3.8 Planes & Intersections

Consider the planes `Pi_1` and `Pi_2` with the following equations.

`Pi_1:3x + 2y + z = 6`

`Pi_2:x - 2y + z = 4`

(a) Find a Cartesian equation of the plane `Pi_3` which is perpendicular to `Pi_1` and `Pi_2` and passes through the origin (0,0,0). 

(b) Find the coordinates of the point where `Pi_1`, `Pi_2` and `Pi_3` intersect. 

(a) attempt to find a vector perpendicular to `Pi_1` and `Pi_2` 

using a cross product

`((3),(2),(1)) × ((1),(-2),(1)) = (2 - (-2))bbi + (1 - 3)bbj + (-6 - 2)bbk`

`=((4), (-2), (-8)) \ (= 2((2), (-1), (-4)))`

equation is `4x-2y-8z=0 \ (=>2x-y-4z=0)`

(b) attempt to solve 3 simultaneous equations in 3 variables 

`(41/21, -10/21, 23/21) \ (= (1.95, -0.476, 1.10))`

Three points `A`(3, 0, 0) , `B`(0, −2, 0) and `C`(1, 1, −7) lie on the plane `Pi_1`.

(a)

(i) Find the vector `vec(AB)` and the vector `vec(AC)`.

(ii) Hence find the equation of `Pi_1`, expressing your answer in the form `ax+by+cz=d`, where `a,b,c,d in ZZ`.

Plane `Pi_2` has equation `3x−y+2z=2`.

(b) The line `L` is the intersection of `Pi_1` and `Pi_2`. Verify that the vector equation of `L` can be written as `bbr = ((0), (-2), (0)) + λ((1), (1), (-1))`.

(c) The plane `Pi_3` is given by `2x − 2z = 3`. The line `L` and the plane `Pi_3` intersect at the point P. 

(i) Show that at the point P, `lambda=3/4`.

(ii) Hence find the coordinates of P.

(d) The point `B`(0, −2,0) lies on `L`. 

(i) Find the reflection of the point `B` in the plane `Pi_3`. 

(ii) Hence find the vector equation of the line formed when `L` is reflected in the plane `Pi_3`. 

(a)

(i) attempts to find either `vec{AB}` or `vec{AC}`

`vec(AB) = ((-3), (-2), (0))` and `vec{AC}=((-2), (1), (-7))`

(ii) attempts to find `vec{AB} times vec{AC}`

`vec{AB} times vec{AC}=((14), (-21), (-7))`

equation of plane is of the form `14x - 21y - 7z = d \ \ \ (2x - 3y - z = d)`

substitutes a valid point e.g. (3,0,0) to obtain a value of `d`

`d=42 \ \ \ (d=6)`

`14x - 21y - 7z = 42 \ \ \ (2x - 3y - z = 6)`

(b) attempts to solve `2x − 3y − z = 6` and `3x − y + 2z = 2` 

for example, `x = -λ, y = -2 - λ, z = λ`

so the vector equation of `L` can be written as `bbr = ((0), (-2), (0)) + λ((1), (1), (-1))`

(c) 

(i) substitutes the equation of `L` into the equation of `Pi_3` 

`2λ + 2λ = 3 => 4λ = 3`

`λ = 3/4`

(ii) P has coordinates `(3/4, -5/4, -3/4)`

(d)

(i) normal to `Pi_3` is `bbn = ((2), (0), (-2))`

considers the line normal to `Pi_3` passing through `B(0, −2,0)`

`bbr = ((0), (-2), (0)) + μ((2), (0), (-2))`

finding the point on the normal line that intersects `Pi_3`

attempts to solve simultaneously with plane `2x - 2z = 3`

`4μ + 4μ = 3 ⇒ μ = 3/8`

point is `(3/4, -2, -3/4)`

so, another point on the reflected line is given by

`bbr = ((0), (-2), (0)) + (3/4)((2), (0), (-2))`

`⇒ B' (3/2, -2, -3/2)`

(ii) attempts to find the direction vector of the reflected line using their P and `B'`

`vec(BP') = ((3/4), (-3/4), (-3/4))` 

`bbr = ((3/2), (-2), (-3/2)) + λ((3/4), (-3/4), (-3/4))` (or equivalent)

(a) attempts to solve `2x − y + 2z = 6` and `4x + 3y − z = 2` 

for example, `x=\lambda, \ y=2-2lambda, \ z=4-2lambda`

so the vector equation of 𝐿 can be written as `bbr = ((0), (2), (4)) + λ((1), (-2), (-2))`

(b) the position vector for point P nearest to the origin is perpendicular to the direction of `L` 

`((λ), (2 - 2λ), (4 - 2λ)) · ((1), (-2), (-2)) = 0`

`λ - 2(2 - 2λ) - 2(4 - 2λ) = 0`

`9λ - 12 = 0`

`λ = 4/3`

`P(4/3, -2/3, 4/3) \ \ \ (P(1.33, -0.667, 1.33))`

The angle between a line and a plane is `alpha`, where `alpha in RR, 0 < alpha < pi/2`.

