

2

Solution

$\begin{array}{l} \bar{R} \times \bar{B} = \bar{C} \times \bar{B} \\ \Rightarrow \bar{A} \times (\bar{R} \times \bar{B}) = \bar{A} \times (\bar{C} \times \bar{B}) \\ \Rightarrow (\bar{A} . \bar{B}) \bar{R} - (\bar{A} . \bar{R}) \bar{B} = (\bar{A} . \bar{B}) \bar{C} - (\bar{A} . \bar{C}) \bar{B} \\ \Rightarrow (\bar{A} . \bar{C}) \bar{B} = (\bar{A} . \bar{B}) \bar{C} - (\bar{A} . \bar{B}) \bar{R} = (\bar{A} . \bar{B}) (\bar{C} - \bar{R}) (∵ \bar{A} . \bar{R} = 0) \\ ∴ \frac{|\bar{B}|}{|\bar{C} - \bar{R}|} = \frac{|\bar{A} . \bar{B}|}{|\bar{A} . \bar{C}|} = \frac{38}{19} = 2 \end{array}$

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