JEE Mathematics 2023 Question with Solution

Given: Four distinct points with position vectors $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ are coplanar.

Find: The value of $[\vec{a}\vec{b}\vec{c}]$ in terms of scalar triple products involving $\vec{d}$.

For coplanar points, the vectors joining consecutive points satisfy

$$[\vec{b}-\vec{a}, \vec{c}-\vec{b}, \vec{d}-\vec{c}] = 0$$

Expanding the scalar triple product,

$$(\vec{b}-\vec{a}) \cdot \left((\vec{c}-\vec{b}) \times (\vec{d}-\vec{c})\right) = 0$$

Using distributive expansion and rewriting in scalar triple product form,

$$[\vec{b}\vec{c}\vec{d}] - [\vec{b}\vec{c}\vec{a}] - [\vec{b}\vec{a}\vec{d}] - [\vec{a}\vec{c}\vec{d}] = 0$$

Rearranging the terms gives

$$[\vec{a}\vec{b}\vec{c}] = [\vec{d}\vec{c}\vec{a}] + [\vec{b}\vec{d}\vec{a}] + [\vec{c}\vec{d}\vec{b}]$$

Therefore, the correct option is A.

The solution labels the option as D, but its final derived expression matches Option A in the given options.

Given: The four points are coplanar.

Find: An identity for $[\vec{a}\vec{b}\vec{c}]$.

A standard coplanarity condition is to take three displacement vectors in the same plane:

$$\vec{b}-\vec{a}, \quad \vec{c}-\vec{b}, \quad \vec{d}-\vec{c}$$

Hence,

$$[\vec{b}-\vec{a}, \vec{c}-\vec{b}, \vec{d}-\vec{c}] = 0$$

Now expand this determinant linearly. The extracted solution simplifies this to

$$[\vec{b}\vec{c}\vec{d}] - [\vec{b}\vec{c}\vec{a}] - [\vec{b}\vec{a}\vec{d}] - [\vec{a}\vec{c}\vec{d}] = 0$$

Move the last three terms to the other side and use cyclic properties of scalar triple products to rewrite them in the required form:

$$[\vec{a}\vec{b}\vec{c}] = [\vec{d}\vec{c}\vec{a}] + [\vec{b}\vec{d}\vec{a}] + [\vec{c}\vec{d}\vec{b}]$$

Thus the required value is the expression listed in Option A.