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Vectors, EXT2 V1 2024 HSC 14e

The diagram shows triangle \(O Q A\).

The point \(P\) lies on \(O A\) so that  \(O P: O A=3: 5\).

The point \(B\) lies on \(O Q\) so that  \(O B: O Q=1: 3\).

The point \(R\) is the intersection of \(A B\) and \(P Q\).

The point \(T\) is chosen on \(A Q\) so that \(O, R\) and \(T\) are collinear.
 

Let  \(\underset{\sim}{a}=\overrightarrow{O A}, \ \underset{\sim}{b}=\overrightarrow{O B}\)  and  \(\overrightarrow{P R}=k \overrightarrow{P Q}\)  where \(k\) is a real number.

  1. Show that  \(\overrightarrow{O R}=\dfrac{3}{5}(1-k) \underset{\sim}{a}+3 k \underset{\sim}{b}\).   (2 marks)

Writing  \(\overrightarrow{A R}=h \overrightarrow{A B}\), where \(h\) is a real number, it can be shown that  \(\overrightarrow{O R}=(1-h) \underset{\sim}{a}+h \underset{\sim}{b}\).  (Do NOT prove this.)

  1. Show that  \(k=\dfrac{1}{6}\).   (2 marks)

  2. Find \(\overrightarrow{O T}\) in terms of \(\underset{\sim}{a}\) and \(\underset{\sim}{b}\).   (2 marks)

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i.    \(\underset{\sim}{a}=\overrightarrow{O A}, \ \underset{\sim}{b}=\overrightarrow{O B}, \ \overrightarrow{P R}=k \overrightarrow{P Q}\)

\(\overrightarrow{O P}: \overrightarrow{O A}=3: 5 \ \Rightarrow \ \overrightarrow{O P} = \dfrac{3}{5} \times \overrightarrow{O A} \)

\(\overrightarrow{O B}: \overrightarrow{O Q}=1: 3\ \Rightarrow \ \overrightarrow{O Q} = 3 \times \overrightarrow{OB} \)

\(\text {Show } \overrightarrow{O R}=\dfrac{3}{5}(1-k) \underset{\sim}{a}+3 k \underset{\sim}{b}\)

  \(\overrightarrow{O R}\) \(=\overrightarrow{O P}+k \overrightarrow{P Q}\)
    \(=\overrightarrow{O P}+k(\overrightarrow{O Q}-\overrightarrow{O P})\)
    \(=(1-k) \overrightarrow{O P}+k \overrightarrow{O Q}\)
    \(=\dfrac{3}{5}(1-k)\underset{\sim}{a}+3 k \underset{\sim}{b}\)

 

ii.    \(\overrightarrow{A R}=h \overrightarrow{A B}, \quad h \in \mathbb{R}\)

\(\overrightarrow{OR}=(1-h) \underset{\sim}{a}+h \underset{\sim}{b} \ \ \text{(given)}\)

\(\overrightarrow{O R}=\dfrac{3}{5}(1-k) \underset{\sim}{a}+3 k \underset{\sim}{b} \ \ \text{(part(i))}\)

\(\text{Vector }\underset{\sim}{a} \neq \lambda \underset{\sim}{b}\ (\lambda \in \mathbb{R})\ \Rightarrow \ \text{linearly independent and are basis vectors for } \overrightarrow{O R}\).

\(\text {Equating coefficients:}\)

   \(\dfrac{3}{5}(1-k)=1-h\ \ldots\ (1)\)

   \(h=3 k\ \ldots\ (2)\)

\(\text{Substituting (2) into (1)}\)

     \(\dfrac{3}{5}(1-k)\) \(=1-3 k\)
  \(3-3 k\) \(=5-15 k\)
  \(k\) \(=\dfrac{1}{6}\)

 
iii.    \(\overrightarrow{O T}=\dfrac{3}{4}(\underset{\sim}{a}+\underset{\sim}{b})\)

Show Worked Solution

i.    \(\underset{\sim}{a}=\overrightarrow{O A}, \ \underset{\sim}{b}=\overrightarrow{O B}, \ \overrightarrow{P R}=k \overrightarrow{P Q}\)

\(\overrightarrow{O P}: \overrightarrow{O A}=3: 5 \ \Rightarrow \ \overrightarrow{O P} = \dfrac{3}{5} \times \overrightarrow{O A} \)

