SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

Vectors, EXT2 V1 2025 HSC 15a

The adjacent sides of a parallelogram are represented by the vectors  \(\underset{\sim}{a}=4 \underset{\sim}{i}+3 \underset{\sim}{j}-\underset{\sim}{k}\)  and  \(\underset{\sim}{b}=2 \underset{\sim}{i}-\underset{\sim}{j}+3 \underset{\sim}{k}\).

Show that the area of the parallelogram is \(6 \sqrt{10}\) square units.   (4 marks)

Show Answers Only

\(A=\dfrac{1}{2}\abs{a}\abs{b} \sin \theta\)

\(\underset{\sim}{a}=4\underset{\sim}{i}+3\underset{\sim}{j}-\underset{\sim}{k} \  \Rightarrow \ \abs{\underset{\sim}{a}}=\sqrt{16+9+1}=\sqrt{26}\)

\(\underset{\sim}{b}=2\underset{\sim}{i}-\underset{\sim}{j}+3 \underset{\sim}{k} \ \Rightarrow \ \abs{\underset{\sim}{b}}=\sqrt{4+1+9}=\sqrt{14}\)

\(\underset{\sim}{a} \cdot \underset{\sim}{b}=\left(\begin{array}{c}4 \\ 3 \\ -1\end{array}\right)\left(\begin{array}{c}2 \\ -1 \\ 3\end{array}\right)=8-3-3=2\)

\(\cos \theta=\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\abs{\underset{\sim}{a}}\abs{\underset{\sim}{b}}}=\dfrac{2}{\sqrt{26} \sqrt{14}}=\dfrac{1}{\sqrt{91}}\)
 

\(\sin \theta=\dfrac{\sqrt{90}}{\sqrt{91}}\)

\(\therefore A=\dfrac{1}{2} \times \sqrt{26} \times \sqrt{14} \times \dfrac{\sqrt{90}}{\sqrt{91}}=6 \sqrt{10} \ \text{units}^2\)

Show Worked Solution

\(A=\dfrac{1}{2}\abs{a}\abs{b} \sin \theta\)

\(\underset{\sim}{a}=4\underset{\sim}{i}+3\underset{\sim}{j}-\underset{\sim}{k} \  \Rightarrow \ \abs{\underset{\sim}{a}}=\sqrt{16+9+1}=\sqrt{26}\)

\(\underset{\sim}{b}=2\underset{\sim}{i}-\underset{\sim}{j}+3 \underset{\sim}{k} \ \Rightarrow \ \abs{\underset{\sim}{b}}=\sqrt{4+1+9}=\sqrt{14}\)

\(\underset{\sim}{a} \cdot \underset{\sim}{b}=\left(\begin{array}{c}4 \\ 3 \\ -1\end{array}\right)\left(\begin{array}{c}2 \\ -1 \\ 3\end{array}\right)=8-3-3=2\)

\(\cos \theta=\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\abs{\underset{\sim}{a}}\abs{\underset{\sim}{b}}}=\dfrac{2}{\sqrt{26} \sqrt{14}}=\dfrac{1}{\sqrt{91}}\)
 

\(\sin \theta=\dfrac{\sqrt{90}}{\sqrt{91}}\)

\(\therefore A=\dfrac{1}{2} \times \sqrt{26} \times \sqrt{14} \times \dfrac{\sqrt{90}}{\sqrt{91}}=6 \sqrt{10} \ \text{units}^2\)

Vectors, EXT2 V1 2024 HSC 16a

Consider the function  \(y=\cos (k x)\), where  \(k>0\). The value of \(k\) has been chosen so that a circle can be drawn, centred at the origin, which has exactly two points of intersection with the graph of the function and so that the circle is never above the graph of the function. The point  \(P(a, b)\)  is the point of intersection in the first quadrant, so  \(a>0\)  and  \(b>0\),  as shown in the diagram.

The vector joining the origin to the point \(P(a, b)\) is perpendicular to the tangent to the graph of the function at that point. (Do NOT prove this.)

