Question

Find:

If \[ (3\hat{i}-2\hat{j}+5\hat{k})\times(4\hat{i}+p\hat{j}+q\hat{k})=\vec{0} \] then find the values of \(p\) and \(q\).

• A. \(p = -\frac{2}{3}, \, q = \frac{5}{3}\)
• B. \(p = -\frac{8}{3}, \, q = \frac{20}{3}\)
• C. \(p = \frac{20}{3}, \, q = -\frac{8}{3}\)
• D. \(p = 0, \, q = 0\)

Hint:

Whenever the cross product of two vectors is zero, avoid expanding the entire determinant! Save time by directly writing the ratio of their corresponding components: \[ \frac{a_x}{b_x} = \frac{a_y}{b_y} = \frac{a_z}{b_z} \]

Answer 1

The Correct Option is B

Concept: The vector product (or cross product) of two non-zero vectors \( \vec{a} \) and \( \vec{b} \) is equal to the zero vector \( \vec{0} \) if and only if the two vectors are collinear or parallel to each other. If \( \vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k} \) and \( \vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k} \), then their cross product can be computed using the determinant method: \[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} = \vec{0} \] Alternatively, since parallel vectors have proportional components, we can apply the condition: \[ \frac{a_1}{b_1} = \frac{a_2}{b_2} = \frac{a_3}{b_3} \]

Step 1: Identifying the vector components.
Let us define the two given vectors explicitly: \[ \vec{a} = 3\hat{i} - 2\hat{j} + 5\hat{k} \] \[ \vec{b} = 4\hat{i} + p\hat{j} + q\hat{k} \] Here, the corresponding coefficients are: a_1 &= 3, a_2 = -2, a_3 = 5
b_1 &= 4, b_2 = p, b_3 = q

Step 2: Setting up the collinearity condition.
Since \( \vec{a} \times \vec{b} = \vec{0} \), the components of the vectors must be directly proportional: \[ \frac{3}{4} = \frac{-2}{p} = \frac{5}{q} \]

Step 3: Solving for \(p\).
By equating the first two ratios: \[ \frac{3}{4} = \frac{-2}{p} \] Cross-multiplying to solve for \(p\): \[ 3 \times p = -2 \times 4 \implies 3p = -8 \implies p = -\frac{8}{3} \]

Step 4: Solving for \(q\).
By equating the first and third ratios: \[ \frac{3}{4} = \frac{5}{q} \] Cross-multiplying to solve for \(q\): \[ 3 \times q = 5 \times 4 \implies 3q = 20 \implies q = \frac{20}{3} \] Thus, the values are \(p = -\frac{8}{3}\) and \(q = \frac{20}{3}\), which matches option (B).