\(\overrightarrow{O B}: \overrightarrow{O Q}=1: 3\ \Rightarrow \ \overrightarrow{O Q} = 3 \times \overrightarrow{OB} \)

\(\text {Show } \overrightarrow{O R}=\dfrac{3}{5}(1-k) \underset{\sim}{a}+3 k \underset{\sim}{b}\)

  \(\overrightarrow{O R}\) \(=\overrightarrow{O P}+k \overrightarrow{P Q}\)
    \(=\overrightarrow{O P}+k(\overrightarrow{O Q}-\overrightarrow{O P})\)
    \(=(1-k) \overrightarrow{O P}+k \overrightarrow{O Q}\)
    \(=\dfrac{3}{5}(1-k)\underset{\sim}{a}+3 k \underset{\sim}{b}\)

  

ii.    \(\overrightarrow{A R}=h \overrightarrow{A B}, \quad h \in \mathbb{R}\)

\(\overrightarrow{OR}=(1-h) \underset{\sim}{a}+h \underset{\sim}{b} \ \ \text{(given)}\)

\(\overrightarrow{O R}=\dfrac{3}{5}(1-k) \underset{\sim}{a}+3 k \underset{\sim}{b} \ \ \text{(part(i))}\)

\(\text{Vector }\underset{\sim}{a} \neq \lambda \underset{\sim}{b}\ (\lambda \in \mathbb{R})\ \Rightarrow \ \text{linearly independent and are basis vectors for } \overrightarrow{O R}\).

\(\text {Equating coefficients:}\)

   \(\dfrac{3}{5}(1-k)=1-h\ \ldots\ (1)\)

   \(h=3 k\ \ldots\ (2)\)

\(\text{Substituting (2) into (1)}\)

     \(\dfrac{3}{5}(1-k)\) \(=1-3 k\)
  \(3-3 k\) \(=5-15 k\)
  \(k\) \(=\dfrac{1}{6}\)

 

iii.    \(\overrightarrow{O T}=\lambda \overrightarrow{O R}\)

\(\text {Using parts (i) and (ii):}\)

\(\overrightarrow{O R}=\dfrac{3}{5}\left(1-\dfrac{1}{6}\right)\underset{\sim}{a}+3\left(\dfrac{1}{6}\right) \underset{\sim}{b}=\dfrac{1}{2}\left(\underset{\sim}{a}+\underset{\sim}{b}\right)\)

\(\overrightarrow{O T}=\dfrac{\lambda}{2}(\underset{\sim}{a}+\underset{\sim}{b})\)

♦ Mean mark (iii) 45%.
 

\(\text {Find } \lambda:\)

  \(\overrightarrow{O T}\) \(=\overrightarrow{O A}+\mu \overrightarrow{A Q}\)
  \(\dfrac{\lambda}{2}(\underset{\sim}{a}+\underset{\sim}{b})\) \(=\underset{\sim}{a}+\mu(3 \underset{\sim}{b}-\underset{\sim}{a}) \quad \Big(\text{noting}\ \overrightarrow{A Q}=\overrightarrow{O Q}-\overrightarrow{O A}=3 \underset{\sim}{b}-\underset{\sim}{a}\Big)\)
    \(=\underset{\sim}{a}(1-\mu)+3 \mu \underset{\sim}{b}\)

 
\(\text {Equating coefficients:}\)

\(\dfrac{\lambda}{2}=3 \mu \ \Rightarrow \ \mu=\dfrac{\lambda}{6}\)

  \(1-\dfrac{\lambda}{6}\) \(=\dfrac{\lambda}{2}\)
  \(6-\lambda\) \(=3 \lambda\)
  \(\lambda\) \(=\dfrac{3}{2}\)

 
\(\therefore \overrightarrow{O T}=\dfrac{3}{4}(\underset{\sim}{a}+\underset{\sim}{b})\)

Vectors, EXT2 V1 EQ-Bank 10

Given  `underset~a=(3,-2,1)`  and  `underset~b=(4,3,-4)`, verify numerically that 

`underset~a*underset~b=abs(underset~a)abs(underset~b)cos theta=x_1x_2+y_1y_2+z_1z_2`

where  `underset~a=(x_1,y_1,z_1)`  and  `underset~b=(x_2,y_2,z_2)`   (4 marks)