Show that  \(k>1\).   (4 marks)

Show Answers Only

\(y=\cos (k x) \ \Rightarrow \ \dfrac{dy}{dx}=-k \, \sin (k x)\)

\(P(a, b)=P(a, \cos (k a))\)

\(\text{Let}\ \  \underset{\sim}{p}=\overrightarrow{OP}.\)

\(\text{At}\ \ x=a, \ m_{\text{tang}} \perp \underset{\sim}{p} :\)

  \(m_{\text{tang}}\) \(=-k \, \sin (ka)\)
  \(m_{\overrightarrow{OP}}\) \(=\dfrac{\cos(ka)}{a}\)

 
\(-k \, \sin(ka) \times \dfrac{\cos(ka)}{a}=-1\)

  \(k\) \(=\dfrac{a}{\sin(ka) \cos(ka)}\)
    \(=\dfrac{2a}{\sin(2ka)}\)

 

\(\text{For} \ \ \theta>0, \ \sin \theta<\theta\)

\(\Rightarrow k=\dfrac{2a}{\sin (2ka)}>\dfrac{2a}{2ka}>1\).

Show Worked Solution

\(y=\cos (k x) \ \Rightarrow \ \dfrac{dy}{dx}=-k \, \sin (k x)\)

\(P(a, b)=P(a, \cos (k a))\)

\(\text{Let}\ \  \underset{\sim}{p}=\overrightarrow{OP}.\)

\(\text{At}\ \ x=a, \ m_{\text{tang}} \perp \underset{\sim}{p} :\)

  \(m_{\text{tang}}\) \(=-k \, \sin (ka)\)
  \(m_{\overrightarrow{OP}}\) \(=\dfrac{\cos(ka)}{a}\)
♦♦♦ Mean mark 8%.

\(-k \, \sin(ka) \times \dfrac{\cos(ka)}{a}=-1\)

  \(k\) \(=\dfrac{a}{\sin(ka) \cos(ka)}\)
    \(=\dfrac{2a}{\sin(2ka)}\)

 

\(\text{For} \ \ \theta>0, \ \sin \theta<\theta\)

\(\Rightarrow k=\dfrac{2a}{\sin (2ka)}>\dfrac{2a}{2ka}>1\).

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)

Show Answers Only

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 2023 HSC 11d

The quadrilaterals \(A B C D\) and \(A B E F\) are parallelograms.

By considering \(\overrightarrow{A B}\), show that \(C D F E\) is also a parallelogram.  (2 marks)

Show Answers Only

\(\text{Proof (See Worked Solutions)} \)

Show Worked Solution

\(\text{Parrallelogram}\ \Rightarrow\ \text{Show opposite sides are equal} \)

\( \Big{|}\overrightarrow{A B} \Big{|} = \Big{|}\overrightarrow{CD} \Big{|} \ \ (ABCD\ \text{is a parallelogram}) \)

\( \Big{|}\overrightarrow{A B} \Big{|} = \Big{|}\overrightarrow{EF} \Big{|} \ \ (ABEF\ \text{is a parallelogram}) \)

\( \Rightarrow \Big{|}\overrightarrow{CD} \Big{|} = \Big{|}\overrightarrow{EF} \Big{|} \)
 

\( \Big{|}\overrightarrow{CE} \Big{|} = \Big{|}\overrightarrow{BE} \Big{|}-\Big{|}\overrightarrow{BC} \Big{|} \)

\(\text{Similarly,} \)

\( \Big{|}\overrightarrow{DF} \Big{|} = \Big{|}\overrightarrow{AF} \Big{|}-\Big{|}\overrightarrow{AD} \Big{|} \)

\( \text{Since}\ \ \Big{|}\overrightarrow{BE} \Big{|} = \Big{|}\overrightarrow{AF} \Big{|}\ \ \text{and}\ \ \Big{|}\overrightarrow{BC} \Big{|} = \Big{|}\overrightarrow{AD} \Big{|}\)

\( (ABCD\ \text{and}\ ABEF\ \text{are parallelograms}) \)

\( \Rightarrow\ \Big{|}\overrightarrow{DF} \Big{|} = \Big{|}\overrightarrow{CE} \Big{|} \)

\(\therefore CDFE\ \text{is a parallelogram} \)

Vectors, EXT2 V1 EQ-Bank 11

Point `O` is the circumcentre of triangle `ABC` which is the centre of the circle that passes through each of its vertices.

Point `P` is the centroid of triangle `ABC` where the bisectors of each angle within the triangle intersect.

Point `Q` is such that `vec(OQ)=3vec(OP)`.
 

