JEE Mathematics 2023 Question with Solution

Given: The position vectors of four points are $\vec{a} = 3\hat{i} - 4\hat{j} + 2\hat{k}$, $\vec{b} = \hat{i} + 2\hat{j} - \hat{k}$, $\vec{c} = -2\hat{i} - \hat{j} + 3\hat{k}$, and $\vec{d} = 5\hat{i} - 2\alpha\hat{j} + 4\hat{k}$.

Find: The value of $\alpha$ for which the four points are coplanar.

For four points to be coplanar, the volume of the tetrahedron formed by them must be zero. This is equivalent to the scalar triple product of vectors $\overrightarrow{AB}$, $\overrightarrow{AC}$, and $\overrightarrow{AD}$ being zero.

1. Find Vectors:

$$\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = (\mathbf{i} + 2\mathbf{j} - \mathbf{k}) - (3\mathbf{i} - 4\mathbf{j} + 2\mathbf{k}) = -2\mathbf{i} + 6\mathbf{j} - 3\mathbf{k}$$ $$\overrightarrow{AC} = \mathbf{c} - \mathbf{a} = (-2\mathbf{i} - \mathbf{j} + 3\mathbf{k}) - (3\mathbf{i} - 4\mathbf{j} + 2\mathbf{k}) = -5\mathbf{i} + 3\mathbf{j} + \mathbf{k}$$ $$\overrightarrow{AD} = \mathbf{d} - \mathbf{a} = (5\mathbf{i} - 2a\mathbf{j} + 4\mathbf{k}) - (3\mathbf{i} - 4\mathbf{j} + 2\mathbf{k}) = 2\mathbf{i} - (2a - 4)\mathbf{j} + 2\mathbf{k}$$

2. Compute the Scalar Triple Product:

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

First, find $\overrightarrow{AC} \times \overrightarrow{AD}$:

$$\overrightarrow{AC} \times \overrightarrow{AD} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -5 & 3 & 1 \\ 2 & -(2a - 4) & 2 \end{vmatrix}$$ $$= \mathbf{i}(3 \cdot 2 - 1 \cdot (-(2a - 4))) - \mathbf{j}(-5 \cdot 2 - 1 \cdot 2) + \mathbf{k}(-5 \cdot (-(2a - 4)) - 3 \cdot 2)$$ $$= \mathbf{i}(6 + 2a - 4) - \mathbf{j}(-10 - 2) + \mathbf{k}(10a - 20 - 6)$$ $$= \mathbf{i}(2a + 2) - \mathbf{j}(-12) + \mathbf{k}(10a - 26)$$ $$= (2a + 2)\mathbf{i} + 12\mathbf{j} + (10a - 26)\mathbf{k}$$

Now, compute the dot product with $\overrightarrow{AB}$:

$$\overrightarrow{AB} \cdot (\overrightarrow{AC} \times \overrightarrow{AD}) = (-2) \cdot (2a + 2) + 6 \cdot 12 + (-3) \cdot (10a - 26) = -4a - 4 + 72 - 30a + 78 = -34a + 146$$

Setting the scalar triple product to zero:

$$-34a + 146 = 0 \Rightarrow -34a = -146 \Rightarrow a = \frac{146}{34} = \frac{73}{17}$$

Thus, $\alpha = \frac{73}{17}$.

The solution working gives $\alpha = \frac{73}{17}$, but the solution labels the correct option as B, which does not match the listed options. The defensible correct option by value is A.

Given: Points $A(3,-4,2)$, $B(1,2,-1)$, $C(-2,-1,3)$, and $D(5,-2\alpha,4)$.

Find: The value of $\alpha$ using the coplanarity determinant.

Since the four points are coplanar,

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

That is,

$$\begin{vmatrix} -2 & 6 & -3 \\ -5 & 3 & 1 \\ 2 & -2\alpha+4 & 2 \end{vmatrix} = 0$$

This is the same scalar triple product condition used in the standard method. Evaluating it gives

$$-34\alpha + 146 = 0$$

Hence,

$$\alpha = \frac{146}{34} = \frac{73}{17}$$

Therefore, the correct option by value is A.