MHT CET · Maths · Three Dimensional Geometry
If the line \(\frac{x-3}{2}=\frac{y+2}{-1}=\frac{z+4}{3}\) lies in the plane \(\ell x+\mathrm{m} y-\mathrm{z}=9\), then \(\ell^2+\mathrm{m}^2\) is
- A 1
- B 4
- C 2
- D 5
Answer & Solution
Correct Answer
(C) 2
Step-by-step Solution
Detailed explanation
Line is perpendicular to normal of plane
\(\begin{aligned}
& \Rightarrow(2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \cdot(l \hat{\mathrm{i}}+\mathrm{m} \hat{\mathrm{j}}-\hat{\mathrm{k}})=0 \\
& \Rightarrow 2 l-\mathrm{m}-3=0
\end{aligned}\)
\((3,-2,-4)\) lies on the plane \(l x+m y-z=9\)
\(\begin{array}{ll}
\therefore \quad & 3 l-2 \mathrm{~m}+4=9 \\
& \Rightarrow 3 l-2 \mathrm{~m}=5
\end{array}\)
Solving (i) and (ii), we get
\(\begin{array}{ll}
& l=1, \mathrm{~m}=-1 \\
\therefore \quad & l^2+\mathrm{m}^2=2
\end{array}\)
\(\begin{aligned}
& \Rightarrow(2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \cdot(l \hat{\mathrm{i}}+\mathrm{m} \hat{\mathrm{j}}-\hat{\mathrm{k}})=0 \\
& \Rightarrow 2 l-\mathrm{m}-3=0
\end{aligned}\)
\((3,-2,-4)\) lies on the plane \(l x+m y-z=9\)
\(\begin{array}{ll}
\therefore \quad & 3 l-2 \mathrm{~m}+4=9 \\
& \Rightarrow 3 l-2 \mathrm{~m}=5
\end{array}\)
Solving (i) and (ii), we get
\(\begin{array}{ll}
& l=1, \mathrm{~m}=-1 \\
\therefore \quad & l^2+\mathrm{m}^2=2
\end{array}\)
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