JEE Mathematics 2023 Question with Solution

Given: $\vec{b} \cdot \vec{c} = 0$, $\vec{a} \times \vec{b} = \frac{\vec{b} - \vec{c}}{2}$ and $\vec{b} \cdot \vec{d} = \vec{a} \cdot \vec{b}$.

Find: The value of $(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d})$.

From the given relation,

$$(\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c} = \frac{\vec{b} - \vec{c}}{2}$$

Comparing coefficients of $\vec{b}$ and $\vec{c}$, we get

$$\vec{a} \cdot \vec{c} = \frac{1}{2}, \qquad \vec{a} \cdot \vec{b} = \frac{1}{2}$$

Hence,

$$\vec{b} \cdot \vec{d} = \frac{1}{2}$$

Now use the identity

$$(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = (\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{d}) - (\vec{a} \cdot \vec{d})(\vec{b} \cdot \vec{c})$$

Since $\vec{b} \cdot \vec{c} = 0$,

$$(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = (\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{d})$$

Substituting the values,

$$(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$$

Therefore, the expression equals $\frac{1}{4}$.

The solution marks D as correct, but the worked steps shown give $\frac{1}{4}$, which matches option B.

Given: $\vec{a} \times \vec{b} = \frac{\vec{b} - \vec{c}}{2}$, $\vec{b} \cdot \vec{c} = 0$ and $\vec{b} \cdot \vec{d} = \vec{a} \cdot \vec{b}$.

Find: $(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d})$.

Start with

$$(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = \left(\frac{\vec{b} - \vec{c}}{2}\right) \cdot (\vec{c} \times \vec{d})$$

Using the scalar product of cross products,

$$(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = (\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{d}) - (\vec{a} \cdot \vec{d})(\vec{b} \cdot \vec{c})$$

Since $\vec{b} \cdot \vec{c} = 0$,

$$(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = (\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{d})$$

Now from

$$\vec{a} \times \vec{b} = \frac{\vec{b} - \vec{c}}{2}$$

compare with the vector triple product form used in the provided working:

(\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c} = \frac\vec{b}2 - \frac\vec{c}2

So,

$$\vec{a} \cdot \vec{c} = \frac{1}{2}, \qquad \vec{a} \cdot \vec{b} = \frac{1}{2}$$

Therefore,

$$\vec{b} \cdot \vec{d} = \vec{a} \cdot \vec{b} = \frac{1}{2}$$

Hence,

$$(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$$

Therefore, the defensible answer from the displayed working is option B.