Question

If \(\overset{̄}{a} = 2\hat{i}+\hat{j}-\hat{k}\), \(\overset{̄}{b} = \hat{i}+3\hat{k}\) and \(\overset{̄}{c}\) is a unit vector, then the maximum value of the scalar triple product \([\overset{̄}{a} \overset{̄}{b} \overset{̄}{c}]\) is

• A. \(\sqrt{10}+\sqrt{6}\)
• B. \(\sqrt{10}\)
• C. \(\sqrt{6}\)
• D. \(\sqrt{59}\)

Hint:

The triple product equals (a x b) dot c, which is largest when c is along a x b.

Answer 1

The Correct Option is D

Step 1: Compute a cross b
\(\vec a\times\vec b = \begin{vmatrix}\hat i&\hat j&\hat k\\2&1&-1\\1&0&3\end{vmatrix} = (3-0)\hat i-(6+1)\hat j+(0-1)\hat k = 3\hat i-7\hat j-\hat k\).

Step 2: Maximise
\([\vec a\ \vec b\ \vec c] = (\vec a\times\vec b)\cdot\vec c = |\vec a\times\vec b|\cos\theta\), which is largest for \(\cos\theta = 1\).

Step 3: Magnitude
\(|\vec a\times\vec b| = \sqrt{9+49+1} = \sqrt{59}\). Option (D).

Step 4: Why not the others
The sum \(\sqrt{10}+\sqrt6\) or the individual lengths \(\sqrt{10}\), \(\sqrt6\) are not the magnitude of the cross product.

Final Answer:
The maximum value is root 59. \[ \boxed{\text{(D)}\ \sqrt{59}} \]