SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

Vectors, SPEC2 2024 VCAA 13 MC

If the angle between the vectors  \(2 \underset{\sim}{ i }-\underset{\sim}{ j }+2 \underset{\sim}{ k }\)  and  \(2 \underset{\sim}{ i }+m \underset{\sim}{ j }+6 \underset{\sim}{ k }\)  is  \(\cos ^{-1}\left(\dfrac{13}{21}\right)\), then the value of \(m\), where  \(m \in R^{+}\), is

  1. \(2\)
  2. \(3\)
  3. \(4\)
  4. \(5\)
Show Answers Only

\(B\)

Show Worked Solution

\(\underset{\sim}{a}=\left(\begin{array}{c}2 \\ -1 \\ 2\end{array}\right)  \Rightarrow \abs{\underset{\sim}{a}}=\sqrt{9}=3\)

\(\underset{\sim}{b}=\left(\begin{array}{c}2 \\ m \\ 6\end{array}\right)  \Rightarrow \abs{\underset{\sim}{a}}=\sqrt{2^2+m^2+6^2}=\sqrt{40+m^2}\)

\(\underset{\sim}{a} \cdot \underset{\sim}{b}\) \(=\abs{\underset{\sim}{a}}\abs{\underset{\sim}{b}} \cos \theta\)  
\(16-m\) \(=3 \sqrt{40+m^2} \times \cos \left(\cos ^{-1}\left(\dfrac{13}{21}\right)\right)\)  

 
\(\text{Solve for \(m\) (by CAS):} \ 16-m=3 \sqrt{40+m^2} \times \dfrac{13}{21}\)

\(m=-\dfrac{241}{15}, 3\)

\(\Rightarrow B\)

Vectors, SPEC1 2024 VCAA 4

Consider the vectors  \(\underset{\sim}{ a }=3 \underset{\sim}{ j }+3 \underset{\sim}{ k }, \ \underset{\sim}{ b }=2 \underset{\sim}{ i }-\underset{\sim}{ j }-2 \underset{\sim}{ k }\)  and  \(\underset{\sim}{ c }=n \underset{\sim}{ i }+2 \underset{\sim}{ j }+\underset{\sim}{ k }\),  where  \(n \in Z\).

  1. Find the angle between \(\underset{\sim}{ a }\) and \(\underset{\sim}{ b }\).   (2 marks)

  2. Find all possible values of \(n\) such that the dot product of \(\underset{\sim}{ a }\) and \(\underset{\sim}{ c }\) is equal to the magnitude of the cross product of \(\underset{\sim}{ a }\) and \(\underset{\sim}{c}\).   (2 marks)

Show Answers Only

a.      \(\theta=\dfrac{3 \pi}{4}\left(\text{or} \ 135^{\circ}\right)\)

b.    \(n= \pm 2\)

Show Worked Solution

a.  \(\underset{\sim}{a}=\left(\begin{array}{l}0 \\ 3 \\ 3\end{array}\right) \Rightarrow \abs{\underset{\sim}{a}}=\sqrt{18}=3 \sqrt{2}\)

     \(\underset{\sim}{b}=\left(\begin{array}{c}2 \\ -1 \\ -2\end{array}\right) \Rightarrow\abs{\underset{\sim}{b}}=\sqrt{9}=3\)

     \(\underset{\sim}{c}=\left(\begin{array}{l}n \\ 2 \\ 1\end{array}\right) \Rightarrow \abs{\underset{\sim}{c}}=\sqrt{n^2+5}\)

     \(\cos \theta=\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\abs{\underset{\sim}{a}} \cdot \abs{\underset{\sim}{b}}}=\dfrac{-3-6}{3 \sqrt{2} \times 3}=-\dfrac{1}{\sqrt{2}}\)

    \(\therefore \theta=\cos ^{-1}\left(-\dfrac{1}{\sqrt{2}}\right)=\dfrac{3 \pi}{4}\left(\text{or} \ 135^{\circ}\right)\)
 

b.  \(\underset{\sim}{a} \times \underset{\sim}{c}=\left|\begin{array}{ccc}\underset{\sim}{i} & \underset{\sim}{j} & \underset{\sim}{k} \\ 0 & 3 & 3 \\ n & 2 & 1\end{array}\right|=-3\underset{\sim}{i}+3 n\underset{\sim}{j}-3 n \underset{\sim}{k}\)

