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Vectors, EXT2 V1 2023 HSC 15b

On the triangular pyramid \(A B C D, L\) is the midpoint of \(A B, M\) is the midpoint of \(A C, N\) is the midpoint of \(A D, P\) is the midpoint of \(C D, Q\) is the midpoint of \(B D\) and \(R\) is the midpoint of \(B C\).
 

Let  \(\underset{\sim}{b}=\overrightarrow{A B}, \underset{\sim}{c}=\overrightarrow{A C}\)  and  \(\underset{\sim}{d}=\overrightarrow{A D}\).

  1. Show that \(\overrightarrow{L P}=\dfrac{1}{2}(-\underset{\sim}{b}+\underset{\sim}{c}+\underset{\sim}{d})\).  (1 mark)

  2. It can be shown that

\(\overrightarrow{M Q}=\dfrac{1}{2}(\underset{\sim}{b}-\underset{\sim}{c}+\underset{\sim}{d})\)  and

\(\overrightarrow{N R}=\dfrac{1}{2}(\underset{\sim}{b}+\underset{\sim}{c}-\underset{\sim}{d})\).   (Do NOT prove these.) 

  1. Prove that

  \( \Big{|}\overrightarrow{A B}\Big{|}^2+\Big{|}\overrightarrow{A C}\Big{|}^2+\Big{|}\overrightarrow{A D}\Big{|}^2+\Big{|}\overrightarrow{B C}\Big{|}^2+\Big{|}\overrightarrow{B D}\Big{|}^2+\Big{|}\overrightarrow{C D}\Big{|}^2 \)

\(=4\left(\Big{|}\overrightarrow{L P}\Big{|}^2+\Big{|}\overrightarrow{M Q}\Big{|}^2+\Big{|}\overrightarrow{N R}\Big{|}^2\right)\)  (3 marks)

Show Answers Only
  1. \(\text{See Worked Solutions}\)
  2. \(\text{Proof (See Worked Solutions)}\)
Show Worked Solution
i.     \(\overrightarrow{LP}\) \(=\overrightarrow{LA}+\overrightarrow{AC}+\overrightarrow{CP} \)
    \(=\dfrac{1}{2} \overrightarrow{BA}+\overrightarrow{AC}+\dfrac{1}{2}\overrightarrow{CD} \)
    \(=-\dfrac{1}{2} \underset{\sim}b +\underset{\sim}c + \dfrac{1}{2}(\underset{\sim}d-\underset{\sim}c) \)
    \(=-\dfrac{1}{2} \underset{\sim}b +\underset{\sim}c + \dfrac{1}{2}\underset{\sim}d-\dfrac{1}{2}\underset{\sim}c\)
    \(=\dfrac{1}{2} (-\underset{\sim}b+\underset{\sim}c+\underset{\sim}d) \)

 

ii.   \(\overrightarrow{M Q}=\dfrac{1}{2}(\underset{\sim}{b}-\underset{\sim}{c}+\underset{\sim}{d})\)

\(\overrightarrow{N R}=\dfrac{1}{2}(\underset{\sim}{b}+\underset{\sim}{c}-\underset{\sim}{d})\).

