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If the volume of the parallelopiped formed by three non-coplanar vectors $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ is 4 cu units, then $[\mathbf{a} \times \mathbf{b} \mathbf{b} \times \mathbf{c} \mathbf{c} \times \mathbf{a}]$ is equal to
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Verified Answer
The correct answer is:
16
In scalar triple product, the position of dot and cross can be changed provided the cyclic order is maintained i.e.,
$$
[\mathbf{a} \mathbf{b} \mathbf{c}]=(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}=\mathbf{a} \cdot(\mathbf{b} \times \mathbf{c})
$$
Put $\mathbf{c} \times \mathbf{a}=\mathbf{n}$
$\therefore[\mathbf{a} \times \mathbf{b} \mathbf{b} \times \mathbf{c} \mathbf{c} \times \mathbf{a}]=(\mathbf{a} \times \mathbf{b}) \cdot\{(\mathbf{b} \times \mathbf{c}) \times \mathbf{n}\}$
$=(\mathbf{a} \times \mathbf{b}) \cdot\{(\mathbf{n} \cdot \mathbf{b}) \mathbf{c}-(\mathbf{n} \cdot \mathbf{c}) \mathbf{b}\}$
$=(\mathbf{a} \times \mathbf{b}) \cdot[\{(\mathbf{c} \times \mathbf{a}) \cdot \mathbf{b}\} \mathbf{c}-\{(\mathbf{c} \times \mathbf{a}) \cdot \mathbf{c}\} \mathbf{b}]$
$=(\mathbf{a} \times \mathbf{b}) \cdot\{[\mathbf{c} \mathbf{a} \mathbf{b}] \mathbf{c}-[\mathbf{c} \mathbf{a} \mathbf{c}] \mathbf{b}\}$
$=[(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}][\mathbf{c} \mathbf{a} \mathbf{b}]-0$
$=[\mathbf{a b c}]$ [abc]
$=[\mathbf{a b c}]^{2}=(4)^{2}=16$
Alternate Method
By properties of Scalar triple product,
$[\mathbf{a} \times \mathbf{b} \mathbf{b} \times \mathbf{c} \mathbf{c} \times \mathbf{a}]=\left[\begin{array}{lll}\mathbf{a} & \mathbf{b} & \mathbf{c}^{2}\end{array}\right.$
$$
=(4)^{2}=16
$$
$$
[\mathbf{a} \mathbf{b} \mathbf{c}]=(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}=\mathbf{a} \cdot(\mathbf{b} \times \mathbf{c})
$$
Put $\mathbf{c} \times \mathbf{a}=\mathbf{n}$
$\therefore[\mathbf{a} \times \mathbf{b} \mathbf{b} \times \mathbf{c} \mathbf{c} \times \mathbf{a}]=(\mathbf{a} \times \mathbf{b}) \cdot\{(\mathbf{b} \times \mathbf{c}) \times \mathbf{n}\}$
$=(\mathbf{a} \times \mathbf{b}) \cdot\{(\mathbf{n} \cdot \mathbf{b}) \mathbf{c}-(\mathbf{n} \cdot \mathbf{c}) \mathbf{b}\}$
$=(\mathbf{a} \times \mathbf{b}) \cdot[\{(\mathbf{c} \times \mathbf{a}) \cdot \mathbf{b}\} \mathbf{c}-\{(\mathbf{c} \times \mathbf{a}) \cdot \mathbf{c}\} \mathbf{b}]$
$=(\mathbf{a} \times \mathbf{b}) \cdot\{[\mathbf{c} \mathbf{a} \mathbf{b}] \mathbf{c}-[\mathbf{c} \mathbf{a} \mathbf{c}] \mathbf{b}\}$
$=[(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}][\mathbf{c} \mathbf{a} \mathbf{b}]-0$
$=[\mathbf{a b c}]$ [abc]
$=[\mathbf{a b c}]^{2}=(4)^{2}=16$
Alternate Method
By properties of Scalar triple product,
$[\mathbf{a} \times \mathbf{b} \mathbf{b} \times \mathbf{c} \mathbf{c} \times \mathbf{a}]=\left[\begin{array}{lll}\mathbf{a} & \mathbf{b} & \mathbf{c}^{2}\end{array}\right.$
$$
=(4)^{2}=16
$$
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