First slide

Vector triple product

Question

$\vec{a} and \vec{b}$ are two non-collinear unit vectors, and $\vec{u} = \vec{a} - (\vec{a} \cdot \vec{b}) \vec{b} and \vec{v} = \vec{a} \times \vec{b} . Then | \vec{v} |$ is

Moderate

A

$| \vec{u} |$
B

$| \vec{u} | + | \vec{u} \cdot \vec{b} |$
C

$| \vec{u} | + | \vec{u} \cdot \vec{a} |$
D

none of these

Solution

$\begin{array}{l} \vec{u} = \vec{a} - (\vec{a} \cdot \vec{b}) \vec{b} \\ = \vec{a} (\vec{b} \cdot \vec{b}) - (\vec{a} \cdot \vec{b}) \vec{b} \\ = \vec{b} \times (\vec{a} \times \vec{b}) \end{array}$
$\begin{array}{l} \Rightarrow | \vec{u} | = | \vec{b} \times (\vec{a} \times \vec{b}) | \\ = | \vec{b} ∥ \vec{a} \times \vec{b} | \sin 90^{\circ} \\ = | \vec{b} ∥ \vec{a} \times \vec{b} | \\ = | \vec{v} | \\ Also \vec{u} \cdot \vec{b} = \vec{b} \cdot \vec{b} \times (\vec{a} \times \vec{b}) \\ = [\vec{b} \vec{b} \vec{a} \times \vec{b}] \\ = 0 \\ \Rightarrow | \vec{v} | = | \vec{u} | + | \vec{u} \cdot \vec{b} | \end{array}$

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