KCET · Maths · Three Dimensional Geometry
The point of intersection of the line \(x+1=\frac{y+3}{3}=\frac{-z+2}{2}\) with the plane \(3 x+4 y+5 z=10\) is
- A \((2,-6,-4)\)
- B \((2,6,-4)\)
- C \((2,6,4)\)
- D \((-2,6,-4)\)
Answer & Solution
Correct Answer
(B) \((2,6,-4)\)
Step-by-step Solution
Detailed explanation
Let \(\frac{x+1}{1}=\frac{y+3}{3}=\frac{-z+2}{2}=k\)
Thus, any point on this line will have co-ordinates.
\(x=k-1, y=3 k-3, z=-2 k+2\)
This line intersects the plane \(3 x+4 y+5 z=10\)
Since, the value of \(x, y, z\) must satisfy equation of plane.
\(\begin{array}{ll}\Rightarrow & 3(k-1)+4(3 k-3)+5(-2 k+2)=10 \\ \Rightarrow & 3 k-3+12 k-12+10-10 k=10 \\ \Rightarrow & 5 k-5=10 \\ \Rightarrow & k=3\end{array}\)
\(\therefore\) The point of intersection is \((x, y, z)\)
\(\begin{aligned} & \Rightarrow x=3-1, y=3 \times 3-3, z=-2 \times 3+2 \\ & \Rightarrow(x, y, z)=(2,6,-4)\end{aligned}\)
Thus, any point on this line will have co-ordinates.
\(x=k-1, y=3 k-3, z=-2 k+2\)
This line intersects the plane \(3 x+4 y+5 z=10\)
Since, the value of \(x, y, z\) must satisfy equation of plane.
\(\begin{array}{ll}\Rightarrow & 3(k-1)+4(3 k-3)+5(-2 k+2)=10 \\ \Rightarrow & 3 k-3+12 k-12+10-10 k=10 \\ \Rightarrow & 5 k-5=10 \\ \Rightarrow & k=3\end{array}\)
\(\therefore\) The point of intersection is \((x, y, z)\)
\(\begin{aligned} & \Rightarrow x=3-1, y=3 \times 3-3, z=-2 \times 3+2 \\ & \Rightarrow(x, y, z)=(2,6,-4)\end{aligned}\)
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