The equation of the line is `(x - 1)/3 = (y + 2)/2 = 5 - z`, and the equation of the plane is `4x + (cos alpha)y + (sin alpha)z = 1`.

Find the value of `alpha`.  

(a) direction vector of the line is `((3), (2), (-1))` (seen anywhere) 

(b) normal vector of the plane is `((4), (cos alpha), (sin alpha))` (seen anywhere) 

correct scalar product `12 + 2 cos alpha - sin alpha` (seen anywhere) 

one correct magnitude (seen anywhere)

`sqrt(16 + cos^2 α + sin^2 α) \ (= sqrt(17)), \ sqrt(9+4+1) \ (=sqrt 14)`

recognizing angle between normal and direction vector is `pi/2-alpha` (seen anywhere)

attempt to substitute into the formula for the angle between two vectors to form an equation in `alpha`

`12 + 2 cos alpha - sin alpha = sqrt(17)sqrt(14) cos (pi/2 - alpha)`

`alpha = 0.932389... ≈ 0.932`

Find a vector that is normal to the plane containing the lines `L_1` and `L_2`, whose equations are:

`L_1: bbr = bbi + bbk + λ(2bbi + bbj - 2bbk)`

`L_2: bbr = 3bbi + 2bbj + 2bbk + mu(bbj + 3bbk)`

A vector that is normal to the plane is given by the vector product `bbd_1 × bbd_2` where `bbd_1` and `bbd_2` are the direction vectors of the lines `L_1` and `L_2` respectively.

`bbd_1 times bbd_2=(([bbi,bbj,bbk]), ([\ \ \ \2,1,-2]), ([0,1,3])) = 5bbi - 6bbj + 2bbk` (or any multiple)

Find the equation of the line of intersection of the two planes `−4x + y + z = −2` and `3x − y + 2z = −1`.

For the line of intersection:

`-4x + y + z = -2`

`3x - y + 2z = -1`

`⇒ -x + 3z = -3`

`-8x + 2y + 2z = -4`

`3x - y + 2z = -1`

`⇒ 11x - 3y = 3`

The equation of the line of intersection is `x = (3y + 3)/11 = 3z + 3` (or equivalent)

The point A is the foot of the perpendicular from the point (1,1, 9) to the plane `2x + y − z = 6`. Find the coordinates of A.

Find the angle between the plane `3x − 2y + 4z = 12` and the `z`-axis. Give your answer to the nearest degree.

Consider the three planes

`Π_1: 2x - y + z = 4`

`Π_2: x - 2y + 3z = 5`

`Π_3: -9x + 3y - 2z = 32`

(a) Show that the three planes do not intersect. 

(b)

(i) Verify that the point `P(1, -2, 0)` lies on both `Π_1` and `Π_2`.

(ii) Find a vector equation of `L`, the line of intersection of `Pi_1` and `Pi_2`. 

(c) Find the distance between `L` and `Pi_3`.

(a) attempt to eliminate a variable

obtain a pair of equations in two variables

`-3x + z = -3` and

`-3x + z = 44`

the two lines are parallel (`−3 ne 44`)

hence the three planes do not intersect

(b) 

(i) `Π_1: 2 + 2 + 0 = 4` and `Π_2: 1 + 4 + 0 = 5`

(ii) attempt to find the vector product of the two normals 

`((2), (-1), (1)) × ((1), (-2), (3)) = ((-1), (-5), (-3))`

`bbr = ((1), (-2), (0)) + λ((1), (5), (3))`

(c) the line connecting `L` and `Pi_3` is given by `L_1` 

attempt to substitute position and direction vector to form `L_1`

`bbs = ((1), (-2), (0)) + t((-9), (3), (-2))`

substitute `(1 - 9t, -2 + 3t, -2t)` in `Pi_3`

`-9(1 - 9t) + 3(-2 + 3t) - 2(-2t) = 32`

`94t = 47 ⇒ t = 1/2`

attempt to find distance between `(1, −2,0)` and their point `(-7/2, -1/2, -1)`

`=| ((1), (-2), (0)) + (1/2)((-9), (3), (-2)) - ((1), (-2), (0)) |=1/2 sqrt((-9)^2+3^2+(-2)^2)=\sqrt94 /2`

Three planes have equations: `2x−y+z=5`, `x+3y−z=4` and `3x−5y+az=b`, where `a,b in RR`.

Find the set of values of `a` and `b` such that the three planes have no points of intersection.

attempt to eliminate a variable (or attempt to find `det A`)

`([2,-1,1,|,5],[1,3,-1,|,4],[3,-5,a,|,b]) -> ([2,-1,1,|,5],[0,7,-3,|,3],[0,-14,a+3,|,b-12])`

(or `det A=14(a-3)`)

(or two correct equations in two variables)

`=> ([2,-1,1,|,5],[0,7,-3,|,3],[0,0,a-3,|,b-6])`

(or solving `det A=0`)

(or attempting to reduce to one variable, e.g. `(a − 3)z = b − 6`) 

`a=3, b ne 6`