H2 Maths 2008 P1 Q11
The equations of three planes, p1, p2, p3 are
2x – 5y + 3z = 3,
3x + 2y – 5z = -5,
5x + λy + 17z = μ,
respectively, where λ and μ are constants. When λ = -20.9 and μ =16.6, find the coordinates of the point at which these planes meet. [2]
The planes p1 and p2 intersect in a line l.
(i) Find a vector equation of l. [4]
(ii) Given that all three planes meet in the line l, find λ and μ. [3]
(iii) Given instead that the three planes have no point in common, what can be said about the values of λ and μ? [2]
(iv) Find the Cartesian equation of the plane which contains l and the point (1, -1, 3). [4]
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Answer:
Use Graphic calculator to solve for 3 unknowns
x = -4/11, y=-4/11, z = 6/11
(i) 2x – 5y + 3z = 3 ---- (1)
3x + 2y – 5z = -5 ---- (2)
line parallel to both planes = (2, -5, 3) x (3, 2, -5)
= (19,19,19) = 19 (1,1,1)
Let z = 0
2x - 5y = 3
3x + 2y = -5
Solving, x = -1, y = -1
(-1,-1,0) is a point on both planes
So, equation of vector l = (-1,-1,0) + t (1,1,1)
Note: Alternatively, use the GC to solve.
(ii) (5,λ,17) dot (1,1,1) = 0
5 + λ + 17 = 0
λ = -22
5x + -22y + 17z = μ
Sub (0,0,1)
μ = 17
(iii) Three planes have no points in common, means p3 does not intersect vector l
This means p3 must be parallel to vector l, but have no points on vector l
Thus, λ will still be -22, and μ can be anything except 17
(iv) Plane must also contain points (0,0,1) and (1,-1,3)
and be perpendicular to (1,1,1)
Vector linking points (0,0,1) and (1,-1,3) is (1,-1,2).
So plane must be perpendicular to both (1,1,1) and (1,-1,2)
Using cross product,
(1,1,1) x (1,-1,2) = (3,-1,-2)
so, cartesian equation of plane is 3x -y -2z = D
sub in (0,0,1)
D = -2
So final answer: 3x -y -2z = -2
