KCET · Maths · Determinants
Equation of the plane perpendicular to the line \( \frac{x}{1}=\frac{y}{2}=\frac{z}{3} \) and passing through the point \( (2, \),
\( 3,4) \) is
- A \( x+2 y+3 z=9 \)
- B \( x+2 y+3 z=20 \)
- C \( 2 x+3 y+z=17 \)
- D \( 3 x+2 y+z=16 \)
Answer & Solution
Correct Answer
(B) \( x+2 y+3 z=20 \)
Step-by-step Solution
Detailed explanation
Given that, \(\frac{x}{1}=\frac{y}{2}=\frac{z}{3} \rightarrow(1)\)
And point is \(\mathrm{P}(2,3,4)\).
Now, direction ratio's of normal to the plane is \(1,2,3\).
So, equation is \(1 x+2 y+3 z+d=0\)
It passes through point \(\mathrm{P}(2,3,4)\)
So, \(2+6+12+d=0\)
\(\Rightarrow d=-20\)
Therefore, required equation is \(x+2 y+3 z=20\)
And point is \(\mathrm{P}(2,3,4)\).
Now, direction ratio's of normal to the plane is \(1,2,3\).
So, equation is \(1 x+2 y+3 z+d=0\)
It passes through point \(\mathrm{P}(2,3,4)\)
So, \(2+6+12+d=0\)
\(\Rightarrow d=-20\)
Therefore, required equation is \(x+2 y+3 z=20\)
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