Question

If $\vec{a}=\hat{i}+\hat{j}+2\hat{k}$, $\vec{b}=\hat{i}+2\hat{j}+2\hat{k}$ and $|\vec{c}|=1$ such that $[\vec{a}\times\vec{b}\ \vec{b}\times\vec{c}\ \vec{c}\times\vec{a}]$ has maximum value, then the value of $|(\vec{a}\times\vec{b})\times\vec{c}|^2$ is

• A. 0
• B. 1
• C. 4/3
• D. none of these

Answer 1

Answer: (A)

0

Answer 2

Solution:

1. Express the given scalar triple product in a different form:

We are given

$$[\vec{a}\times\vec{b}\ \vec{b}\times\vec{c}\ \vec{c}\times\vec{a}],$$

which is the scalar triple product of $\vec{a}\times\vec{b}$, $\vec{b}\times\vec{c}$, and $\vec{c}\times\vec{a}$. Using the vector identity

$$\vec{p}\cdot (\vec{q}\times \vec{r}) = [\vec{p},\vec{q},\vec{r}],$$

and after some manipulation it can be shown that

$$[\vec{a}\times\vec{b}\ \vec{b}\times\vec{c}\ \vec{c}\times\vec{a}] = [\vec{a},\vec{b},\vec{c}]^2,$$

where

$$[\vec{a},\vec{b},\vec{c}] = \vec{a}\cdot(\vec{b}\times\vec{c}) = \vec{c}\cdot(\vec{a}\times\vec{b}).$$

2. Maximization condition:

Since $|\vec{c}| = 1$, we have

$$[\vec{a},\vec{b},\vec{c}] = \vec{c}\cdot(\vec{a}\times\vec{b}) = |\vec{a}\times\vec{b}|\cos\theta.$$

To maximize $[\vec{a},\vec{b},\vec{c}]^2$, we require $|\cos\theta| = 1$. This happens when $\vec{c}$ is parallel (or antiparallel) to $\vec{a}\times\vec{b}$.

3. Evaluating $|(\vec{a}\times\vec{b})\times\vec{c}|^2$:

When $\vec{c}$ is parallel to $\vec{a}\times\vec{b}$, the vectors $\vec{a}\times\vec{b}$ and $\vec{c}$ are parallel. The cross product of two parallel vectors is zero. Hence,

$$(\vec{a}\times\vec{b})\times\vec{c} = \vec{0}.$$

Therefore,

$$|(\vec{a}\times\vec{b})\times\vec{c}|^2 = 0.$$