Question:
If the points with position vectors
\[
(\mathbf{a}i + 10j + 13k), \quad (6i + 11j + 11k), \quad \left(\frac{9}{2}i + \beta j - 8k\right)
\]
are collinear, then evaluate \( (19a - 6\beta)^2 \).
If the points with position vectors
\[
(\mathbf{a}i + 10j + 13k), \quad (6i + 11j + 11k), \quad \left(\frac{9}{2}i + \beta j - 8k\right)
\]
are collinear, then evaluate \( (19a - 6\beta)^2 \).
Show Hint
To check collinearity, equate vector ratios and solve for unknowns.
Updated On: May 25, 2025
- 16
- 36
- 25
- 49
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The Correct Option is B
Solution and Explanation
Step 1: Condition for Collinearity
For three points to be collinear, the vectors formed by them must be proportional. Finding direction vectors: \[ \overrightarrow{AB} = (6i + 11j + 11k) - (\mathbf{a}i + 10j + 13k) \] \[ = (6 - \mathbf{a})i + (11 - 10)j + (11 - 13)k \] \[ = (6 - \mathbf{a})i + j - 2k \] \[ \overrightarrow{BC} = \left(\frac{9}{2}i + \beta j - 8k \right) - (6i + 11j + 11k) \] \[ = \left(\frac{9}{2} - 6 \right)i + (\beta - 11)j + (-8 - 11)k \] \[ = \left(-\frac{3}{2}\right)i + (\beta - 11)j - 19k \] Step 2: Equating Proportions
Since the vectors must be proportional: \[ \frac{6 - a}{-3/2} = \frac{1}{\beta - 11} = \frac{-2}{-19} \] Solving for \( (19a - 6\beta)^2 \): \[ (19a - 6\beta)^2 = 36 \] Thus, the correct answer is \( 36 \).
For three points to be collinear, the vectors formed by them must be proportional. Finding direction vectors: \[ \overrightarrow{AB} = (6i + 11j + 11k) - (\mathbf{a}i + 10j + 13k) \] \[ = (6 - \mathbf{a})i + (11 - 10)j + (11 - 13)k \] \[ = (6 - \mathbf{a})i + j - 2k \] \[ \overrightarrow{BC} = \left(\frac{9}{2}i + \beta j - 8k \right) - (6i + 11j + 11k) \] \[ = \left(\frac{9}{2} - 6 \right)i + (\beta - 11)j + (-8 - 11)k \] \[ = \left(-\frac{3}{2}\right)i + (\beta - 11)j - 19k \] Step 2: Equating Proportions
Since the vectors must be proportional: \[ \frac{6 - a}{-3/2} = \frac{1}{\beta - 11} = \frac{-2}{-19} \] Solving for \( (19a - 6\beta)^2 \): \[ (19a - 6\beta)^2 = 36 \] Thus, the correct answer is \( 36 \).
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