First slide

Vector triple product

Question

$\vec{a} and \vec{b}$ are given given vectors. with these vectors as adjacent sides, a parallelogram is constructed. The vector which is the altitude of the parallelogram and which is perpendicular to $\vec{a}$ is

Moderate

A

$\frac{(\vec{a} \cdot \vec{b})}{| \vec{a} |^{2}} \vec{a} - \vec{b}$
B

$\frac{1}{| \vec{a} |^{2}} \{| \vec{a} |^{2} \vec{b} - (\vec{a} \cdot \vec{b}) \vec{a}\}$
C

$\frac{\vec{a} \times (\vec{a} \times \vec{b})}{| \vec{a} |^{2}}$
D

$\frac{\vec{a} \times (\vec{b} \times \vec{a})}{| \vec{b} |^{2}}$

Solution

We have
$\begin{array}{l} AM = projection of \vec{b} on \vec{a} = \frac{\vec{a} \cdot \vec{b}}{| \vec{a} |} \\ ∴ \vec{AM} = (\frac{\vec{a} \cdot \vec{b}}{| \vec{a} |^{2}}) \vec{a} \end{array}$
$Now, in △ A D M$
$\begin{array}{l} \vec{AD} = \vec{AM} + \vec{MD} or \vec{DM} = \vec{AM} - \vec{AD} \\ \Rightarrow \vec{DM} = \frac{(\vec{a} \cdot \vec{b}) \vec{a}}{| \vec{a} |^{2}} - \vec{b} \\ = \frac{1}{| \vec{a} |^{2}} [(\vec{a} \cdot \vec{b}) \vec{a} - | \vec{a} |^{2} \vec{b}] \\ \Rightarrow \vec{MD} = \frac{1}{| \vec{a} |^{2}} [| \vec{a} |^{2} \vec{b} - (\vec{a} \cdot \vec{b}) \vec{a}] \\ now, \frac{\vec{a} \times (\vec{a} \times \vec{b})}{| \vec{a} |^{2}} = \frac{1}{| \vec{a} |^{2}} [(\vec{a} \cdot \vec{b}) \vec{a} - (\vec{a} \cdot \vec{a}) \vec{b}] \\ = \vec{DM} \end{array}$

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