smc-1177-20-Pyramids

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 Answers Only

`a = 6, b = 2, c = −3`
`4/9`
`2sqrt65 u^2`
`text(See Worked Solutions)`
`24\ text(u³)`

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)`

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 Answers Only

`A`

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`

Vectors, SPEC2 2019 VCAA 4

Vectors, SPEC2-NHT 2017 VCAA 13 MC