The equation of the plane through the points $ (2, 2,1) $ and $ (9, 3, 6) $ and perpendicular to the plane $ 2x + 6y + 6 z - 1 = 0 $ is
- $ 3x + 4y + 5z + 9 = 0 $
- $ 3x + 4y - 5z + 9 = 0 $
- $ 3x - 4y + 5z + 9 = 0 $
\(3x\ + 4y\ -5z\ -9\ =\ 0\)
The Correct Option is D
Solution and Explanation
The equation of plane through the point (2,2,1) is
a(x-2) + b(y-2) + c(z-1) = 0 ………(1)
Since this line passes through (9,3,6)
a(9-2) + b(3-2) + c(6-1) = 0
7a+b+5c = 0 ………(2)
Since plane (1) is perpendicular to the plane 2x+6y+6z = 9
a(2) + b(6) + c(6) = 0
2a + 6b +6c = 0
a + 3b + 3c = 0 ………(3)
from eq (2) and eq (3)
\(\frac {a}{3-15}\) = \(\frac {b}{5-21}\) = \(\frac {c}{21-1}\)
\(\frac {a}{-12}\) = \(\frac {b}{-16}\) = \(\frac {c}{20}\)
\(\frac a3\)= \(\frac {b}{4}\) = \(\frac {c}{-5}\)
Let \(\frac a3\) = \(\frac b4\) = \(\frac {c}{-5}\) = k
Then a = 3k, b= 4k and C= -5k
From eq (1)
3k(x-2) + 4k(y-2) + (-5k)(z-1) = 0
3x - 6 +4y - 8 - 5z + 5 = 0
3x + 4y - 5z - 9 = 0
This is the required equation of plane.
So, the correct answer is (D): 3x + 4y - 5z - 9 = 0
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Concepts Used:
Equation of a Line in Space
In a plane, the equation of a line is given by the popular equation y = m x + C. Let's look at how the equation of a line is written in vector form and Cartesian form.
Vector Equation
Consider a line that passes through a given point, say ‘A’, and the line is parallel to a given vector '\(\vec{b}\)‘. Here, the line ’l' is given to pass through ‘A’, whose position vector is given by '\(\vec{a}\)‘. Now, consider another arbitrary point ’P' on the given line, where the position vector of 'P' is given by '\(\vec{r}\)'.
\(\vec{AP}\)=𝜆\(\vec{b}\)
Also, we can write vector AP in the following manner:
\(\vec{AP}\)=\(\vec{OP}\)–\(\vec{OA}\)
𝜆\(\vec{b}\) =\(\vec{r}\)–\(\vec{a}\)
\(\vec{a}\)=\(\vec{a}\)+𝜆\(\vec{b}\)
\(\vec{b}\)=𝑏1\(\hat{i}\)+𝑏2\(\hat{j}\) +𝑏3\(\hat{k}\)