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Vectors, SPEC1 2022 VCAA 6b

`OPQ` is a semicircle of radius `a` with equation  `y=sqrt(a^(2)-(x-a)^(2))`. `P(x,y)` is a point on the semicircle `OPQ`, as shown below.

  1. Express the vectors `vec(OP)` and `vec(QP)` in terms of  `a`, `x`, `y`, `underset~i` and `underset~j`, where `underset~i` is a unit vector in the direction of the positive `x`-axis and `underset~j` is a unit vector in the direction of the positive `y`-axis.   (1 mark)

  2. Hence, using the vector scalar (dot) product, determine whether `vec(OP)` is perpendicular to `vec(QP)`.   (3 marks)

Show Answers Only

i.    `vec(OP)= xunderset~i + sqrt(a^2-(x-a)^2 underset~j`

`vec(QP)= (x-2a) underset~i + sqrt(a^2-(x-a)^2) underset~j`
 

ii.   `vec(OP)* vec(QP)` `=x(x-2a)+a^(2)-(x-a)^(2)`
  `=x^(2)-2ax+a^(2)-x^(2)+2ax-a^(2)`
  `= 0`

 
`:. vec(OP)\ \text{and}\ vec(QP)\ \text{are perpendicular.}`

Show Worked Solution

i.    `vec(OP)= xunderset~i + sqrt(a^2-(x-a)^2 underset~j`

`vec(QP)= (x-2a) underset~i + sqrt(a^2-(x-a)^2) underset~j`
 

ii.   `vec(OP)* vec(QP)` `=x(x-2a)+a^(2)-(x-a)^(2)`
  `=x^(2)-2ax+a^(2)-x^(2)+2ax-a^(2)`
  `= 0`

 
`:. vec(OP)\ \text{and}\ vec(QP)\ \text{are perpendicular.}`

Vectors, SPEC2 2023 VCAA 5

The points with coordinates \(A(1,1,2), B(1,2,3)\) and \(C(3,2,4)\) all lie in a plane \(\Pi\).

  1. Find the vectors \(\overrightarrow{A B}\) and \(\overrightarrow{A C}\), and hence show that the area of triangle \(A B C\) is 1.5 square units.  (2 marks)

  2. Find the shortest distance from point \(B\) to the line segment \(A C\).  (2 marks)

A second plane, \(\psi\), has the Cartesian equation  \(2 x-2 y-z=-18\).

  1. At what acute angle does the line given by \(\underset{\sim}{ r }(t)=3 \underset{\sim}{ i }+2 \underset{\sim}{ j }+4 \underset{\sim}{ k }+t(\underset{\sim}{ i }-2 \underset{\sim}{ j }+2\underset{\sim}{ k }), t \in R\), intersect the plane \(\psi\) ? Give your answer in degrees correct to the nearest degree.  (2 marks)

A line \(L\) passes through the origin and is normal to the plane \(\psi\). The line \(L\) intersects \(\psi\) at a point \(D\).

  1. Write down an equation of the line \(L\) in parametric form.  (1 mark)

  2. Find the shortest distance from the origin to the plane \(\psi\).  (2 marks)

  3. Find the coordinates of point \(D\).  (2 marks)

Show Answers Only

a.     \(\overrightarrow{A B}  =\overrightarrow{O B}-\overrightarrow{O A}=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)-\left(\begin{array}{l}1 \\ 1 \\ 2\end{array}\right)=\left(\begin{array}{l}0 \\ 1 \\ 1\end{array}\right)\)
  \(\overrightarrow{A C}  =\overrightarrow{O C}-\overrightarrow{O A}=\left(\begin{array}{l}3 \\ 2 \\ 4\end{array}\right)-\left(\begin{array}{l}1 \\ 1 \\ 2\end{array}\right)=\left(\begin{array}{l}2 \\ 1 \\ 2\end{array}\right)\)