\(\text{RHS}\) \(=4\left(\Big{|}\overrightarrow{L P}\Big{|}^2+\Big{|}\overrightarrow{M Q}\Big{|}^2+\Big{|}\overrightarrow{N R}\Big{|}^2\right)\)  
  \(=4\left(\overrightarrow{L P}\cdot \overrightarrow{L P} +\overrightarrow{MQ}\cdot \overrightarrow{MQ} +\overrightarrow{NR}\cdot \overrightarrow{NR}\right)\)  
  \(=4\Bigg{(}\dfrac{1}{4}(-\underset{\sim}b+\underset{\sim}c+\underset{\sim}d) \cdot(-\underset{\sim}b+\underset{\sim}c+\underset{\sim}d) +  \)  
  \( \dfrac{1}{4}(\underset{\sim}{b}-\underset{\sim}{c}+\underset{\sim}{d}) \cdot(\underset{\sim}{b}-\underset{\sim}{c}+\underset{\sim}{d}) +  \)  
  \( \dfrac{1}{4}(\underset{\sim}{b}+\underset{\sim}{c}-\underset{\sim}{d}) \cdot(\underset{\sim}{b}+\underset{\sim}{c}-\underset{\sim}{d}) \Bigg{)}\)  
  \(=(\underset{\sim}c+(\underset{\sim}d-\underset{\sim}b)) \cdot(\underset{\sim}c+(\underset{\sim}d-\underset{\sim}b))+(\underset{\sim}d+(\underset{\sim}b-\underset{\sim}c))\cdot (\underset{\sim}d+(\underset{\sim}b-\underset{\sim}c)) + \)  
  \( (\underset{\sim}b+(\underset{\sim}c-\underset{\sim}d))\cdot (\underset{\sim}b+(\underset{\sim}c-\underset{\sim}d))  \)  
  \(=|\underset{\sim}c|^2+2\underset{\sim}c(\underset{\sim}d-\underset{\sim}b) + |\underset{\sim}d-\underset{\sim}b|^2 + \)  
  \(|\underset{\sim}d|^2+2\underset{\sim}d(\underset{\sim}b-\underset{\sim}c) + |\underset{\sim}b-\underset{\sim}c|^2 + \)  
  \( |\underset{\sim}b|^2+2\underset{\sim}b(\underset{\sim}c-\underset{\sim}d) + |\underset{\sim}c-\underset{\sim}d|^2 \)  
  \(=|\underset{\sim}b|^2+|\underset{\sim}c|^2+|\underset{\sim}d|^2+|\underset{\sim}b-\underset{\sim}c|^2+|\underset{\sim}d-\underset{\sim}b|^2+|\underset{\sim}c-\underset{\sim}d|^2 + \)  
  \( 2(\underset{\sim}c \cdot\underset{\sim}d-\underset{\sim}c \cdot\underset{\sim}b+\underset{\sim}d \cdot\underset{\sim}b-\underset{\sim}d \cdot\underset{\sim}c+\underset{\sim}b \cdot\underset{\sim}c-\underset{\sim}b \cdot\underset{\sim}d) \)  
  \(=\Big{|}\overrightarrow{A B}\Big{|}^2+\Big{|}\overrightarrow{A C}\Big{|}^2+\Big{|}\overrightarrow{A D}\Big{|}^2+\Big{|}\overrightarrow{B C}\Big{|}^2+\Big{|}\overrightarrow{B D}\Big{|}^2+\Big{|}\overrightarrow{C D}\Big{|}^2 + 0\)  
  \(=\ \text{LHS} \)  

♦♦ Mean mark (ii) 34%.

Vectors, EXT2 V1 EQ-Bank 13

In a rectangular prism, `M` is the midpoint of `AD`.

Use vector methods to find

  1. `angle HBD` to 1 decimal place  (2 marks)

  2. `angle HBM` to 1 decimal place  (2 marks)

Show Answers Only
  1. `40.5°`
  2. `51.5°`
Show Worked Solution

i.   `text{Consider the position vectors of points using}\ B\ text{as origin:}`

`vec(BH)=((0),(9),(6)), \ abs(vec(BH))=sqrt(9^2+6^2)=sqrt117`

`vec(BD)=((4),(9),(0)), \ abs(vec(BD))=sqrt(4^2+9^2)=sqrt97`

`cos angleHBD` `=(vec(BH)*vec(BD))/(abs(vec(BH))*abs(vec(BD)))`  
  `=(0xx4+9xx9+6xx0)/(sqrt117 xx sqrt97)`  
`angleHBD` `=cos^(-1)(81/(sqrt117 xx sqrt97))`  
  `=40.5°`  

 

ii.   `vec(BM)=((4),(9/2),(0)), \ abs(vec(BM))=sqrt(4^2+(9/2)^2)=sqrt(145/4)`

`cos angleHBM` `=(vec(BH) * vec(BM))/(abs(vec(BH))*abs(vec(BM)))`  
  `=(0xx4+9xx9/2+6xx0)/(sqrt117 xx sqrt(145/4))`  
`angleHBM` `=cos^(-1)(40.5/(sqrt117 xx sqrt(145/4)))`  
  `=51.5°\ \text{(1 d.p.)}`  

Vectors, EXT2 V1 EQ-Bank 9

A parallelogram is formed by joining the points  `P(-2,1,4), Q(1,4,5), R(0,2,3)` and `S(a,b,c)`.

