Question

For three vectors \( \vec{A} = (-xi - 6j - 2k), \) \( \vec{B} = (-i + 4j + 3k) \) and \( \vec{C} = (-8i - j + 3k), \) if \( \vec{A} \cdot (\vec{B} \times \vec{C}) = 0, \) then value of x is ___________.

Answer 1

Correct Answer: 4

First, compute \(\vec{B} \times \vec{C}\):

\[ \vec{B} \times \vec{C} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 4 & 3 \\ -8 & -1 & 3 \end{vmatrix} \]

\[ \vec{B} \times \vec{C} = \hat{i}(12 + 3) - \hat{j}(-3 + 24) + \hat{k}(-1 - (-32)) \]

\[ \vec{B} \times \vec{C} = 15\hat{i} - 21\hat{j} + 33\hat{k} \]

Now, compute \(\vec{A} \cdot (\vec{B} \times \vec{C})\):

\[ \vec{A} \cdot (\vec{B} \times \vec{C}) = (-x\hat{i} - 6\hat{j} - 2\hat{k}) \cdot (15\hat{i} - 21\hat{j} + 33\hat{k}) \]

\[ \vec{A} \cdot (\vec{B} \times \vec{C}) = (-x)(15) + (-6)(-21) + (-2)(33) \]

\[ \vec{A} \cdot (\vec{B} \times \vec{C}) = -15x + 126 - 66 \]

\[ \vec{A} \cdot (\vec{B} \times \vec{C}) = -15x + 60 \]

Since \(\vec{A} \cdot (\vec{B} \times \vec{C}) = 0\), we get:

\[ -15x + 60 = 0 \]

\[ 15x = 60 \]

\[ x = 4 \]

Answer 2

The problem provides three vectors, \(\vec{A}\), \(\vec{B}\), and \(\vec{C}\), and states that their scalar triple product, \(\vec{A} \cdot (\vec{B} \times \vec{C})\), is equal to zero. We are asked to find the value of the unknown component 'x' in vector \(\vec{A}\).

Concept Used:

The scalar triple product of three vectors \(\vec{A} = A_x\hat{i} + A_y\hat{j} + A_z\hat{k}\), \(\vec{B} = B_x\hat{i} + B_y\hat{j} + B_z\hat{k}\), and \(\vec{C} = C_x\hat{i} + C_y\hat{j} + C_z\hat{k}\) can be calculated using a determinant:

\[ \vec{A} \cdot (\vec{B} \times \vec{C}) = \begin{vmatrix} A_x & A_y & A_z \\ B_x & B_y & B_z \\ C_x & C_y & C_z \end{vmatrix} \]

When the scalar triple product is zero, it implies that the three vectors are coplanar. We can use this property to set up an equation and solve for the unknown value \(x\).

Step-by-Step Solution:

Step 1: Identify the components of the three given vectors.

For \(\vec{A} = -x\hat{i} - 6\hat{j} - 2\hat{k}\), the components are \(A_x = -x, A_y = -6, A_z = -2\).

For \(\vec{B} = -1\hat{i} + 4\hat{j} + 3\hat{k}\), the components are \(B_x = -1, B_y = 4, B_z = 3\).

For \(\vec{C} = -8\hat{i} - 1\hat{j} + 3\hat{k}\), the components are \(C_x = -8, C_y = -1, C_z = 3\).

Step 2: Set up the determinant for the scalar triple product and equate it to zero as per the given condition.

\[ \vec{A} \cdot (\vec{B} \times \vec{C}) = \begin{vmatrix} -x & -6 & -2 \\ -1 & 4 & 3 \\ -8 & -1 & 3 \end{vmatrix} = 0 \]

Step 3: Expand the determinant along the first row to solve for \(x\).

\[ (-x) \begin{vmatrix} 4 & 3 \\ -1 & 3 \end{vmatrix} - (-6) \begin{vmatrix} -1 & 3 \\ -8 & 3 \end{vmatrix} + (-2) \begin{vmatrix} -1 & 4 \\ -8 & -1 \end{vmatrix} = 0 \]

Step 4: Calculate the values of the 2x2 determinants.

\[ -x((4)(3) - (3)(-1)) + 6((-1)(3) - (3)(-8)) - 2((-1)(-1) - (4)(-8)) = 0 \]

Step 5: Simplify the expression.

\[ -x(12 - (-3)) + 6(-3 - (-24)) - 2(1 - (-32)) = 0 \] \[ -x(12 + 3) + 6(-3 + 24) - 2(1 + 32) = 0 \] \[ -x(15) + 6(21) - 2(33) = 0 \] \[ -15x + 126 - 66 = 0 \]

Final Computation & Result:

Solve the linear equation for \(x\).

\[ -15x + 60 = 0 \] \[ 60 = 15x \] \[ x = \frac{60}{15} \] \[ x = 4 \]

The value of x is 4.