Let the lines l1: \(\frac{x+5}{3} = \frac{y+4}{1}=\frac{z−α}{−2}\) and l2: 3x + 2y + z – 2 = 0 = x – 3y + 2z - 13 be coplanar. If the point P(a, b, c) on l1 is nearest to the point Q(- 4, -3, 2), then |a| + |b|+|c| is equal to
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For problems involving lines and planes, remember the conditions for intersection, parallelism, and perpendicularity.
The scalar triple product is zero for coplanar vectors.
The shortest distance between a point and a line is along the perpendicular.
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- 10
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- 14
The Correct Option is B
Solution and Explanation
Given:
\( (3x + 2y + z - 2) + \mu(x - 3y + 2z - 13) = 0 \)
Substituting values:
\( 3(3 + \mu) + 1 \cdot (2 - 3\mu) - 2(1 + 2\mu) = 0 \)
\( 9 - 4\mu = 0 \)
Solving for \( \mu \):
\( \mu = \frac{9}{4} \)
Next:
\( 4(-15 - 8 + \alpha - 2) + 9(-5 + 12 + 2\alpha - 13) = 0 \)
\( -100 + 4\alpha - 54 + 18\alpha = 0 \)
Simplifying:
\( \Rightarrow \alpha = 7 \)
Let:
\( P \equiv (3\lambda - 5, \lambda - 4, -2\lambda + 7) \)
Direction ratios of PQ:
\( (3\lambda - 1, \lambda - 1, -2\lambda + 5) \)
Since \( PQ \perp \ell_1 \):
\( 3(3\lambda - 1) + 1 \cdot (\lambda - 1) - 2(-2\lambda + 5) = 0 \)
Solving:
\( \lambda = 1 \)
Substituting \( \lambda = 1 \) into \( P \):
\( P = (-2, -3, 5) \)
Finally:
\( |a| + |b| + |c| = 10 \)
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