     \(\abs{\underset{\sim}{a} \times \underset{\sim}{c}}=\sqrt{9+9n^2+9n^2}=\sqrt{18 n^2+9}\)

     \(\underset{\sim}{a} \cdot \underset{\sim}{c}=0 \times n + 3 \times 2 + 3 \times 1=9\)

     \(\text{Find \(n\) given:}\)

  \(\sqrt{18 n^2+9}\) \(=9\)
  \(18 n^2+9\) \(=81\)
  \(n^2\) \(=4\)
  \(n\) \(= \pm 2\)

Vectors, SPEC1 2022 VCAA 6a

Find the cosine of the acute angle between the vectors  `underset~a=2underset~i-3underset~j+6underset~k`  and  `underset~b=underset~i+2underset~j+2underset~k`.   (2 marks)

Show Answers Only

`8/21`

Show Worked Solution
`cos \ theta` `=(underset~a*underset~b)/(|underset~a||underset~b|)`  
  `=(2-6+12)/(sqrt(4+9+36)\ sqrt(1+4+4))`  
  `=8/(7 xx 3)`  
  `=8/21`  

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 2021 VCAA 12 MC

Consider the vectors  `underset~a = x underset~i + underset~j, \ underset~b = underset~i - underset~j`  and  `underset~c = underset~i + x underset~j`.

Given that  `theta`  is the angle between  `underset~a`  and  `underset~b`,  and  `phi`  is the angle between  `underset~b`  and  `underset~c, cos(theta) cos (phi)`  is

  1. `(2(1 + x^2))/(1 - x^2)`
  2. `(sqrt2(1 - x^2))/(1 + x^2)`
  3. `-((x + 1)^2)/(2(1 + x^2))`
  4. `-((x - 1)^2)/(2(1 + x^2))`
  5. `(sqrt2(1 + x^2))/(1 - x^2)`
Show Answers Only

`D`

Show Worked Solution

`underset~a = x underset~i – underset~j, \ underset~b = underset~i – underset~j, \ underset~c = underset~i + x underset~j`

`underset~a · underset~b = |underset~a||underset~b|costheta`

`costheta = (x – 1)/(sqrt(x^2 + 1)sqrt2)`

`cos phi = (underset~b · underset~c)/(|underset~b||underset~c|) = (1 – x)/(sqrt2 sqrt(1 – x^2))`

`costheta · cos phi` `= ((x – 1)(1 – x))/(2(1 + x^2))`
  `= -((x – 1)^2)/(2(1 + x^2))`

 
`=>\ D`

Vectors, SPEC2 2020 VCAA 16 MC

Let  `underset~a = underset~i + 2underset~j + 2underset~k`  and  `underset~b = 2underset~i - 4underset~j + 4underset~k`, where the acute angle between these vectors is  `theta`.

The value of  `sin(2theta)`  is

  1. `1/9`
  2. `(4sqrt5)/9`
  3. `(4sqrt5)/81`
  4. `(8sqrt5)/81`
  5. `(2sqrt46)/25`
Show Answers Only

`D`

Show Worked Solution

`underset~a = ((1),(2),(2)), |underset~a| = sqrt(1 + 4 + 4) = 3`

`underset~b = ((2),(−4),(4)) , |underset~b| = sqrt(4 + 16 + 16) = 6`

`underset~a · underset~b = ((1),(2),(2))((1),(−4),(4)) = 2 – 8 + 8 = 2`

`costheta = (underset~a · underset~b)/(|underset~a||underset~b|) = 2/(3 xx 6) = 1/9`
 

