Question

Let b =2i+j-k and c =i+3k if a is a unit vector then find the maximum value of [abc]

Answer 1

Answer:

Here, a=1 i or 1 j or 1 k

b=2 i +j-k

c=i +3 k

The vector triple product is given by [a b c]=(a×b).c

In terms of Determinant , we can represent as

$\begin{vmatrix}1 & 0 &3 \\ 1&  0& 0\\  2& 1 &  -1\end{vmatrix}$

                                    -------------------------------(1)

or

$\begin{vmatrix}1 & 0 &3 \\ 0&  1& 0\\  2& 1 &  -1\end{vmatrix}$

                                 -----------------------------(2)

or

$\begin{vmatrix}1 & 0 &3 \\ 0&  0& 1\\  2& 1 &  -1\end{vmatrix}$

                                -------------------------------(3)

As, triple product is equal to Volume of Parallelepiped having edges represented by vectors a, b and c.

So, Value of determinant 1 is, if we interchange first two rows =,

| 1(0-3)+0+0|=3

So, Value of determinant 2 is, if we interchange first two rows =,

|0-1(-1 -6)+0 |=7

So, Value of determinant 3 is, if we interchange first two rows =,

|0+0+1(1-0) |=1

So, Maximum value of the determinant among 3 is Value of determinant 2, which is = 7

So, maximum value of [a b c ], when a is unit vector, b=2 i +j-k, c=i + 3 k is 7.