3.6 Vectors

Two lines `L_1` and `L_2` are given by the following equations, where `p in RR`.

`L_1: r = ((2), (p + 9), (-3)) + λ((p), (2p), (4))`

`L_2: r = ((14), (7), (p + 12)) + μ((p + 4), (4), (-7))`

It is known that `L_1` and `L_2` are perpendicular.

(a) Find the possible value(s) for `p`. 

(b) In the case that `p < 0`, determine whether the lines intersect. 

(a) setting a dot product of the direction vectors equal to zero 

`((p), (2p), (4)) · ((p + 4), (4), (-7)) = 0`

`p(p + 4) + 8p - 28 = 0`

`p^2 + 12p - 28 = 0`

`(p + 14)(p - 2) = 0`

`p = -14, p = 2`

(b) `p=-14 =>`

`L_1: r = ((2), (-5), (-3)) + λ((-14), (-28), (4))`

`L_2: r = ((14), (7), (-2)) + μ((-10), (4), (-7))`

a common point would satisfy the equations

`2 - 14λ = 14 - 10μ`

`-5 - 28λ = 7 + 4μ`

`-3 + 4λ = -2 - 7μ`

attempting to solve the equations using a GDC

GDC indicates no solution

so lines do not intersect

A garden has a triangular sunshade suspended from three points `A`(2, 0, 2), `B`(8, 0, 2) and `C`(5, 4, 3), relative to an origin in the corner of the garden. All distances are measured in metres.

(a)

(i) Find `vec(CA)`.

(ii) Find `vec(CB)`.

(b) Find `vec(CA) times vec(CB)`.

(c) Hence find the area of the triangle ABC. 

(a)

(i) `vec(CA) = ((-3), (-4), (-1))`

(ii) `vec(CB) = ((3), (-4), (-1))`

(b) `vec(CA) × vec(CB) = ((0), (-6), (24))`

(c) area is `1/2 sqrt(6^2 + 24^2) = 12.4 \ m^2 \ \ (12.3693..., 3sqrt17)`

Consider the vectors `bba` and `bb b` such that `bba = ((12), (-5))` and `|bb b| = 15`.

(a) Find the possible range of values for `|bba + bb b|`.

(a) `|bba| = sqrt(12^2 + (-5)^2) = 13`

`2 <= |bba + bb b| <= 28` (accept min 2 and max 28) 

(b) recognition that `bbp` or `bb b` is a negative multiple of `bba` 

`bbp = -2hat(bba)` OR `bb b = -15/13 bba = -15/13((12), (-5))`

`bbp = -2/13((12), (-5)) = ((-1.85), (0.769))`

(c) `bbq` is perpendicular to `((12), (-5))`

`=> bbq` is in the direction `((5), (12))`

`bbq = k((5), (12))`

`|bbq| = sqrt((5k)^2 + (12k)^2) = 15`

`k = 15/13`

`bbq = 15/13((5), (12)) = ((75/13), (180/13)) = ((5.77), (13.8))`

A vertical pole stands on a sloped platform. The bottom of the pole is used as the origin, O, of a coordinate system in which the top, F, of the pole has coordinates (0, 0, 5.8). All units are in metres.

The pole is held in place by ropes attached at F.

One of these ropes is attached to the platform at point `A(3.2, 4.5, −0.3)`. The rope forms a straight line from A to F.

(a) Find `vec(AF)`.

(b) Find the length of the rope. 

(c) Find `hat(FAO)`, the angle the rope makes with the platform. 

(a) `((-3.2), (-4.5), (6.1))`

(b) `sqrt((-3.2)^2 + (-4.5)^2 + 6.1^2) = 8.22800...`

A triangular cover is positioned over a walled garden to provide shade. It is anchored at points A and C, located at the top of a 2 m wall, and at a point B, located at the top of a 1 m vertical pole fixed to a top corner of the wall.

The three edges of the cover can be represented by the vectors `vec(AB) = ((0), (6), (1))`, `vec(AC) = ((7), (3), (0))` and `vec(BC) = ((7), (-3), (-1))`, where distances are measured in metres. 

(a) Calculate the vector product `vec(AB) × vec(AC)`.

(b) Hence find the area of the triangular cover. 

The point X on [AC] is such that [BX] is perpendicular to [AC].

(c) Use your answer to part (b) to find the distance BX. 

(d) Find the angle the cover makes with the horizontal plane. 

(a) attempt to find the vector product (e.g. one term correct) 

`((0), (6), (1)) × ((7), (3), (0)) = ((-3), (7), (-42))`

(b) attempt to use the vector product formula for the area of triangle 

(condone incorrect signs and missing `1/2`)

area `= (1/2)sqrt(3^2 + 7^2 + 42^2) = 21.3 \ (m^2) \ \ \ (21.3424... , 1/2sqrt1822)`

(c) `AC = 7.61577... (= sqrt58)`

setting the area formula `1/2 times` base `times` height equal to their part (b) 

`BX = (2 × 21.3424...) / sqrt58 = 5.60 \ (5.60480...)`

(d) attempting to set up a trig ratio 

angle is `arcsin (1/(BX))`

`10.3^@ \ \ \ (10.2776...^@, 0.179378... \ radians)`

A straight line `L` has vector equation `bbr = ((1), (3), (0)) + lambda ((1), (1), (2))` and point Q has coordinates `(11, −1, 3)`. 