\(\text{Area}=\dfrac{1}{2}\ \bigg|\overrightarrow{AB} \times \overrightarrow{AC}\bigg|=\dfrac{1}{2}\ \left|\begin{array}{ccc}\underset{\sim}{i} & \underset{\sim}{j} & \underset{\sim}{k} \\ 0 & 1 & 1 \\ 2 & 1 & 2\end{array}\right|=\dfrac{1}{2}\ \bigg|\underset{\sim}{i}+2 \underset{\sim}{j}-2 \underset{\sim}{k}\bigg|=\dfrac{3}{2}\)

b.    \(\text {Shortest distance }=1 \text { unit }\)

c.     \(26^{\circ}\)

d.    \(\underset{\sim}{r}(t)=\underset{\sim}{n} \times t=2 t \underset{\sim}{i}-2 t \underset{\sim}{j}-t \underset{\sim}{k}\)

\(x=2 t, y=-2 t, z=-t \quad \text{(also accepted)}\)

e.    \(\text {Shortest distance }=6\)

f.    \(D(-4,4,2)\)

Show Worked Solution

a.     \(\overrightarrow{A B}  =\overrightarrow{O B}-\overrightarrow{O A}=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)-\left(\begin{array}{l}1 \\ 1 \\ 2\end{array}\right)=\left(\begin{array}{l}0 \\ 1 \\ 1\end{array}\right)\)
  \(\overrightarrow{A C}  =\overrightarrow{O C}-\overrightarrow{O A}=\left(\begin{array}{l}3 \\ 2 \\ 4\end{array}\right)-\left(\begin{array}{l}1 \\ 1 \\ 2\end{array}\right)=\left(\begin{array}{l}2 \\ 1 \\ 2\end{array}\right)\)

\(\text{Area}=\dfrac{1}{2}\ \bigg|\overrightarrow{AB} \times \overrightarrow{AC}\bigg|=\dfrac{1}{2}\ \left|\begin{array}{ccc}\underset{\sim}{i} & \underset{\sim}{j} & \underset{\sim}{k} \\ 0 & 1 & 1 \\ 2 & 1 & 2\end{array}\right|=\dfrac{1}{2}\ \bigg|\underset{\sim}{i}+2 \underset{\sim}{j}-2 \underset{\sim}{k}\bigg|=\dfrac{3}{2}\)

 
b.    \(\text {In } \triangle ABC \text {, shortest distance of } B \text { to } A C\ \text {is the} \perp \text {distance }\)

\(|\overrightarrow{A C}|= \displaystyle{\sqrt{2^2+1^2+2^2}}=3\)

\(\dfrac{3}{2}=\dfrac{1}{2}\ |\overrightarrow{A C}| \times h \ \Rightarrow \ h=1\)

\(\therefore \text { Shortest distance }=1 \text { unit }\)
 

♦♦ Mean mark (b) 35%.

c.    \(\text {Vector} \perp \text {to plane}\ \psi\  \text {is}\ \ \underset{\sim}{n}=2 \underset{\sim}{i}-2 \underset{\sim}{j}-\underset{\sim}{k}\)

\(\text{Parallel line to}\ \underset{\sim}{r}(t) \ \text{is}\ \ \underset{\sim}{m}=\underset{\sim}{i}-2 \underset{\sim}{j}+2\underset{\sim}{k}\)

\(\text{Find angle } \alpha \text{ between }\underset{\sim}{n} \text{ and } \underset{\sim}{m},\)

\(\text{Solve for } \alpha:\)

\((2 \underset{\sim}{i}-2 \underset{\sim}{j}-\underset{\sim}{k})(\underset{\sim}{i}-2 \underset{\sim}{j}+\underset{\sim}{2 k})=3 \times 3 \times \cos \, \alpha\)

\(\Rightarrow \alpha=64^{\circ} \text { (nearest degree) }\)
 

\(\text{Find angle } \theta \text{ between } \underset{\sim}{m} \text{ and plane } \psi :\)

\(\theta=90-64=26^{\circ}\)
 

d.    \(\underset{\sim}{r}(t)=\underset{\sim}{n} \times t=2 t \underset{\sim}{i}-2 t \underset{\sim}{j}-t \underset{\sim}{k}\)

\(x=2 t, y=-2 t, z=-t \quad \text{(also accepted)}\)
 

♦ Mean mark (d) 52%.

e.    \(|\overrightarrow{OD}|=\text{shortest distance from } O \text{ to plane } \psi\).

