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JEE Mains · Maths · STD 12 - 11. three dimension geometry
If the line \(\frac{{x - 3}}{2} = \frac{{y + 2}}{{ - 1}} = \frac{{z + 4}}{3}\) lies in the plane \(lx + my - z = 9\) then \({l^2} + {m^2} = \;.\;.\;.\;.\;.\;\)
- A \(5\)
- B \(2\)
- C \(26\)
- D \(18\)
Answer & Solution
Correct Answer
(B) \(2\)
Step-by-step Solution
Detailed explanation
\(\frac{x-3}{2}=\frac{y+2}{-1}=\frac{z+4}{3}\) and Given plane is \(\ell \mathrm{x}+\mathrm{my}-\mathrm{z}=9\) Now, it is given that line lies on plane \(\therefore 2 \ell-\mathrm{m}-3=0 \Rightarrow 2 \ell-\mathrm{m}=3\) ....\((1)\) Also, \((3,-2,-4)\) lies on plane…
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