JEE Mathematics 2023 Question with Solution

Given: The position vectors are

• $\mathbf{A} = \mathbf{a} - \mathbf{b} + \mathbf{c}$
• $\mathbf{B} = \lambda \mathbf{a} - 3\mathbf{b} + 4\mathbf{c}$
• $\mathbf{C} = -\mathbf{a} + 2\mathbf{b} - 3\mathbf{c}$
• $\mathbf{D} = 2\mathbf{a} - 4\mathbf{b} + 6\mathbf{c}$

Find: The value of $\lambda$ such that $\mathbf{AB}$, $\mathbf{AC}$ and $\mathbf{AD}$ are coplanar.

For three vectors to be coplanar, their scalar triple product must be zero:

$$\mathbf{AB} \cdot (\mathbf{AC} \times \mathbf{AD}) = 0$$

Now,

$$\mathbf{AB} = \mathbf{B} - \mathbf{A} = (\lambda - 1)\mathbf{a} - 2\mathbf{b} + 3\mathbf{c}$$ $$\mathbf{AC} = \mathbf{C} - \mathbf{A} = -2\mathbf{a} + 3\mathbf{b} - 4\mathbf{c}$$ $$\mathbf{AD} = \mathbf{D} - \mathbf{A} = \mathbf{a} - 3\mathbf{b} + 5\mathbf{c}$$

Substituting in the coplanarity condition,

$$\left((\lambda - 1)\mathbf{a} - 2\mathbf{b} + 3\mathbf{c}\right) \cdot \left(\left(-2\mathbf{a} + 3\mathbf{b} - 4\mathbf{c}\right) \times \left(\mathbf{a} - 3\mathbf{b} + 5\mathbf{c}\right)\right) = 0$$

After expanding and simplifying, we get

$$\lambda = 2$$

Therefore, the required value is $2$.

Given: $\mathbf{AB} = (\lambda - 1)\mathbf{a} - 2\mathbf{b} + 3\mathbf{c}$, $\mathbf{AC} = -2\mathbf{a} + 3\mathbf{b} - 4\mathbf{c}$, and $\mathbf{AD} = \mathbf{a} - 3\mathbf{b} + 5\mathbf{c}$.

Find: $\lambda$.

Since $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$ are non-coplanar, they are linearly independent. Hence coplanarity of $\mathbf{AB}$, $\mathbf{AC}$ and $\mathbf{AD}$ means the determinant of their coefficients with respect to $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$ must be zero:

$$\begin{vmatrix} \lambda - 1 & -2 & 3 \\ -2 & 3 & -4 \\ 1 & -3 & 5 \end{vmatrix} = 0$$

The extracted solution states that simplifying this condition gives

$$\lambda = 2$$

Therefore, the required numerical value is $2$.