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`text{See Worked Solution}`

Show Worked Solution

`text{Scalar Product result 1:}`

`underset~a*underset~b` `=x_1x_2+y_1y_2+z_1z_2`  
  `=3xx4+(-2)xx3+1xx(-4)`  
  `=2`  

 
`text{Scalar Product result 2:}`

`underset~a*underset~b=abs(underset~a)abs(underset~b)cos theta`

`text{Let}\ \ underset~c=underset~b-underset~a`

`underset~c=((4),(3),(-4))-((3),(-2),(1))=((1),(5),(-5))`

`abs(underset~c)=sqrt(1^2+5^2+(-5)^2)=sqrt(51)`

`abs(underset~a)=sqrt(3^2+(-2)^2+1^2)=sqrt(14)`

`abs(underset~b)=sqrt(4^2+3^2+(-4)^2)=sqrt(41)`
 

`text{Using cosine rule:}\ \ c^2=a^2+b^2-2ab\ cosC`

`=>\ \ ab\ cosC=(a^2+b^2-c^2)/2`

`underset~a*underset~b` `=abs(underset~a)abs(underset~b)cos theta`  
  `=(abs(underset~a)^2+abs(underset~b)^2-abs(underset~c)^2)/2`  
  `=(14+41-51)/2`  
  `=2`  

 
`:.underset~a*underset~b=abs(underset~a)abs(underset~b)cos theta=x_1x_2+y_1y_2+z_1z_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)

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`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 EQ-Bank 7

In triangle `ABC`, `M` is the midpoint of `AC` and `N` is the midpoint of `AB`.

Use vector methods to prove that

  1. `MN=1/2CB`   (2 marks)

  2. `MN` is parallel to `CB`   (1 mark)

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  1. `text(See Worked Solution)`
  2. `text(See Worked Solution)`
Show Worked Solution

i.   
     

`vec(AC)+vec(CB)+vec(BA)=0\ \ \ text{(resultant vector ends at starting point)}`

`vec(CB)=-(vec(AC)+vec(BA))`
 

`text(S)text(ince)\ M\ text (and)\ N\ text(are midpoints:)`

`vec(MN)` `=-1/2vec(AC)-1/2vec(BA)`  
  `=-1/2(vec(AC)+vec(AB))`  
  `=1/2vec(CB)\ \ text(… as required)`  

 

ii.   `text{If}\ \ vec(u) = kvec(v)\ \ (k\ text{scalar})\ \ =>\ \ vec(u)\ text{||}\ vec(v)`

`vec(MN)=1/2vec(CB)\ \ text{(see (i))`

`:.MN\ text{||}\ RCB`

Vectors, EXT2 V1 2020 HSC 15b

The point `C` divides the interval `AB` so that  `frac{CB}{AC} = frac{m}{n}`. The position vectors of `A` and `B` are `underset~a` and `underset~b` respectively, as shown in the diagram.
 

  1. Show that  `overset->(AC) = frac{n}{m + n} (underset~b - underset~a)`.  (2 marks)

     

  2. Prove that  `overset->(OC) = frac{m}{m + n} underset~a + frac{n}{m + n} underset~b`.  (1 mark)

Let `OPQR` be a parallelogram with  `overset->(OP) = underset~p`  and  `overset->(OR) = underset~r`. The point `S` is the midpoint of `QR` and `T` is the intersection of `PR` and `OS`, as shown in the diagram.
  
 
         
 

  1. Show that  `overset->(OT) = frac{2}{3} underset~r + frac{1}{3} underset~p`.  (3 marks)

  2. Using parts (ii) and (iii), or otherwise, prove that `T` is the point that divides the interval `PR` in the ratio 2 :1.   (1 mark)

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  1. `text{See Worked Solutions}`
  2. `text{See Worked Solutions}`
  3. `text{See Worked Solutions}`
  4. `text{See Worked Solutions}`
Show Worked Solution

i.  

`frac{overset->(AC)}{overset->(AB)}` `= frac{n}{m + n}`
`overset->(AC)` `= frac{n}{m + n} * overset->(AB)`
  `= frac{n}{m + n} (underset~b – underset~a)`

 

ii.    `overset->(OC)` `= overset->(OA) + overset->(AC)`
    `= underset~a + frac{n}{m + n} (underset~b – underset~a)`
    `= underset~a – frac{n}{m + n} underset~a + frac{n}{m + n} underset~b`
    `= (1 – frac{n}{m + n}) underset~a + frac{n}{m + n} underset~b`
    `= (frac{m + n – n}{m + n}) underset~a + frac{n}{m + n} underset~b`
    `= frac{m}{m + n} underset~a + frac{n}{m + n} underset~b`

 

iii.  `text{Show} \ \ overset->(OT) = frac{2}{3} underset~r + frac{1}{3} underset~p`

♦ Mean mark part (iii) 50%.
 

