Question

If \(\left[\vec{a}\times\vec{b}\ \ \vec{b}\times\vec{c}\ \ \vec{c}\times\vec{a}\right]=\lambda\left[\vec{a}\ \ \vec{b}\ \ \vec{c}\right]^2\) then \(\lambda\) is equal to :

• A. 0
• B. 1
• C. 2
• D. 3

Hint:

The scalar triple product of the three cross products equals the square of \(\left[\vec a\ \vec b\ \vec c\right]\).

Answer 1

The Correct Option is B

Use \((\vec{b}\times\vec{c})\times(\vec{c}\times\vec{a})=\left[\vec{b}\ \vec{c}\ \vec{a}\right]\vec{c}-\left[\vec{b}\ \vec{c}\ \vec{c}\right]\vec{a}=\left[\vec{a}\ \vec{b}\ \vec{c}\right]\vec{c}\).
Then \(\left[\vec{a}\times\vec{b}\ \ \vec{b}\times\vec{c}\ \ \vec{c}\times\vec{a}\right]=(\vec{a}\times\vec{b})\cdot\left[(\vec{b}\times\vec{c})\times(\vec{c}\times\vec{a})\right]=\left[\vec{a}\ \vec{b}\ \vec{c}\right](\vec{a}\times\vec{b})\cdot\vec{c}=\left[\vec{a}\ \vec{b}\ \vec{c}\right]^2\).
Comparing with \(\lambda\left[\vec{a}\ \vec{b}\ \vec{c}\right]^2\) gives \(\lambda=1\).
Answer: (2) 1