A cube with side length 3 units is pictured below.
- Calculate the magnitude of vector `vec(AG)`. (1 mark)
- Find the acute angle between the diagonals `vec(AG)` and `vec(BH)`. (3 marks)
Vectors, SPEC2 2019 VCAA 4
The base of a pyramid is the parallelogram `ABCD` with vertices at points `A(2,−1,3), B(4,−2,1), C(a,b,c)` and `D(4,3,−1)`. The apex (top) of the pyramid is located at `P(4,−4,9)`.
- Find the values of `a, b` and `c`. (2 marks)
- Find the cosine of the angle between the vectors `overset(->)(AB)` and `overset(->)(AD)`. (2 marks)
- Find the area of the base of the pyramid. (2 marks)
- Show that `6underset~i + 2underset~j + 5underset~k` is perpendicular to both `overset(->)(AB)` and `overset(->)(AD)`, and hence find a unit vector that is perpendicular to the base of the pyramid. (3 marks)
- Find the volume of the pyramid. (2 marks)
Show Worked Solution
| a. | `overset(->)(AB)` | `= (4 – 2)underset~i + (−2 + 1)underset~j + (1 – 3)underset~k` |
| `= 2underset~i – underset~j – 2underset~k` |
`text(S)text(ince)\ ABCD\ text(is a parallelogram)\ => \ overset(->)(AB)= overset(->)(DC)`
`overset(->)(DC) = (a – 4)underset~i + (b – 3)underset~j + (c + 1)underset~k`
`a – 4 = 2 \ => \ a = 6`
`b – 3 = −1 \ => \ b = 2`
`c + 1 = −2 \ => \ c = −3`
b. `overset(->)(AB) = 2underset~i – underset~j – 2underset~k`
`overset(->)(AD) = 2underset~i + 4underset~j – 4underset~k`
| `cos angleBAD` | `= (overset(->)(AB) · overset(->)(AD))/(|overset(->)(AB)| · |overset(->)(AD)|)` |
| `= (4 – 4 + 8)/(sqrt(4 + 1 + 4) · sqrt(4 + 16 + 16))` | |
| `= 4/9` |
| c. | `1/2 xx text(Area)_(ABCD)` | `= 1/2 ab sin c` |
| `text(Area)_(ABCD)` | `= |overset(->)(AB)| · |overset(->)(AD)| *sin(cos^(−1)\ 4/9)` |
| `:. text(Area)_(ABCD)` | `= 3 · 6 · sqrt65/9` |
| `= 2sqrt65\ text(u²)` |
d. `text(Let)\ \ underset~r = 6underset~i + 2underset~j + 5underset~k`
| `underset~r · overset(->)(AB)` | `= 6 xx 2 + 2 xx −1 + 5 xx −2 = 0` |
| `underset~r · overset(->)(AD)` | `= 6 xx 2 + 2 xx 4 + 5 xx −2 = 0` |
`:. underset~r\ \ text(is ⊥ to)\ \ overset(->)(AB)\ \ text(and)\ \ overset(->)(AD)`
`text(Let)\ \ hatr\ = text(unit vector ⊥ to pyramid base)`
`underset~overset^r = 1/sqrt(6^2 + 2^2 + 5^2) *underset~r = 1/sqrt65(6underset~i + 2underset~j + 5underset~k)`
e. `text(Find height)\ (h)\ text(of pyramid:)`
| `overset(->)(AP)` | `= (4 – 2)underset~i + (−4 + 1)underset~j + (9 – 3)underset~k` |
| `= 2underset~i – 3underset~j + 6underset~k` |
| `h` | `= overset(->)(AP) · underset~overset^r` |
| `= (2 xx 6/sqrt65) – (3 xx 2/sqrt65) + (6 xx 5/sqrt65)` | |
| `= 36/sqrt65` |
| `:. V` | `= 1/3 b xx h` |
| `= 1/3 xx 2sqrt65 xx 36/sqrt65` | |
| `= 24\ text(u³)` |
Vectors, SPEC2-NHT 2017 VCAA 13 MC
Let `OABCD` be a right square pyramid where `underset ~a = vec(OA),\ underset ~b = vec(OB),\ underset ~c = vec(OC)` and `underset ~d = vec(OD)`.
An equation correctly relating these vectors is
A. `underset ~a + underset ~c = underset ~b + underset ~d`
B. `(underset ~a - underset ~c) ⋅ (underset ~d - underset ~c) = 0`
C. `underset ~a + underset ~d = underset ~b + underset ~c`
D. `(underset ~a - underset ~d) ⋅ (underset ~c - underset ~b) = 0`
E. `underset ~a + underset ~b = underset ~c + underset ~d`
Show Worked Solution
`|underset ~a| = |underset ~b| = |underset ~c| = |underset ~d|`
| `text(Let)\ \ underset ~a` | `= alpha underset ~i + beta underset ~j + gamma underset ~k` |
| `underset ~b` | `= -alpha underset ~i + beta underset ~j + gamma underset ~k` |
| `underset ~c` | `= -alpha underset ~i + beta underset ~j – gamma underset ~k` |
| `underset ~d` | `= alpha underset ~i + beta underset ~j – gamma underset ~k` |
| `A:\ \ underset ~a + underset ~ c` | `= 2 beta underset ~ j` |
| `underset ~b + underset ~d` | `= 2 beta underset ~j` |
`=> A`