Point P is the point on `L` closest to Q.

(a) Find the coordinates of P.

(b) Find a vector that is perpendicular to both `L` and the line passing through points P and Q. 

(a) vector from Q to any point in `L` or vice versa 

`= ((1 + lambda), (3 + lambda), (2lambda)) - ((11), (-1), (3)) = ((-10 + lambda), (4 + lambda), (2lambda - 3))`

attempt to use distance formula

minimizing `(-10 + lambda)^2 + (4 + lambda)^2 + (-3 + 2lambda)^2`

`lambda=2`

(a) valid approach to find `vec(AB)`

e.g. `vec(OB) - vec(OA), A - B`

`vec(AB) = ((8), (m - 1), (-8))`

(b) valid approach 

e.g. `L = ((9), (m), (-6)) = ((-3), (-19), (24)) + s ((2), (4), (-5))`

one correct equation

e.g. `-3 + 2s = 9, -6 = 24 - 5s`

correct value for `s`

e.g. `s=6`

substituting their `s` value into their expression/equation to find `m`

e.g. `−19 + 6 × 4`

`m=5`

(c) valid approach 

e.g. `vec(BC) = ((9p), (-6), (3)), C = 9bbu + B, vec(BC) = ((x - 9), (y - 5), (z + 6))`

correct working to find C

e.g. `vec(OC) = ((9p + 9), (-1), (-3)), C = 9 ((p), (-2/3), (1/3)) + ((9), (5), (-6)), y = -1 and z = -3`

correct approach to find `|bbu|` (seen anywhere)

e.g. `p^2 + (-2/3)^2 + (1/3)^2, sqrt(p^2 + 4/9 + 1/9)`

recognizing unit vector has magnitude of 1

e.g. `|bbu| = 1, sqrt(p^2 + (-2/3)^2 + (1/3)^2) = 1, p^2 + 5/9 = 1`

correct working

`p^2 = 4/9, p = +- 2/3`

`p = 2/3`

substituting their value of `p`

e.g. `((x - 9), (y - 5), (z + 6)) = ((6), (-6), (3)), C = ((6), (-6), (3)) + ((9), (5), (-6)), C = 9((2/3), (-2/3), (1/3)) + ((9), (5), (-6)), x - 9 = 6`

`C(15, -1, -3)` (accept `((15), (-1), (-3))`)

Consider the vectors `bb a = ((0),(3),(p))` and `bb b = ((0),(6),(18))`.

Find the value of `p` for which `bba` and `bb b` are

(a) parallel; 

(b) perpendicular. 

(a) valid approach

e.g. `bb b = 2bba, bba = k bb b, cos theta = 1, bba * bb b = -|bba| |bb b|, 2p = 18`

`p=9`

(b) evidence of scalar product

e.g. `bba * bb b, (0)(0) + (3)(6) + (p)(18)`

recognizing `bba * bb b=0` (seen anywhere) 

correct working

e.g. `18 + 18p = 0, 18p = -18`

`p=-1`

The magnitudes of two vectors, `bbu` and `bbv`, are 4 and `sqrt 3` respectively. The angle between `bbu` and `bbv` is `pi/6`.

Let `bbw=bbu-bbv`. Find the magnitude of `bbw`.

valid approach, in terms of `bbu` and `bbv` (seen anywhere)

e.g. `|bbw|^2 = (bbu - bbv) * (bbu +bbv), |bbw|^2 = bbu * bbu - 2bbu * bbv + bbv * bbv,`

`|bbw|^2 = (u_1 - v_1)^2 + (u_2 - v_2)^2,`

`|bbw| = sqrt((u_1 - v_1)^2 + (u_2 - v_2)^2 + (u_3 - v_3)^2)`

correct value for `bbu * bbu` (seen anywhere) 

e.g. `|bbu|^2 = 16, bbu * bbu = 16, u_1^2 + u_2^2 = 16`

correct value for `bbv * bbv` (seen anywhere)

e.g. `|bbv|^2 = 3, bbv * bbv = 3, v_1^2 + v_2^2 = 3`

`cos(pi/6) = sqrt(3)/2` (seen anywhere) 

`bbu * bbv = |bbu| |bbv| cos(pi/6) = 4 * sqrt(3) * sqrt(3)/2 \ (= 6)` (seen anywhere) 

correct substitution into `bbu * bbu - 2bbu * bbv + bbv * bbv` or `u_1^2 + u_2^2 + v_1^2 + v_2^2 - 2(u_1 v_1 + u_2 v_2)` (2 or 3 dimensions)

e.g. `16 - 2(6) + 3 \ (= 7)`

`|bbw| = sqrt(7)`

Consider the vectors `bba = ((3),(2p))` and `bb b = ((p+1),(8))`.

recognizing parallel vectors are multiples of each other

e.g. `bba = kbb b, ((3), (2p)) = k((p + 1), (8)), 3k=p + 1 and 2kp=8`

correct working (must be quadratic)

e.g. `3k^2 - k = 4, 3k^2 - k - 4 , 4k^2 = 3 - k`

one correct value for `k`

e.g. `k = -1, k = 4/3, k = 3/4`

substituting their value(s) of `k`

e.g. `((3), (2p)) = 3/4((p + 1), (8)), 3(4/3) = p + 1 and 2(4/3)p = 8, (-1)((3), (2p)) = ((p + 1), (8))`

`p = -4, p = 3`