First slide

Vector triple product

Question

Given that $\vec{a}, \vec{b}, \vec{p}, \vec{q}$ are four vectors such that $\vec{a} + \vec{b} = μ \vec{p}, \vec{b} \cdot \vec{q} = 0 and (\vec{b})^{2} = 1$ where $μ$ is a scalar. Then $| (\vec{a} \cdot \vec{q}) \vec{p} - (\vec{p} \cdot \vec{q}) \vec{a} |$ is equal to

Moderate

A

$2 | \vec{p} \cdot \vec{q} |$
B

$(1 / 2) | \vec{p} \cdot \vec{q} |$
C

$| \vec{p} \times \vec{q} |$
D

$| \vec{p} . \vec{q} |$

Solution

$\begin{array}{l} \vec{a} + \vec{b} = μ \vec{p} \vec{b} \cdot \vec{q} = 0, | \vec{b} |^{2} = 1 \\ ∵ \vec{a} + \vec{b} = μ \vec{p} \\ \Rightarrow (\vec{a} + \vec{b}) \times \vec{a} = μ \vec{p} \times \vec{a}, \vec{b} \times \vec{a} = μ \vec{p} \times \vec{a} \\ \Rightarrow \vec{q} \times (\vec{b} \times \vec{a}) = μ \vec{q} \times (\vec{p} \times \vec{a}) \\ \Rightarrow (\vec{q} \cdot \vec{a}) \vec{b} - (\vec{q} \cdot \vec{b}) \vec{a} = μ \vec{q} \times (\vec{p} \times \vec{a}) \\ \Rightarrow (\vec{q} \cdot \vec{a}) \vec{b} = μ \vec{q} \times (\vec{p} \times \vec{a}) \\ ∵ \vec{a} + \vec{b} = μ \vec{p} \\ \Rightarrow \vec{q} \cdot (\vec{a} + \vec{b}) = μ \vec{q} \cdot \vec{p} \\ \Rightarrow \vec{q} \cdot \vec{a} + \vec{q} \cdot \vec{b} = μ \vec{p} \cdot \vec{q} \\ \Rightarrow μ = \frac{\vec{q} \cdot \vec{a}}{\vec{p} \cdot \vec{q}} \\ \Rightarrow (\vec{q} \cdot \vec{a}) \vec{b} = \frac{\vec{q} \cdot \vec{a}}{\vec{p} \cdot \vec{q}} [(\vec{q} \cdot \vec{a}) \cdot \vec{p} - (\vec{q} \cdot \vec{p}) \vec{a}] \\ \Rightarrow | (\vec{q} \cdot \vec{a}) \vec{p} - (\vec{q} \cdot \vec{p}) \vec{a} | \\ = | (\vec{p} \cdot \vec{q}) \vec{b} | = | (\vec{p} \cdot \vec{q}) | \cdot | \vec{b} | \\ \Rightarrow | (\vec{q} \cdot \vec{a}) \vec{p} - (\vec{q} \cdot \vec{p}) \vec{a} | = | \vec{p} \cdot \vec{q} | \end{array}$

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