First slide

Vector triple product

Question

$If \vec{a} = 5 \hat{i} - \hat{j} + \hat{k}, \vec{b} = 2 \hat{i} - 3 \hat{j} - \hat{k}, \vec{c} = - 3 \hat{i} + \hat{j} + \hat{k} and \vec{d} = 2 \hat{j} + \hat{k}, then the value of \vec{d} \cdot (\vec{a} \times {\vec{b} \times (\vec{c} \times \vec{d})}) equals$

Moderate

A

98
B

99
C

100
D

101

Solution

$Given vectors are \vec{a} = 5 \hat{i} - \hat{j} + \hat{k}, \vec{b} = 2 \hat{i} - 3 \hat{j} - \hat{k}, \vec{c} = - 3 \hat{i} + \hat{j} + \hat{k}$
$\begin{array}{l} and \vec{d} = 2 \hat{j} + \hat{k} \\ We have \vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} \\ Here \vec{b} \cdot \vec{d} = - 7 and \vec{b} \cdot \vec{c} = - 10 \\ Now, \vec{b} \times (\vec{c} \times \vec{d}) = (\vec{b} \cdot \vec{d}) \vec{c} - (\vec{b} \cdot \vec{c}) \vec{d} \end{array}$
$\begin{array}{l} = - 7 \vec{c} + 10 \vec{d} \\ \vec{a} \times (\vec{b} \times (\vec{c} \times \vec{d})) = \vec{a} \times (10 \vec{d} - 7 \vec{c}) \\ = 10 \vec{a} \times \vec{d} - 7 \vec{a} \times \vec{c} \\ Now, \vec{d} \cdot (\vec{a} \times (\vec{b} \times (\vec{c} \times \vec{d}))) = \vec{d} \cdot (10 \vec{a} \times \vec{d} - 7 \vec{a} \times \vec{c}) \\ = 10 [\vec{d} \vec{a} \vec{d}] - 7 [\vec{d} \vec{a} \vec{c}] \\ = 10 (0) - 7 |\begin{array}{ccc} 0 & 2 & 1 \\ 5 & - 1 & 1 \\ - 3 & 1 & 1 \end{array}| \\ = - 7 [0 - 2 (5 + 3) + 1 (5 - 3)] \\ = 98 \end{array}$
Therefore, the correct answer is (1).

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