find the equation of the plane passing through the line of intersection of planes 2x+y-z=3 and 5x-3y+4z+9=0 and parallel to the line x-1/2=y-3/4=z-5/5

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Solution

$Equation of any plane through the intersection of the plane 2 x + y - z = 3 and 5 x - 3 y + 4 z + 9 = 0 is (2 x + y - z - 3) + λ (5 x - 3 y + 4 z + 9) = 0 \Rightarrow (2 + 5 λ) x + (1 - 3 λ) y + (4 λ - 1) z - (3 - 9 λ) = 0 . . . . (i) and this is parallel to the line \frac{x - 1}{2} = \frac{y - 3}{4} = \frac{z - 5}{5} ∴ (2 + 5 λ) . 2 + (1 - 3 λ) . 4 + (4 λ - 1) . 5 = 0 \Rightarrow 18 λ + 3 = 0 \Rightarrow λ = \frac{- 1}{6} From (i), we have (2 - \frac{5}{6}) x + (1 + \frac{1}{2}) y + (\frac{- 2}{3} - 1) z - (3 + \frac{3}{2}) = 0 \Rightarrow 7 x + 9 y - 10 z - 27 = 0 This is the required equation of the plane .$

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Find the equation of the plane passing through the line of intersection of the planes 2x + y - z = 3 and 5x - 3y + 4z + 9 = 0 and is parallel to the line $\frac{x - 1}{2} = \frac{y - 3}{4} = \frac{5 - z}{- 5} .$

Q. Find the equation of the plane which passes through the line of intersection of the planes

2 x + 3 y - 4 z + 5 = 0

and

x - 5 y + z + 6 = 0

and is parallel to the line

x + y - z + 8 = 0 = x - y - z + 2

Q. Find the equation of the plane passing through the line of intersection of the planes

x + y + z = 1

and

2 x + 3 y + 4 z = 5

which is perpendicular to the plane

x - y + z = 0

Q. Find the vector equation of the through the line of intersection of the planes

x + y + z = 1

and

2 x + 3 y + 4 z = 5

which is perpendicular to plane

x - y + z = 0

. Hence find whether the plane thus contains the line

\frac{x - 2}{5} = \frac{y - 3}{4} = \frac{z}{5}

or not.

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