Use vector methods to find `a,b` and `c`.  (2 marks)

Show Answers Only

`a=-3, b=1, c=2`

Show Worked Solution

`text{Opposite sides of a parallelogram are equal and parallel.}`

`=> vec(PQ)=vec(SR)`
 

`vec(PQ)=((1+2),(4-3),(5-4))=((3),(1),(1))`
 

`vec(SR)=((-a),(2-b),(3-c))`
 

`text{Equating coordinates:}`

`-a=3\ \ =>\ \ a=-3`

`2-b=1\ \ =>\ \ b=1`

`3-c=1\ \ =>\ \ c=2`

Vectors, EXT2 V1 EQ-Bank 8

Classify the triangle formed by joining the points  `A(3,1,0), B(-2,4,3)` and `C(3,3,-2)`.  (4 marks)

Show Answers Only

`text{ΔABC is a right-angled (scalene) triangle with a right-angle at B.}`

`text{(See Worked Solutions)}`

Show Worked Solution

`text{Calculating the side lengths:}`

`abs(AB)=sqrt((3+2)^2+(1-4)^2+(0-3)^2)=sqrt43`

`abs(BC)=sqrt((-2-3)^2+(4-3)^2+(3+2)^2)=sqrt51`

`abs(AC)=sqrt((3-3)^2+(1-3)^2+(0+2)^2)=sqrt8`

`=>\ text{Triangle is scalene.}`
 

`text{From above,}`

`abs(BC)^2=abs(AB)^2+abs(AC)^2\ \ (angleBAC=90°)`
 

`text{Consider scalar product of direction vectors:}`

`vec(AB)=((3),(1),(0))+lambda((-2-3),(4-1),(3-0))=((3),(1),(0))+lambda((-5),(3),(3))`
 

`vec(AC)=((3),(1),(0))+mu((3-3),(3-1),(-2-0))=((3),(1),(0))+mu((0),(2),(-2))`
 

`vec(AB)*vec(AC)=((-5),(3),(3))((0),(2),(-2))=0+6-6=0`

`:.vec(AB) ⊥ vec(AC)`

`:.\ text{ΔABC is a right-angled (scalene) triangle with a right-angle at A.}`

Vectors, EXT2 V1 2022 HSC 14a

  1. The two non-parallel vectors `\vec{u}` and `\vec{v}` satisfy  `lambda vec(u)+mu vec(v)= vec(0)`  for some real numbers `\lambda` and `\mu`.
  2. Show that  `\lambda=\mu=0`.  (2 marks)

  3. The two non-parallel vectors `\vec{u}` and `\vec{v}` satisfy  `\lambda_1 \vec{u}+\mu_1 \vec{v}=\lambda_2 \vec{u}+\mu_2 \vec{v}`  for some real numbers `\lambda_1, \lambda_2, \mu_1` and `\mu_2`.
  4. Using part (i), or otherwise, show that  `\lambda_1=\lambda_2`  and  `\mu_1=\mu_2`. (1  mark)

The diagram below shows the tetrahedron with vertices `A, B, C` and `S`.

The point `K` is defined by  `vec(SK)=(1)/(4) vec(SB)+(1)/(3) vec(SC)`, as shown in the diagram.

The point `L` is the point of intersection of the straight lines `S K` and `B C`.
 
                       
 

  1. Using part (ii), or otherwise, determine the position of `L` by showing that  `vec(BL)=(4)/(7) vec(BC)`.  (2 marks)

  2. The point `P` is defined by  `vec(AP)=-6 vec(AB)-8 vec(AC)`.
  3. Does `P` lie on the line `A L`? Justify your answer.  (2 marks)

Show Answers Only
  1. `text{Proof (See Worked Solution)}`
  2. `text{Proof (See Worked Solution)}`
  3. `text{Proof (See Worked Solution)}`
  4. `P\ text{lies on}\ vec(AL)\ \ text{(See Worked Solution)}`
Show Worked Solution
i.    `lambda vecu+mu vecv` `=0`
  `lambda vecu` `=-mu vecv`

 
`lambda=0\ \ text{or}\ \ vecu=-(mu/lambda)vecv=k vecv\ \ (kinRR)`

`text{S}text{ince}\ \ vecu and vecv\ \ text{are not parallel}`

`=> lambda=mu=0`
 

ii.  `\lambda_1 \vec{u}+\mu_1 \vec{v}=\lambda_2 \vec{u}+\mu_2 \vec{v}`

`\lambda_1 \vec{u}-\lambda_2 \vec{u}+\mu_1 \vec{v}-\mu_2 \vec{v}` `=vec0`  
`(\lambda_1-\lambda_2)\vec{u}+(\mu_1-\mu_2)\vec{v}` `=vec0`  

 

`text{Using part (i):}`

`(\lambda_1-\lambda_2)=0 and (\mu_1-\mu_2)=0`

`:.\lambda_1=\lambda_2  and  \mu_1=\mu_2\ …\ text{as required}`


Mean mark (ii) 56%.
iii.   `vec(BL)` `=lambda vec(BC)`
    `=lambda(vec(BS)+vec(SC))\ \ \ …\ (1)`

 

`vec(BL)` `=vec(BS)+mu vec(SK)`  
  `=-vec(SB)+mu(1/4vec(SB)+1/3vec(SC))`  
  `=-vec(SB)+mu/4vec(SB)+mu/3vec(SC)`  
  `=mu/3vec(SC)+(mu/4-1)vec(SB)\ \ \ …\ (2)`  


♦♦♦ Mean mark (iii) 25%.

