JEE Mains · Maths · STD 12 - 11. three dimension geometry
If the equation of the plane containing the line \(x+2 y+3 z-4=0=2 x+y-z+5\) and perpendicular to the plane \(\overrightarrow{ r }=(\hat{ i }-\hat{ j })+\lambda(\hat{ i }+\hat{ j }+\hat{ k })+\mu(\hat{ i }-2 \hat{ j }+3 k )\) \(a x+b y+c z=4\), then \((a-b+c)\) is equal to
- A \(20\)
- B \(24\)
- C \(22\)
- D \(18\)
Answer & Solution
Correct Answer
(C) \(22\)
Step-by-step Solution
Detailed explanation
D.R's of line \(\overrightarrow{ n }_1=-5 \hat{i}+7 \hat{j}-3 \hat{k}\) D.R's of normal of second plane \(\vec{n}_2=5 \hat{i}-2 \hat{j}-3 \hat{k}\) \(\vec{n}_1 \times \vec{n}_2=-27 \hat{i}-30 \hat{j}-25 \hat{k}\) A point on the required plane is…
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