\(\Rightarrow D \text{ is on } L \text{ and } \psi\)

\(\text{Solve for } t: \  2(2 t)-2(-2 t)-(t)=-18\)

\(\Rightarrow t=-2\)

\(|\overrightarrow{OD}|=-4 \underset{\sim}{i}+4 \underset{\sim}{j}+2 \underset{\sim}{k}\)

\(\text {Shortest distance }=\displaystyle{\sqrt{(-4)^2+4^2+2^2}}=6\)
 

f.    \(\overrightarrow{OD}=-4 \underset{\sim}{i}+4 \underset{\sim}{j}+2 \underset{\sim}{i}\)

\(D(-4,4,2)\)

♦ Mean mark (f) 41%.

Vectors, SPEC2 2023 VCAA 15 MC

If the sum of two unit vectors is a unit vector, then the magnitude of the difference of the two vectors is

  1. \(0\)
  2. \(\dfrac{1}{\sqrt{2}}\)
  3. \(\sqrt{2}\)
  4. \(\sqrt{3}\)
  5. \(\sqrt{5}\)
Show Answers Only

\(D\)

Show Worked Solution

\(\text{Vectors can be drawn as two sides of an equilateral triangle.}\)

\(\text{Using the cosine rule for the difference between the two vectors:}\)

\(c^2\) \(=a^2+b^2-2ab\, \cos C\)  
  \(=1+1-2\times -\dfrac{1}{2} \)  
  \(=3\)  
\(c\) \(=\sqrt{3}\)  

 
\(\Rightarrow D\)

Vectors, SPEC1 2023 VCAA 9

A plane contains the points \( A(1,3,-2), B(-1,-2,4)\) and \( C(a,-1,5)\), where \(a\) is a real constant. The plane has a \(y\)-axis intercept of 2 at the point \(D\).

  1. Write down the coordinates of point \(D\).   (1 mark)
  1. Show that \(\overrightarrow{A B}\) and \(\overrightarrow{A D}\) are \(-2 \underset{\sim}{\text{i}}-5 \underset{\sim}{\text{j}}+6 \underset{\sim}{\text{k}}\)  and  \(-\underset{\sim}{\text{i}}-\underset{\sim}{\text{j}}+2 \underset{\sim}{\text{k}}\), respectively.   (1 mark)
  1. Hence find the equation of the plane in Cartesian form.  (2 marks)
  1.  Find \(a\).   (1 mark)
  1.  \(\overline{A B}\) and \(\overline{A D}\) are adjacent sides of a parallelogram. Find the area of this parallelogram.   (1 mark)
Show Answers Only

a.    \(D(0,2,0)\)

b.    \(\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}=\left(\begin{array}{c}
-1 \\
-2 \\
4
\end{array}\right)-\left(\begin{array}{c}
1 \\
3 \\
-2
\end{array}\right)=\left(\begin{array}{c}
-2 \\
-5 \\
6
\end{array}\right)=-2 \underset{\sim}{\text{i}}-5\underset{\sim}{\text{j}}+6 \underset{\sim}{\text{k}}\)

\(\overrightarrow{A D}=\overrightarrow{O D}-\overrightarrow{O A}=\left(\begin{array}{c}0 \\ 2 \\ 0\end{array}\right)-\left(\begin{array}{c}1 \\ 3 \\ -2\end{array}\right)=\left(\begin{array}{c}-1 \\ -1 \\ 2\end{array}\right)=-\underset{\sim}{\text{i}}-\underset{\sim}{\text{j}}+2\underset{\sim}{\text{k}}\)

c.    \(4 x+2 y+3 z =4\)

d.    \(a=-\dfrac{9}{4}\)

e.    \(A=\sqrt{29}\)

Show Worked Solution

a.    \(D(0,2,0)\)

 
b.    \(\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}=\left(\begin{array}{c}
-1 \\
-2 \\
4
\end{array}\right)-\left(\begin{array}{c}
1 \\
3 \\
-2
\end{array}\right)=\left(\begin{array}{c}
-2 \\
-5 \\
6
\end{array}\right)=-2 \underset{\sim}{\text{i}}-5\underset{\sim}{\text{j}}+6 \underset{\sim}{\text{k}}\)

