Question

If the points with position vectors \(a\hat{i} +10\hat{j} +13\hat{k}, 6\hat{i} +11\hat{k} +11\hat{k},\frac{9}{2}\hat{i}+B\hat{j}−8\hat{k}\) are collinear, then (19α-6β)² is equal to

• A. 16
• B. 25
• C. 36
• D. 49

Hint:

For collinear points, use vector cross-product conditions and solve systematically for unknown parameters.

Answer 1

The Correct Option is C

Step 1: Condition for collinearity.
- The points are collinear if the vectors \(\vec{AB}\) and \(\vec{BC}\) are parallel, i.e., \(\vec{AB} \times \vec{BC} = 0\).
Step 2: Find \(\vec{AB}\) and \(\vec{BC}\).
The vector \(\vec{AB}\) is calculated as: \[ \vec{AB} = (6 - \alpha)\hat{i} + (11 - 10)\hat{j} + (11 - 13)\hat{k} = (6 - \alpha)\hat{i} + \hat{j} - 2\hat{k}. \] The vector \(\vec{BC}\) is calculated as: \[ \vec{BC} = \left(\frac{9}{2} - 6\right)\hat{i} + (\beta - 11)\hat{j} + (-8 - 11)\hat{k} = \left(-\frac{3}{2}\right)\hat{i} + (\beta - 11)\hat{j} - 19\hat{k}. \] Step 3: Compute \(\vec{AB} \times \vec{BC}\).
- Using the determinant formula for the cross product: \[ \vec{AB} \times \vec{BC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 6 - \alpha & 1 & -2 \\ -\frac{3}{2} & \beta - 11 & -19 \end{vmatrix}. \] Expanding the determinant, we get the components of the cross product and set each component equal to zero to find \(\alpha\) and \(\beta\). After solving, we get the values: \[ \alpha = \frac{117}{19}, \quad \beta = \frac{41}{2}. \] Step 4: Calculate \((19\alpha - 6\beta)^2\).
Substitute \(\alpha = \frac{117}{19}\) and \(\beta = \frac{41}{2}\) into the expression \(19\alpha - 6\beta\): \[ 19\alpha - 6\beta = 19 \cdot \frac{117}{19} - 6 \cdot \frac{41}{2} = 117 - 123 = -6. \] Now, calculate the square: \[ (19\alpha - 6\beta)^2 = (-6)^2 = 36. \] Final Answer: \((19\alpha - 6\beta)^2 = 36\).