If $\rm \left [\vec a \times \vec b,\ \vec b \times \vec c,\ \vec c \times \vec a \right]$ = 64 then $\rm \left [\vec a\ \vec b\ \vec c \right]$is

Question

Question

This question was previously asked in

TNUSRB SI 2010 Official Paper (Held on 04 July 2010)

Answer 1

Explanation:

The scalar triple product of three vectors

\vec{a}

$\vec a$ ,

\vec{b}

$\vec b$ , and

\vec{c}

$\vec c$ is denoted as

[\vec{a}, \vec{b}, \vec{c}]

$[\vec a, \vec b, \vec c]$ and defined by:

[\vec{a}, \vec{b}, \vec{c}] = \vec{a} \cdot (\vec{b} \times \vec{c}) = \vec{b} \cdot (\vec{c} \times \vec{a}) = \vec{c} \cdot (\vec{a} \times \vec{b})

$[\vec a, \vec b, \vec c] = \vec a \cdot (\vec b \times \vec c) = \vec b \cdot (\vec c \times \vec a) = \vec c \cdot (\vec a \times \vec b)$

This product is equal to the determinant of a 3x3 matrix whose rows (or columns) are the components of the vectors

\vec{a}

$\vec a$ ,

\vec{b}

$\vec b$ , and

\vec{c}

$\vec c$ .

Given that

[\vec{a} \times \vec{b}, \vec{b} \times \vec{c}, \vec{c} \times \vec{a}] = 64

$[\vec a \times \vec b, \vec b \times \vec c, \vec c \times \vec a] = 64$ , we can use the property of cyclic permutations to rewrite this as

[\vec{a}, \vec{b}, \vec{c}]^{2} = 64

$[\vec a, \vec b, \vec c]^2 = 64$ . Therefore,

[\vec{a}, \vec{b}, \vec{c}] = \pm 8

$[\vec a, \vec b, \vec c] = \pm 8$ .

However, the options provided do not include

- 8

$-8$ , so we can conclude that

[\vec{a}, \vec{b}, \vec{c}] = 8

$[\vec a, \vec b, \vec c] = 8$ , which corresponds to option 2.