MHT CET · Maths · Three Dimensional Geometry
The equation of a plane containing the line \(x-2=\frac{y-4}{4}=\frac{z-6}{7}\) and parallel to the
line \(\bar{r}=(\hat{i}+3 \hat{\jmath}+5 \hat{k})+\lambda(3 \hat{\imath}+5 \hat{\jmath}+7 \hat{k})\) is
- A \(x-2 y+z=10\)
- B \(3 x-2 y+z=4\)
- C \(x-2 y+z=9\)
- D \(x-2 y+z=0\)
Answer & Solution
Correct Answer
(D) \(x-2 y+z=0\)
Step-by-step Solution
Detailed explanation
d.r. of given liens are \((1,4,7)\) and \((3,5,7)\). Normal to the plane is perpendicular to them.
\(
\begin{array}{l}
=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & 4 & 7 \\
3 & 5 & 7
\end{array}\right| \\
=\hat{i}(-7)-\hat{j}(-14)+\hat{k}(-7) \\
=-7(\hat{i}-2 \hat{j}+\hat{k})
\end{array}
\)
Hence eq. of plane is \(-7(x-2 y+z)=a\)
This plane plasses through point \((2,4,6)\)
\(\therefore \quad a=-7(2-8+6)=0\)
\(\therefore\) Eq. of plane is \(x-2 y+z=0\)
\(
\begin{array}{l}
=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & 4 & 7 \\
3 & 5 & 7
\end{array}\right| \\
=\hat{i}(-7)-\hat{j}(-14)+\hat{k}(-7) \\
=-7(\hat{i}-2 \hat{j}+\hat{k})
\end{array}
\)
Hence eq. of plane is \(-7(x-2 y+z)=a\)
This plane plasses through point \((2,4,6)\)
\(\therefore \quad a=-7(2-8+6)=0\)
\(\therefore\) Eq. of plane is \(x-2 y+z=0\)
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