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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)`.

  1. Find the values of  `a, b`  and  `c`.  (2 marks)

  2. Find the cosine of the angle between the vectors  `overset(->)(AB)`  and  `overset(->)(AD)`.  (2 marks)

  3. Find the area of the base of the pyramid.  (2 marks)

  4. 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)

  5. Find the volume of the pyramid.  (2 marks)

Show Answers Only

  1. `a = 6, b = 2, c = −3`
  2. `4/9`
  3. `2sqrt65 u^2`
  4. `text(See Worked Solutions)`
  5. `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)`

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 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`