JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let a line with direction ratios \(a,-4 a,-7\) be perpendicular to the lines with direction ratios 3, \(-1,2 b\) and \(b, a,-2\). If the point of intersection of the line \(\frac{x+1}{a^{2}+b^{2}}=\frac{y-2}{a^{2}-b^{2}}=\frac{z}{1}\) and the plane \(x - y + z =0\) is \((\alpha, \beta, \gamma)\), then \(\alpha+\beta+\gamma\) is equal to\(.......\)
- A \(20\)
- B \(10\)
- C \(30\)
- D \(40\)
Answer & Solution
Correct Answer
(B) \(10\)
Step-by-step Solution
Detailed explanation
\(( a ,-4 a ,-7) \perp\) to \((3,-1,2 b )\) \(a =2 b\) \(( a ,-4 a ,-7) \perp\) to \(( b , a ,-2)\) \(3 a +4 a -14 b =0\) \(ab -4 a ^{2}+14=0\) From Equations \((i)\) and \((ii)\) \(2 b ^{2}-16 b ^{2}+14=0\) \(b ^{2}=1\) \(a ^{2}=4 b ^{2}=4\)…
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