KCET · Maths · Three Dimensional Geometry
The value of \( k \) such that the line \( \frac{x-4}{1}=\frac{y-2}{1}=\frac{z-k}{2} \) lies on the plane \( 2 x-4 y+z=7 \) is
- A \( -7 \)
- B \( 4 \)
- C \( -4 \)
- D \( 7 \)
Answer & Solution
Correct Answer
(D) \( 7 \)
Step-by-step Solution
Detailed explanation
Given line,
\[
\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-k}{2} \rightarrow(1)
\]
So, point \( (4,2, \mathrm{k}) \) lies on the line
Given plane, \( 2 x-4 y+z=7 \rightarrow(2) \)
Point \( (4,2, \mathrm{k}) \) satisfies the plane. So,
\[
\begin{array}{l}
2(4)-4(2)+k=7 \\
\Rightarrow 8-8+k=7 \\
\Rightarrow k=7
\end{array}
\]
\[
\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-k}{2} \rightarrow(1)
\]
So, point \( (4,2, \mathrm{k}) \) lies on the line
Given plane, \( 2 x-4 y+z=7 \rightarrow(2) \)
Point \( (4,2, \mathrm{k}) \) satisfies the plane. So,
\[
\begin{array}{l}
2(4)-4(2)+k=7 \\
\Rightarrow 8-8+k=7 \\
\Rightarrow k=7
\end{array}
\]
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