`sintheta = sqrt(81 – 1)/9 = (4sqrt5)/9`

`:. sin(2theta)` `= 2 xx 1/9 xx (4sqrt5)/9`
  `= (8sqrt5)/81`

 
`=>D`

Vectors, SPEC2-NHT 2019 VCAA 12 MC

Given that `theta` is the acute angle between the vectors  `underset~a = −2underset~i + 2underset~j - underset~k`  and  `underset~b = −4underset~i + 4underset~j + 7underset~k`, then  `sin(2theta)` is equal to

  1.  `2sqrt2`
  2.  `(4sqrt2)/9`
  3.  `(2sqrt2)/9`
  4.  `(2sqrt2)/3`
  5.  `(4sqrt2)/3`
Show Answers Only

`B`

Show Worked Solution

`underset~a = ((−2),(2),(−1)), \ |underset~a| = sqrt(4 + 4 + 1) = 3`

`underset~b = ((−4),(4),(7)), \ |underset~b| = sqrt(16 + 16 + 49) = 9`

`underset~a · underset~b = ((−2),(2),(−1))((−4),(4),(7)) = 8 + 8 – 7 = 9`

`costheta = (underset~a · underset~b)/(|underset~a||underset~b|) = 9/(3 xx 9) = 1/3`

`sintheta` `= sqrt8/3`
`sin2theta` `= 2sinthetacostheta`
  `= 2 · sqrt8/3 · 1/3`
  `= (4sqrt2)/9`

`=>\ B`

Vectors, SPEC2 2011 VCAA 12 MC

The angle between the vectors  `3underset~i + 6underset~j - 2underset~k`  and  `2underset~i - 2underset~j + underset~k`, correct to the nearest tenth of a degree, is

A.       2.0°

B.     91.0°

C.   112.4°

D.   121.3°

E.   124.9°

Show Answers Only

`C`

Show Worked Solution

`|3underset~i + 6underset~j – 2underset~k| = sqrt(9 + 36 + 4) = sqrt49 = 7`

`|2underset~i – 2underset~j + underset~k| = sqrt(4 + 4 + 1) = sqrt9 = 3`

`(3underset~i + 6underset~j – 2underset~k) * (2underset~i – 2underset~j + underset~k)`

`= 3 xx 2 + 6 xx (−2) + (−2) xx 1`

`= 6 – 12 – 2`

`= -8\ \ text{(do calculations on CAS)}`
  

`costheta` `= ((3tildei + 6tildej – 2tildek).(2tildei – 2tildej + tildek))/(|\ 3tildei + 6tildej – 2tildek\ ||\ 2tildei – 2tildej + tildek\ |)`
  `= −8/21`
`:. theta` `= cos^(−1)(−8/12)`
  `~~ 112.4^@`

 
`=> C`

Vectors, SPEC2 2014 VCAA 15 MC

If  `theta`  is the angle between  `underset ~a = sqrt 3 underset ~i + 4 underset ~j - underset ~k`  and  `underset ~b = underset ~i - 4 underset ~j + sqrt 3 underset ~k`, then  `cos(2 theta)`  is

  1. `−4/5`
  2.    `7/25`
  3. `−7/25`
  4.    `14/25`
  5. `−24/25`
Show Answers Only

`B`

Show Worked Solution
`underset ~a ⋅ underset ~b` `= sqrt 3 xx 1 + 4 xx (-4) + (-1) xx sqrt 3`
  `= -16`

 
`|\ underset~a\ | = sqrt20,\ \ |\ underset~b\ | = sqrt20,`

`cos(theta)` `=(underset ~a ⋅ underset ~b)/(|\ underset~a\ |\ |\ underset~b\ | `
  `= (-16)/(sqrt 20 xx sqrt 20)`
  `= -4/5`

 

`cos (2 theta)` `= 2cos^2theta – 1`
  `= 2(-4/5)^2 – 1`
  `= 7/25`

 
`=> B`

Vectors, SPEC2 2015 VCAA 17 MC

Points  `A`, `B` and `C` have position vectors  `underset~a = 2underset~i + underset~j`,  `underset~b = 3underset~i - underset~j + underset~k`  and  `underset~c = -3underset~j + underset~k`  respectively.