Prove that  `vec(CQ) ⊥ vec(AB)`   (5 marks)

Show Answers Only

`text{See Worked Solution}` 

Show Worked Solution

`vec(OP)=(vec(OA)+vec(OB)+vec(OC))/3`

`vec(OQ)` `=3\ vec(OP)`  
  `=3((vec(OA)+vec(OB)+vec(OC))/3)`  
  `=vec(OA)+vec(OB)+vec(OC)`  
  `=((x_1+x_2+x_3),(y_1+y_2+y_3))`  

 

`vec(CQ)` `=vec(OQ)-vec(OC)`  
  `=((x_1+x_2+x_3),(y_1+y_2+y_3))-((x_3),(y_3))`  
  `=((x_1+x_2),(y_1+y_2))`  

 

`vec(AB)` `=vec(OB)-vec(OA)`  
  `=((x_2),(y_2))-((x_1),(y_1))`  
  `=((x_2-x_1),(y_2-y_1))`  

 

`text{Test}\ \ vec(CQ)*vec(AB)=0:`

`text{LHS}` `=((x_1+x_2),(y_1+y_2))((x_2-x_1),(y_2-y_1))`  
  `=(((x_1+x_2)(x_2-x_1)),((y_1+y_2)(y_2-y_1)))`  
  `=(x_2)^2-(x_1)^2+(y_2)^2-(y_1)^2`  
  `=(x_2)^2+(y_2)^2-((x_1)^2+(y_1)^2)`  
  `=abs(vec(OB))^2-abs(vec(OA))^2`  
  `=OB^2-OA^2\ \ \ text{(radii)}`  
  `=0`  
 
`:.vec(CQ) ⊥ vec(AB)\ \ text{… as required}` 

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)

Show Answers Only
  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)

Show Answers Only
  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)

Show Answers Only
  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)

Show Answers Only

`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.)`

Vectors, EXT2 V1 SM-Bank 20

`OABD`  is a trapezium in which  `overset(->)(OA) = underset~a`  and  `overset(->)(OB) = underset~b`.

`OC`  is parallel to  `AB`  and  `DC : CB = 1:2`
 


 

Using vectors, express  `overset(->)(DA)`  in terms of  `underset~a`  and  `underset~b`.  (3 marks)

Show Answers Only

`text(See Worked Solution)`

Show Worked Solution

`overset(->)(DA) = overset(->)(DB) + overset(->)(BA)`

`overset(->)(BA) = underset~a – underset~b`
 
`text(S) text(ince) \  OC  ||  AB \ \ text(and) \ \ OA  ||  CB`

`=> OABC \ text(is a parallelogram)`

`overset(->)(OA)` `= overset(->)(CB) = underset~a`
`overset(->)(DC)` `= (1)/(2) \ underset~a `
`overset(->)(DB)` `= overset(->)(DC) + overset(->)(CB)`
  `= (3)/(2) \ underset~a`

 

`:. \ overset(->)(DA)` `= (3)/(2) \ underset~a + (underset~a – underset~b)`
  `= (5)/(2) \ underset~a – underset~b`

Vectors, EXT2 V1 2015 VCE 1

Consider the rhombus  `OABC`  shown below, where  `vec (OA) = a underset ~i`  and  `vec (OC) = underset ~i + underset ~j + underset ~k`, and `a` is a positive real constant.
 

VCAA 2015 spec 1a
 

  1. Find  `a.`  (1 mark)

  2. Show that the diagonals of the rhombus  `OABC`  are perpendicular.  (2 marks)

Show Answers Only
  1. `sqrt 3`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution

i.   `|\ vec(OA)\ | = |\ vec(OC)\ |,`

`:. a` `= sqrt (1^2 + 1^2 + 1^2)`
  `= sqrt 3`

 

ii.   `overset(->)(CB) = overset(->)(OA), \ \ overset(->)(AB) = overset(->)(OC)`

MARKER’S COMMENT: Vector notation was poor in many answers.

`overset(->)(OB) = overset(->)(OC) + overset(->)(CB)`

`= underset~i + underset~j + underset~k + sqrt3 underset~i`

`= (sqrt3 + 1)underset~i + underset~j + underset~k`
 

`overset(->)(AC) = overset(->)(AO) + overset(->)(OC)`

`= −sqrt3underset~i + underset~i + underset~j + underset~k`

`= (1 – sqrt3)underset~i + underset~j + underset~k`
 

`overset(->)(AC) · overset(->)(OB)` `= (1 + sqrt3)(1 – sqrt3) + 1 + 1`
  `= 1 – 3 + 1 + 1`
  `= 0`

 
`:. overset(->)(AC) ⊥ overset(->)(OB)`