If the equation of the plane passing through the line of intersection of the planes 2x – y + z = 3, 4x– 3y +5z + 9 = 0 and parallel to the line \(\frac{x+1}{−2} =\frac{ y+3}{4}= \frac{z−2}{5}\) is ax by+cz+6=0. then a + b + c is equal to
- 12
- 13
- 14
- 15
The Correct Option is C
Solution and Explanation
Plane Intersection and Parallel Line
Let the given planes be P1 : 2x − y + z − 3 = 0 and P2 : 4x − 3y + 5z + 9 = 0. The equation of the family of planes passing through the line of intersection of P1 and P2 is given by:
P1 + λP2 = 0 ⇒ (2x − y + z − 3) + λ(4x − 3y + 5z + 9) = 0.
(2 + 4λ)x + (−1 − 3λ)y + (1 + 5λ)z + (−3 + 9λ) = 0.
The given plane is parallel to the line $\frac{x+1}{-2} = \frac{y+3}{4} = \frac{z-2}{5}$. The direction vector of the line is vL = (−2, 4, 5). The normal vector of the plane is np = (2 + 4λ, −1 − 3λ, 1 + 5λ). Since the plane is parallel to the line, the normal vector of the plane is perpendicular to the direction vector of the line. So, np · vL = 0.
−2(2 + 4λ) + 4(−1 − 3λ) + 5(1 + 5λ) = 0.
−4 − 8λ − 4 − 12λ + 5 + 25λ = 0.
−3 + 5λ = 0 ⇒ λ = $\frac{3}{5}$.
Substituting λ = $\frac{3}{5}$ in the equation of the plane:
$2 + 4 \frac{3}{5} x + -1 - 3\frac{3}{5} y + 1 + 5 \frac{3}{5} z + -3 + 9 \frac{3}{5} = 0$.
$\frac{22}{5}x - \frac{14}{5}y + \frac{20}{5}z + \frac{12}{5} = 0$.
22x − 14y + 20z + 12 = 0.
11x − 7y + 10z + 6 = 0.
Comparing this with ax + by + cz + 6 = 0, we get a = 11, b = −7, and c = 10.
Therefore, a + b + c = 11 − 7 + 10 = 14.
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