Question

If $a,b$ and $c$ are three mutually orthogonal unit vectors, then the triple product $\left[\begin{array}{ccc}a+b+c& a+b& b+c\end{array}\right]$ is equal to

• A. $0$
• B. $1$ or $-1$
• C. $6$
• D. $3$

Answer 1

The correct option is C $6$

$[a+b+ca+bb+c]$ $=$ {axb+c) +bx(b+c)+cx(b+c)}(a+b) + {ax(a+b) + bx(a+b) + cx(a+b)}(b+c)

After simplification we get,

$=a(a\times b)+a(a\times c)+b(a\times b)+b(a\times c)+b(c\times a)+b(c\times b)+c(c\times a)+c(c\times b)$

since a, band c are orthonal unit vectors, $a\times c=|a||b|sin{90}^o$ and so on

$=a^{2}b+a^{2}c+a{b}^2+c{b}^{2}a+a{c}^2+b{c}^2$

substituting the value of 1 for the unit vectors, we get,

=6