Question

Let $a=3j+2k$ and $b=2j+k.$ If c is a unit vector, then the maximum value of $\left[abc\right]$ is:

• A. $\sqrt{59}$
• B. $\sqrt{61}$
• C. $\sqrt{108}$
• D. none

Answer 1

Answer: (B)

$\sqrt{61}$

$a\times b=\left|\begin{array}{ccc}i& j& k\\ 3& 0& 2\\ 0& 2& 1\end{array}\right|=-4i-3j+6k$

$\therefore |a\times b|=\sqrt{16+9+36}=\sqrt{61}$

Now $\left[abc\right]=(a\times b).c=|a\times b|\left|c\right|.\cos \theta$

$\therefore$ Max. value is $\sqrt{61}.1.1=\sqrt{61}$