11. 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]
Damn... I also forgot a lot of vectors... Last time I also damn
power one :(
Also shall try with what I remember
I think you can use graphic calculator to solve... I use my
calculator (not graphic) to solve for 3 unknowns
x = -0.367, y=-0.367, z = 0.636
(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)
Add (1) and (2) gives
5x - 3y - 2z = -2
Let x = 0, y = 0, so z = -1
(0,0,-1) is a point on both planes
so, equation of vector l = (0,0,-1) + t
(1,1,1)
(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
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 -1
(iv) Plane must also contain (0,0,-1)
Contain (1,-1,3)
and be perpendicular to (1,1,1)
Vector between (0,0,-1) and (1,-1,3) is (1,-1,4)
So plane must be perpendicular to both (1,1,1) and
(1,-1,4)
Use cross product
(1,1,1) x (1,-1,4) = (5,-3,-2)
so cartesian equation of plane is 5x -3y -2z = D
sub in (0,0,-1)
D = 2
So final answer: 5x -3y -2z = 2
