JEE PYQ: Vector Algebra - Question ID acf1d673ab32 (JEE Main 2024)

In vector algebra, the scalar triple product of three vectors $\vec{A}$, $\vec{B}$, and $\vec{C}$ is defined as: $\vec{A} \cdot (\vec{B} \times \vec{C})$ This product represents the volume of the parallelepiped formed by the three vectors. If the scalar triple product is zero, it implies that the three vectors are coplanar, meaning they lie in the same plane.

The scalar triple product can also be computed using the determinant of a matrix formed by the components of the vectors: $$\vec{A} \cdot (\vec{B} \times \vec{C}) = \begin{vmatrix} A_x & A_y & A_z \\ B_x & B_y & B_z \\ C_x & C_y & C_z \\ \end{vmatrix}$$ where $\vec{A} = (A_x, A_y, A_z)$, $\vec{B} = (B_x, B_y, B_z)$, and $\vec{C} = (C_x, C_y, C_z)$.

Given the vectors: $\vec{A} = (-x \hat{i} - 6 \hat{j} - 2 \hat{k}), \quad \vec{B} = (-\hat{i} + 4 \hat{j} + 3 \hat{k}), \quad \vec{C} = (-8 \hat{i} - \hat{j} + 3 \hat{k})$ We are given that: $\vec{A} \cdot (\vec{B} \times \vec{C}) = 0$ Using the determinant formula for the scalar triple product: $$\begin{vmatrix} -x & -6 & -2 \\ -1 & 4 & 3 \\ -8 & -1 & 3 \\ \end{vmatrix} = 0$$

Now, compute the determinant step-by-step:

The determinant of a $3 \times 3$ matrix: $$\begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \\ \end{vmatrix} = a(ei - fh) - b(di - fg) + c(dh - eg)$$ Applying this to our matrix: $$\begin{vmatrix} -x & -6 & -2 \\ -1 & 4 & 3 \\ -8 & -1 & 3 \\ \end{vmatrix} = (-x) \left( (4)(3) - (3)(-1) \right) - (-6) \left( (-1)(3) - (3)(-8) \right) + (-2) \left( (-1)(-1) - (4)(-8) \right)$$

Simplify each term: 1. First term: $(-x) \left( 12 + 3 \right) = (-x)(15) = -15x$ 2. Second term: $-(-6) \left( -3 + 24 \right) = 6(21) = 126$ 3. Third term: $(-2) \left( 1 + 32 \right) = (-2)(33) = -66$

Combine the terms: $-15x + 126 - 66 = 0$ Simplify further: $-15x + 60 = 0$ Solve for $x$: $$-15x = -60 \\ x = \frac{-60}{-15} \\ x = 4$$

1. Sign Errors: Students often make mistakes with the signs while expanding the determinant, especially when dealing with negative components. Always double-check the signs of each term in the expansion. 2. Incorrect Determinant Formula: Some students confuse the formula for the scalar triple product with the cross product formula. Remember, the scalar triple product is a determinant, not a cross product. 3. Arithmetic Mistakes: Simple arithmetic errors in multiplication or addition can lead to incorrect answers. Always verify each step carefully. 4. Misinterpretation of Zero Scalar Triple Product: Students sometimes forget that a zero scalar triple product implies coplanarity, not necessarily that any two vectors are parallel or equal.

Final Answer: The value of $x$ is $\boxed{4}$.