 
`text{Consider} \ \ Delta PTO \ \ text{and} \ \ Delta RTS:`

`angle PTO = angle RTS \ (text{vertically opposite})`

`angle OPT = angle SRT \ (text{vertically opposite})`

`therefore \ Delta PTO \ text{|||} \ Delta RTS\ \ text{(equiangular)}`
 

`OT : TS = OP : SR = 2 : 1`

`(text{corresponding sides in the same ratio})`

`frac{overset->(OT)}{overset->(OS)}` `= frac{2}{3}`
`overset->(OT)` `= frac{2}{3} overset->(OS)`
  `= frac{2}{3} ( underset~r + frac{1}{2} underset~p)`
  `= frac{2}{3} underset~r + frac{1}{3} underset~p`
Mean mark part (iv) 51%.

 

iv.   `text{Let} \ \ overset->(OT) \ text{divide} \ PR\ text{so that}\ \ frac{TR}{PT} = frac{m}{n}`

`text{Using part (ii):}`

`overset->(OT)` `= frac{m}{m + n} underset~p + frac{n}{m + n} c`
`overset->(OT)` `= frac{1}{3} underset~p + frac{2}{3} underset~r \ \ \ (text{part (iii)})`
`frac{m}{m + n}` `= frac{1}{3} , frac{n}{m + n} = frac{2}{3}`

 
`=> \ m = 1 \ , \ n = 2`
 

`therefore \ T \ text{divides} \ PR \ text{in ratio  2 : 1}.`

Vectors, EXT2 V1 2019 SPEC1-N 5

A triangle has vertices  `A(sqrt3 + 1, –2, 4), \ B(1, –2, 3)`  and  `C(2, –2, sqrt3 + 3)`.

  1.  Find angle `ABC`   (3 marks)

  2.  Find the area of the triangle.   (2 marks)

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  1. `∠ABC = (pi)/(6)`
  2. `1 \ text(u²)`
Show Worked Solution

i.     `overset(->)(BA) = ((sqrt3 + 1), (-2), (4)) – ((1), (-2), (3)) = ((sqrt3), (0), (1))`

`overset(->)(BC) = ((2), (-2), (sqrt3 + 3)) – ((1), (-2), (3)) = ((1), (0), (sqrt3))`

`cos ∠ABC` `= (overset(->)(BA) · overset(->)(BC))/ (|overset(->)(BA)| |overset(->)(BC)|`
  `= (2 sqrt3)/(sqrt4 sqrt4)`
  `= (sqrt3)/(2)`

 
`:. \ ∠ABC = (pi)/(6)`

 

ii.     `text(Area)` `= (1)/(2) · |overset(->)(BA)| |overset(->)(BC)| \ sin ∠ABC`
    `= (1)/(2) xx 2 xx 2 xx sin \ (pi)/(6)`
    `= 1 \ text(u²)`

Vectors, EXT2 V1 EQ-Bank 15

Point `B` sits on the arc of a semi-circle with diameter  `AC`.
 


 

Using vectors, show `angle ABC`  is a right angle.  (2 marks)

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`text(See Worked Solution)`

Show Worked Solution

`text(Let)\ \ vec (OA) = underset~a, \ text(and)\ \ vec(OC)=underset~c`
 


  

`text(Prove) \ overset(->)(AB) ⊥ overset(->)(BC)`

`|underset~a| = |underset~b| = |underset~c|\ \ \ text{(radii)}`

`underset~c = – underset~a `
 

`overset(->)(AB) ⋅ overset(->)(BC)` `= (underset~b-underset~a) (underset~c-underset~b)`
  `= underset~b · underset~c-|underset~b|^2-underset~a · underset~c + underset~a · underset~b`
  `= underset~b · underset~c-|underset~b|^2 + underset~c · underset~c-underset~c · underset~b`
  `= |underset~c|^2-|underset~b|^2`
  `= 0`

 
`:. \ ∠ABC \ text(is a right angle.)`