KCET · Maths · Determinants
The line \( \frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5} \) is parallel to the plane
- A \( 3 x+4 y+5 z=7 \)
- B \( x+y+z=2 \)
- C \( 2 x+3 y+4 z=0 \)
- D \( 2 x+y-2 z=0 \)
Answer & Solution
Correct Answer
(D) \( 2 x+y-2 z=0 \)
Step-by-step Solution
Detailed explanation
Given that, \(\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5} \rightarrow(1)\)
So, direction ratio's of line are \(3,4,5\)
Since, line and plane are parallel. So, normal to plane and line are perpendicular
Now, \(a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0\)
for plane \(2 x+y-2 z=0\)
Now, \(3(2)+4(1)-2(5)=0\)
Hence, required equation is \(2 x+y-2 z=0\)
So, direction ratio's of line are \(3,4,5\)
Since, line and plane are parallel. So, normal to plane and line are perpendicular
Now, \(a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0\)
for plane \(2 x+y-2 z=0\)
Now, \(3(2)+4(1)-2(5)=0\)
Hence, required equation is \(2 x+y-2 z=0\)
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