Question

The value of 'a' for which the scaler triple product formed by the vectors \(\vec\alpha=\hat{i}+a\hat{j}+\hat{k}\),  \(\vec{\beta}=\hat{j}+a\hat{k}\)  and  \(\vec{\gamma}=a\hat{i}+\hat{k}\)  is maximum, is

• A. 3
• B. -3
• C. \(\frac{1}{\sqrt3}\)
• D. \(-\frac{1}{\sqrt3}\)

Answer 1

The Correct Option is D

\(\begin{vmatrix} 1 &a  &1 \\   0&1  &a \\   a& 0 &1  \end{vmatrix}\)
\(=1-a(-a^2)-a\)
\(=1+a^3-a\)
differentiating we get, \(\frac{\alpha v}{qa}=3a^2-1=0\)
\(q\equiv \frac{-1}{\sqrt3}\)
So, the correct option is (D) : \(\frac{-1}{\sqrt3}\)

Answer 2

Given :
\(α=\hat{i}+a\hat{j}+\hat{k},β=0\hat{i}+\hat{j}+a\hat{k},γ=a\hat{i}+0\hat{j}+\hat{k}\)
\([α,β,γ]=\begin{vmatrix} 1 & a & 1 \\ 0 & 1 & a \\ a & 0 & 1 \end{vmatrix}\)
\(=1(1-0)-0+a(a^2-1)\)
\(=a^3-a+1=f(a)\) (let)
Now, for maximum scalar triple product,
\(f'(a)=0⇒3a^2-1=0\)
\(⇒a=\pm\frac{1}{\sqrt3}\)
Now, \(f''(a)=6a\)
\(⇒f''(\frac{1}{\sqrt3})\gt0,f''(\frac{-1}{\sqrt3})\lt0\)
Therefore, \(a=-\frac{1}{\sqrt3}\) is the point of maxima.
So, the correct option is (D) : \(\frac{-1}{\sqrt3}\)