`text{Using}\ \ (1) = (2):`

`lambda vec(BS)+ lambda vec(SC)=mu/3vec(SC)+(mu/4-1)vec(SB)`

`mu/3=lambda, \ \ 1-mu/4=lambda`

`mu/3` `=1-mu/4`  
`(4mu+3mu)/12` `=1`  
`(7mu)/12` `=1`  
`mu` `=12/7`  

 
`lambda=(12/7)/3=4/7`

`:.vec(BL)=(4)/(7) vec(BC)\ \ text{… as required}`
 

iv.  `vec(AP)=-6 vec(AB)-8 vec(AC)`

`text{If}\ P\ text{lies on}\ AL, ∃k\ text{such that}\ \ vec(AP)=k vec(AL)`


♦♦♦ Mean mark (iv) 23%.
`-6 vec(AB)-8 vec(AC)` `=k vec(AL)`  
`-6 vec(AB)-8 (vec(AB)+vec(BC))` `=k(vec(AB)+vec(BL))`  
`-14 vec(AB)-8vec(BC)` `=k(vec(AB)+4/7vec(BC))\ \ text{(see part (iii))}`  
`-14 vec(AB)-8vec(BC)` `=kvec(AB)+(4k)/7vec(BC)`  

 
`k=-14, \ \ (4k)/7=-8\ \ =>\ \ k=-14`

`:.\ P\ text{lies on}\ \ AL.`

Vectors, EXT2 V1 2021 HSC 12e

The diagram shows the pyramid  `ABCDS`  where  `ABCD`  is a square. The diagonals of the square bisect each other at `H`.
 

  1. Show that  `overset->{HA} + overset->{HB} + overset->{HC} + overset->{HD} = underset~0`   (1 mark)

    Let `G` be the point such that  `overset->{GA} + overset->{GB} + overset->{GC} + overset->{GD} + overset->{GS}  =  underset~0`.

  1. Using part (i), or otherwise, show that  `4 overset->{GH} + overset->{GS}  =  underset~0`.   (2 marks)

  2. Find the value of  `λ`  such that  `overset->{HG} = λ overset->{HS}`   (1 mark)

Show Answers Only
  1. `text{See Worked Solution}`
  2. `text{See Worked Solution}`
  3. `1/5`
Show Worked Solution

i.    `text{S} text{ince diagonal} \ overset->{AC} \ text{is bisected by H:}`

`overset->{HA} =- overset->{HC}`

`text{Similarly for diagonal} \ overset->{BD}`

`overset->{HB} = – overset->{HD}`
 

`:. \ overset->{HA} + overset->{HB} + overset->{HC} + overset->{HD}` `= – overset->{HC} – overset->{HD} +  overset->{HC} + overset->{HD}`
  `= underset~0`
 
 
ii.    `overset->{GA} = overset->{GH} + overset->{HA} \ , \ overset->{GB} = overset->{GH} + overset->{HB}`
`overset->{GC} = overset->{GH} + overset->{HC} \ , \ overset->{GD} = overset->{GH} + overset->{HD}`
 
`overset->{GA} + overset->{GB} + overset->{GC} + overset->{GD} + overset->{GS}`
`= overset->{GH} + overset->{HA} + overset->{GH} + overset->{HB} + overset->{GH} + overset->{HC} + overset->{GH} + overset->{HD} + overset->{GS}`
`= 4 overset->{GH} + (overset->{HA} + overset->{HB} + overset->{HC} + overset->{HD} + overset->{GS})`
`= 4 overset->{GH} + overset->{GS}`
 
`overset->{GA} + overset->{GB} + overset->{GC} + overset->{GD} + overset->{GS} = underset~0\ \ \ text{(given)}`
`:. 4 overset->{GH} + overset->{GS} = underset~0`
 
iii.  `overset->{HS} = overset->{HG} + overset->{GS}`
`overset->{GS} = overset->{HS} + overset->{GH}`
♦ Mean mark 48%.