\(\overrightarrow{A D}=\overrightarrow{O D}-\overrightarrow{O A}=\left(\begin{array}{c}0 \\ 2 \\ 0\end{array}\right)-\left(\begin{array}{c}1 \\ 3 \\ -2\end{array}\right)=\left(\begin{array}{c}-1 \\ -1 \\ 2\end{array}\right)=-\underset{\sim}{\text{i}}-\underset{\sim}{\text{j}}+2\underset{\sim}{\text{k}}\)

 
c.    \(\overrightarrow{A B} \times \overrightarrow{A D}=\left|\begin{array}{ccc}
\underset{\sim}{\text{i}} & \underset{\sim}{\text{j}} & \underset{\sim}{\text{k}} \\ -2 & -5 & 6 \\ -1 & -1 & 2\end{array}\right|\)

\(\ \quad \quad \quad \quad \   \begin{aligned} & =(-10+6)\underset{\sim}{\text{i}}-(-4+6)\underset{\sim}{\text{j}}+(2-5) \underset{\sim}{\text{k}} \\ & =-4 \underset{\sim}{i}-2 \underset{\sim}{j}-3 \underset{\sim}{k}\end{aligned}\)

\(\text{Plane:}\ \ -4 x-2 y-3 z=k\)

\(\text{Substitute}\  D(0,2,0)\  \text{into equation}\ \ \Rightarrow \ \text{k}=-4\)

\begin{aligned}
\therefore-4 x-2 y-3 z & =-4 \\
4 x+2 y+3 z & =4
\end{aligned}

 
d.    \(\text{Find}\ a\ \Rightarrow \ \text{substitute}\ (a, 1,-5)\ \text{into plane equation:}\)

\begin{aligned}
-4 & =-4a+2-15 \\
4 a & =-9 \\
a & =-\dfrac{9}{4}
\end{aligned}

e.     \(\text{Area}\) \(=|\overrightarrow{A B} \times \overrightarrow{A D}|\)
    \(=|-4\underset{\sim}{i}-2\underset{\sim}{j}-3\underset{\sim}{k}|\)
    \(=\sqrt{16+4+9} \)
    \(=\sqrt{29}\)

Vectors, SPEC1-NHT 2019 VCAA 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

a.     `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)`

 

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

Vectors, SPEC2 2009 VCAA 17 MC

Vectors `underset ~a, underset ~b` and `underset ~c` are shown below.
 

VCAA 2009 spec2 17mc
 

From the diagram it follows that

A.   `|\ underset ~c\ |^2 = |\ underset ~a\ |^2 + |\ underset ~b\ |^2`

B.   `|\ underset ~c\ |^2 = |\ underset ~a\ |^2 + |\ underset ~b\ |^2 - |\ underset ~a\ | |\ underset ~b\ |`

C.   `|\ underset ~c\ |^2 = |\ underset ~a\ |^2 + |\ underset ~b\ |^2 + |\ underset ~a * underset ~b\ |`

D.   `|\ underset ~c\ |^2 = |\ underset ~a\ |^2 + |\ underset ~b\ |^2 + |\ underset ~a\ | |\ underset ~b\ |`

E.   `|\ underset ~c\ |^2 = |\ underset ~a\ |^2 + |\ underset ~b\ |^2 - |\ underset ~a * underset ~b\ |`

Show Answers Only

`D`

Show Worked Solution

`underset~ c + underset~ b = underset~ a\ \ => \ underset~ c = underset~ a – underset~ b`

♦♦ Mean mark 30%.

`:. underset~ c * underset~ c` `= (underset~ a – underset~ b) * (underset~ a – underset~ b)`
  `= underset~ a * underset~ a – underset~ a * underset~ b -underset~b*underset~a + underset~ b * underset~ b`
  `= underset~ a * underset~ a  + underset~ b * underset~ b-2underset~b*underset~a `

`:. |\ underset~ c\ |^2 = |\ underset~ a\ |^2 + |\ underset~ b\ |^2 – 2 |\ underset~ a\ | |\ underset~ b\ | cos 120^@`

`:. |\ underset~ c\ |^2 = |\ underset~ a\ |^2 + |\underset~ b\ |^2 + |\ underset~ a\ | |\ underset~ b\ |`

`=>   D`

Vectors, SPEC2 2011 VCAA 10 MC

The diagram below shows a rhombus, spanned by the two vectors  `underset~a`  and  `underset~b`.
 

