First slide

Vector triple product

Question

The condition for equations $\vec{r} \times \vec{a} = \vec{b} and \vec{r} \times \vec{c} = \vec{d}$ to be consistent is

Moderate

A

$\vec{b} \cdot \vec{c} = \vec{a} \cdot \vec{d}$
B

$\vec{a} \cdot \vec{b} = \vec{c} \cdot \vec{d}$
C

$\vec{b} \cdot \vec{c} + \vec{a} \cdot \vec{d} = 0$
D

$\vec{a} \cdot \vec{b} + \vec{c} \cdot \vec{d} = 0$

Solution

$\begin{array}{l} \vec{r} \times \vec{a} = \vec{b} \\ or \vec{d} \times (\vec{r} \times \vec{a}) = \vec{d} \times \vec{b} \end{array}$
$\begin{array}{l} or (\vec{a} \cdot \vec{d}) \vec{r} - (\vec{d} \cdot \vec{r}) \vec{a} = \vec{d} \times \vec{b} - - - - (i) \\ \vec{r} \times \vec{c} = \vec{d} \\ or \vec{b} \times (\vec{r} \times \vec{c}) = \vec{b} \times \vec{d} \\ or (\vec{b} \cdot \vec{c}) \vec{r} - (\vec{b} \cdot \vec{r}) \vec{c} = \vec{b} \times \vec{d - - - (ii)} \end{array}$
Adding (i) and (ii), we get
$(\vec{a} \cdot \vec{d} + \vec{b} \cdot \vec{c}) \vec{r} - (\vec{d} \cdot \vec{r}) \vec{a} - (\vec{b} \cdot \vec{r}) \vec{c} = \vec{0}$
$Now \vec{r} \cdot \vec{d} = 0 and \vec{b} \cdot \vec{r} = 0 as \vec{d} and \vec{r} as well$
$as \vec{b} and \vec{r} are mutually perpendicular.$
Hence $(\vec{b} \cdot \vec{c} + \vec{a} \cdot \vec{d}) \vec{r} = \vec{0}$

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