Question

If $\begin{bmatrix}a \times b&b \times c&c \times a\end{bmatrix} = \lambda \begin{bmatrix}a&b&c\end{bmatrix}^2$. Then find the value of $\lambda $.
A. 0
B. 1
C. 2
D. 3

Answer 1

Hint: Apply triple product on the left side of the given equation. Then apply a cross product of two terms in the expression $\left( {b \times c} \right) \times \left( {c \times a} \right)$. Then use the commutative property of the triple product to get the simplest value of LHS and compare the simplest form with RHS to calculate the value of $\lambda$.

Formula Used:
 $\begin{bmatrix}a&b&c\end{bmatrix} = a \cdot \left( {b \times c} \right)$
$\left( {a \times b} \right) \times \left( {c \times d} \right) = \begin{bmatrix}a&b&d\end{bmatrix}c - \begin{bmatrix}a&b&c\end{bmatrix}d$
$a \cdot \left( {b \times c} \right) = \left( {b \times c} \right) \cdot a$
$\begin{bmatrix}a&a&b\end{bmatrix} = \begin{bmatrix}a&b&b\end{bmatrix} = \begin{bmatrix}a&c&c\end{bmatrix} = 0$
$\begin{bmatrix}a&b&c\end{bmatrix} = \begin{bmatrix}b&c&a\end{bmatrix} = \begin{bmatrix}c&a&b\end{bmatrix}$

Complete step by step solution:
Given equation is $\begin{bmatrix}{a \times b}&{b \times c}&{c \times a}\end{bmatrix} = \lambda {\begin{bmatrix}a&b&c\end{bmatrix}^2}$
Apply the formula $\begin{bmatrix}a&b&c\end{bmatrix} = a \cdot \left( {b \times c} \right)$ on the left side of the equation
$\left( {a \times b} \right) \cdot \left( {\left( {b \times c} \right) \times \left( {c \times a} \right)} \right) = \lambda {\begin{bmatrix}a&b&c\end{bmatrix}^2}$
Now applying $\left( {a \times b} \right) \times \left( {c \times d} \right) = \begin{bmatrix}a&b&d\end{bmatrix}c - \begin{bmatrix}a&b&c\end{bmatrix}d$ in $\left( {b \times c} \right) \times \left( {c \times a} \right)$
$ \Rightarrow \left( {a \times b} \right) \cdot \left( {\begin{bmatrix}b&c&a\end{bmatrix} \cdot c - \begin{bmatrix}b&c&c\end{bmatrix} \cdot a} \right) = \lambda {\begin{bmatrix}a&b&c\end{bmatrix}^2}$
Apply the formula $\begin{bmatrix}a&a&b\end{bmatrix} = \begin{bmatrix}a&b&b\end{bmatrix} = \begin{bmatrix}a&c&c\end{bmatrix} = 0$
$ \Rightarrow \left( {a \times b} \right) \cdot \left( {\begin{bmatrix}b&c&a\end{bmatrix} \cdot c - 0} \right) = \lambda {\begin{bmatrix}a&b&c\end{bmatrix}^2}$
Apply the formula $a \cdot \left( {b \times c} \right) = \left( {b \times c} \right) \cdot a$
$ \Rightarrow \left( {a \times b} \right) \cdot c\begin{bmatrix}b&c&a\end{bmatrix} = \lambda {\begin{bmatrix}a&b&c\end{bmatrix}^2}$
$ \Rightarrow \left( {\left( {a \times b} \right) \cdot c} \right)\begin{bmatrix}b&c&a\end{bmatrix} = \lambda {\begin{bmatrix}a&b&c\end{bmatrix}^2}$
Apply the formulas $\begin{bmatrix}a&b&c\end{bmatrix} = a \cdot \left( {b \times c} \right)$ and $\begin{bmatrix}a&b&c\end{bmatrix} = \begin{bmatrix}b&c&a\end{bmatrix} = \begin{bmatrix}c&a&b\end{bmatrix}$
$ \Rightarrow \begin{bmatrix}a&b&c\end{bmatrix}\begin{bmatrix}a&b&c\end{bmatrix} = \lambda {\begin{bmatrix}a&b&c\end{bmatrix}^2}$
$ \Rightarrow {\begin{bmatrix}a&b&c\end{bmatrix}^2} = \lambda {\begin{bmatrix}a&b&c\end{bmatrix}^2}$
Cancel out ${\begin{bmatrix}a&b&c\end{bmatrix}^2}$
$ \Rightarrow \lambda = 1$

Option ‘A’ is correct

Note: Vector product and cross product can be confusing at times but they mean the same.
Cross Product: We calculate the cross product of two vectors using determinant notation. A $2 \times 2$ determinant is defined by. $\begin{vmatrix} a_1 & b_ 1 \\ a_2 & b_2\end{vmatrix} = a_1 b_2 − b_1 a_2$.
Vector Product: It is the product of the magnitude of the vectors and the sine of the angle between them, $a \times b =|a| |b| \sin \theta$.