 

`text{Using part (ii):}`
`4 overset->{GH} + overset->{HS} + overset->{GH}` `= underset~0`
`5 overset->{GH}` `= overset->{SH}`
`overset->{GH}` `= 1/5 overset->{SH}`
`overset->{HG}` `= 1/5 overset->{HS}`

`:. \ λ = 1/5`

Vectors, EXT2 V1 2019 SPEC2 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)

Show Answers Only
  1. `a = 6, b = 2, c = −3`
  2. `4/9`
  3. `2sqrt65\ text(u²)`
Show Worked Solution
i.    `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`

 

ii.   `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`

 

iii.    `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)`

Vectors, EXT2, V1 SM-Bank 16

 

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)`.

Show that  `underset~a + underset~c = underset~b + underset~d`.   (3 marks)

Show Answers Only

`text(See Worked Solution)`

Show Worked Solution

`text(Let)\ \ A=(p,–p,–k),`

`underset ~a` `= overset(->)(OA) = punderset~i – punderset~j – qunderset~k`
`underset ~b` `= overset(->)(OB) = punderset~i + punderset~j – qunderset~k`
`underset ~c` `= overset(->)(OC) = −punderset~i + punderset~j – qunderset~k`
`underset ~d` `= overset(->)(OA) = −punderset~i – punderset~j – qunderset~k`

 

`underset~a + underset~c` `= −2qunderset~k`
`underset ~b + underset ~d` `= −2qunderset~k`

 
`:. underset~a + underset~c = underset~b + underset~d`

Vectors, EXT2 V1 SM-Bank 23

A cube with side length 3 units is pictured below.
 

     
 

  1. Calculate the magnitude of vector `vec(AG)`.  (1 mark)

  2. Find the acute angle between the diagonals `vec(AG)` and `vec(BH)`.  (3 marks)

Show Answers Only
  1. `3 sqrt 3\ text(units)`
  2. `70^@32′`
Show Worked Solution

i.   `A(3, 0 , 0), \ \ G(0, 3, 3)`

  `vec(AG)` `= ((0), (3), (3)) – ((3), (0), (0)) = ((text{−3}), (3), (3))`
  `|\ vec(AG)\ |` `= sqrt (9 + 9 + 9)`
    `= 3 sqrt 3\ text(units)`

 

ii.    `H (3, 3, 3)`
  `vec(BH) = ((3), (3), (3))`
`vec(AG) ⋅ vec(BH)` `= |\ vec(AG)\ | ⋅ |\ vec(BH)\ |\ cos theta`
`((text{−3}), (3), (3)) ⋅ ((3), (3), (3))` `= sqrt (9 + 9 + 9) ⋅ sqrt (9 + 9 + 9) cos theta`
`-9 + 9 + 9` `= 27 cos theta`
`cos theta` `= 1/3`
`theta` `= 70.52…`
  `= 70^@32′`

Vectors, EXT2 V1 SM-Bank 22

`ABCDEFGH` are the vertices of a rectangular prism.
  


 

  1. Show that the internal diagonals of the prism, `AG` and `DF`, intersect.  (2 marks)

  2. Calculate the acute angle, `theta`, between the diagonals, to the nearest minute.  (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `83^@37′`
Show Worked Solution
i.    `A(2, text{−2}, 0),`   `G(text{−2}, 2, 2)`
  `D(2, 2, 0),`   `F (text{−2}, text{−2}, 2)`

 

`text(Midpoint)\ AG = ((1/2 (2 – 2)),(1/2 (text{−2} + 2)),(1/2 (0 + 2))) = ((0), (0), (1))`

`text(Midpoint)\ DF = ((1/2 (2 – 2)),(1/2 (2 – 2)),(1/2 (0 + 2))) = ((0), (0), (1))`
 

`text(S) text(ince midpoints are the same), AG and DF\ text(intersect.)`

 

ii.    `vec(AG) = ((text{−2}), (2), (2)) – ((2), (text{−2}), (0)) = ((text{−4}), (4), (2))`
  `vec(DF) = ((text{−2}), (text{−2}), (2)) – ((2), (2), (0)) = ((text{−4}), (text{−4}), (2))`

 

`vec (AG) ⋅ vec (DF) = |\ vec (AG)\ | ⋅ |\ vec(DF)\ |\  cos theta`

`((text{−4}), (4), (2)) ⋅ ((text{−4}), (text{−4}), (2)) = sqrt 36 sqrt 36 cos theta`

`16 – 16 + 4` `= 36 cos theta`
`cos theta` `= 1/9`
`theta` `= 83.62…`
  `= 83^@37′`