JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let \(A\) be a point on the line \(\vec r = \left( {1 - 3\mu } \right)\hat i + \left( {\mu - 1} \right)\hat j + \left( {2 + 5\mu } \right)\hat k\) and \(B(3, 2, 6)\) be a point in the space. Then the value of \(\mu \) for which the vector \(\overrightarrow {AB} \) is parallel to the plane \(x -4y +3z = 1\) is
- A \(\frac{1}{4}\)
- B \(\frac{1}{8}\)
- C \(\frac{1}{2}\)
- D \(-\frac{1}{4}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{4}\)
Step-by-step Solution
Detailed explanation
Let \(A\) is \((1-3 \mu, \mu-1,2+5 \mu)\) \(\overrightarrow{\mathrm{AB}}=(3 \mu+2) \mathrm{i}+(3-\mu) \mathrm{j}+(4-5 \mu)\) \({\hat k}\) which is parallel to plane \(x-4 y+3 z=1\) \(\Rightarrow 1(3 \mu+2)-4(3-\mu)+3(4-5 \mu)=0\) \(=-8 \mu+2=0 \Rightarrow \mu=\frac{1}{4}\)
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