SPEC2 2011 VCAA 10 MC
 

It follows that

A.   `underset~a.underset~b = 0`

B.   `underset~a = underset~b`

C.   `(underset~a + underset~b).(underset~a - underset~b) = 0`

D.   `|\ underset~a + underset~b\ | = |\ underset~a - underset~b|`

E.   `2underset~a + 2underset~b = underset~0`

Show Answers Only

`C`

Show Worked Solution

`text(Consider A:)`

`text(If)\ \ underset~a · underset~b = 0\ \ =>\ \ underset~a ⊥ underset~b\ \ text{(incorrect)}`

`text(Consider B:)`

`underset~a != underset~b\ \ text{(incorrect)}`

`text(Consider C:)`

`underset~a + underset~b` `= overset(->)(OC)`
`underset~a – underset~b` `= overset(->)(BA)`

 
`overset(->)(OC) · overset(->)(BA)=0`

`text{The diagonals of a rhombus are perpendicular  (correct)}`

`=> C`

Vectors, SPEC1 2013 VCAA 3

The coordinates of three points are  `A (– 1, 2, 4), \ B(1, 0, 5) and C(3, 5, 2).`

  1. Find  `vec (AB)`  (1 mark)

  2. The points `A, B` and `C` are the vertices of a triangle.
  3. Prove that the triangle has a right angle at `A.`  (2 marks)

  4. Find the length of the hypotenuse of the triangle.  (1 mark)

Show Answers Only

  1. `2underset~i – 2underset~j + underset~k`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `sqrt 38`

Show Worked Solution

a.   `vec(AB)` `=(1 – −1)underset~i + (0 – 2)underset~j + (5 – 4)underset~k`
    `= 2underset~i – 2underset~j + underset~k`

 

b.    `overset(->)(AC)` `= (3 – −1)underset~i + (5 – 2)underset~j + (2 – 4)underset~k`
    `= 4underset~i + 3underset~j – 2underset~k`

 

`overset(->)(AB) · overset(->)(AC)` `= 2 xx 4 + (−2) xx 3 + 1 xx (−2)`
  `= 8 – 6 – 2`
  `= 0`

 
`=>  overset(->)(AB) ⊥ overset(->)(AC)`

`:. DeltaABC\ text(has a right angle at)\ A.`
 

c.    `overset(->)(BC)` `= (3 – 1)underset~i + (5 – 0)underset~j + (2 – 5)underset~k`
    `= 2underset~i + 5underset~j – 3underset~k`

 

`|overset(->)(BC)|` `= sqrt(2^2 + 5^2 + (−3)^2)`
  `= sqrt(4 + 25 + 9)`
  `= sqrt38`

Vectors, SPEC1 2015 VCAA 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

a.   `|\ vec(OA)\ | = |\ vec(OC)\ |,`

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

 

b.   `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)`

Vectors, SPEC2-NHT 2018 VCAA 12 MC


 

In the diagram above, `LOM` is a diameter of the circle with centre `O`.

`N` is a point on the circumference of the circle.

If  `underset~r = vec(ON)`  and  `underset ~s = vec(MN)`, then  `vec(LN)`  is equal to

  1. `2 underset ~r - 2 underset ~s`
  2. `underset ~r - 2 underset ~s`
  3. `underset ~r + 2 underset ~s`
  4. `2 underset ~r + underset ~s`
  5. `2 underset ~r - underset ~s`
Show Answers Only

`E`

Show Worked Solution

`vec(LN)` `= vec(LM) + vec(MN)`
`vec(LN)` `= vec(LO) + vec(ON)`
  `= 1/2\ vec(LM) + vec(ON)`

 

`:. vec(LM) + underset~s` `= 1/2\ vec(LM) + underset~r`
`1/2\ vec(LM)` `= underset ~r – underset~s`
`vec(LM)` `= 2 underset~r – 2 underset~s`

 

`:. vec(LN)` `= 2 underset~r – 2 underset~s + underset~s`
  `= 2 underset~r –  underset~s`

 
`=>  E`