The cosine of angle `ABC` is equal to

  1. `5/(sqrt6sqrt10)`
  2. `7/(sqrt6sqrt13)`
  3. `-1/(sqrt6sqrt13)`
  4. `-7/(sqrt21sqrt6)`
  5. `-2/(sqrt6sqrt13)`
Show Answers Only

`C`

Show Worked Solution

`vec(BA) = vec(OA) – vec(OB) = -tildei + 2tildej – tildek`

♦ Mean mark 48%.

`=>\ |\ vec(BA)\ | = sqrt6`

`vec(BC) = vec(OC) – vec(OB) = -3tildei – 2tildej`

`=>\ |\ vec(BC)\ | = sqrt13`
 

`vec(BA).vec(BC)` `= |\ vec(BA)\ ||\ vec(BC)\ |cos angleABC`
`cos angleABC` `= (vec(BA).vec(BC))/(|\ vec(BA)\ ||\ vec(BC)\ |)`
  `= (-3xx-1 + 2 xx – 2)/(sqrt6sqrt13)`
  `= (-1)/(sqrt6sqrt13)`

`=> C`

Vectors, SPEC1-NHT 2017 VCAA 10

Consider the vectors  `underset ~a = - underset ~i - 2 underset ~j + 3 underset ~k`  and  `underset ~b = 2 underset ~i + c underset ~j + underset ~k`.

Find the value of  `c, \ c in R`, if the angle between  `underset ~a`  and  `underset ~b`  is  `pi/3`.  (4 marks)

Show Answers Only

`c = -3`

Show Worked Solution
`underset ~a ⋅ underset ~b` `= -1 xx 2 + (-2) xx c + 3 xx 1`
  `= -2 – 2c + 3`
  `= 1 – 2c`

 

`1-2c` `= sqrt((-1)^2 + (-2)^2 + 3^3) *sqrt(2^2 + c^2 + 1^2) xx cos (pi/3)`
`1 – 2c` `= 1/2(sqrt 14 ⋅ sqrt(5 + c^2))`
`2 – 4c` `= sqrt(14(5 + c^2))`
`(2 – 4c)^2` `= 14(5 + c^2)`
`4 – 16c + 16c^2` `= 70 + 14c^2`
`2c^2 – 16c – 66` `= 0`
`c^2 – 8c – 33` `= 0`
`(c – 11)(c + 3)` `= 0`

 
`c = 11 or c = -3`

`text(S)text(ince)\ \ 2 – 4c = sqrt(15(5 + c^2))`

`2 – 4c > 0\ \ =>\ \ c<2`

`:. c = -3`

Vectors, SPEC1 2014 VCAA 1

Consider the vector  `underset ~a = sqrt 3 underset ~i - underset ~j - sqrt 2 underset ~k`, where  `underset ~i, underset ~j`  and  `underset ~k`  are unit vectors in the positive directions of the `x, y` and `z` axes respectively.

  1. Find the unit vector in the direction of  `underset ~a`.  (1 mark)

  2. Find the acute angle that  `underset ~a`  makes with the positive direction of the `x`-axis.  (2 marks)

  3. The vector  `underset ~b = 2 sqrt 3 underset ~i + m underset ~j - 5 underset ~k`.
  4. Given that  `underset ~b`  is perpendicular to  `underset ~a,` find the value of `m`.  (2 marks)

Show Answers Only

  1. `1/sqrt 6 (sqrt 3 underset ~i – underset ~j – sqrt 2 underset ~k)`
  2. `theta = 45^@`
  3. `m = 6 + 5 sqrt 2`

Show Worked Solution

a.    `|underset ~a|` `= sqrt((sqrt 3)^2 + (-1)^2 + (-sqrt 2)^2)`
    `= sqrt 6`
`:. hat underset ~a` `= underset ~a/|underset ~a|`
  `= 1/sqrt 6 (sqrt 3 underset ~i – underset ~j – sqrt 2 underset ~k)`

 

Mean mark part (b) 51%.

b.    `underset ~a ⋅ underset ~i` `= sqrt 3 xx 1 = sqrt 3`
  `underset ~a ⋅ underset ~i` `= |underset ~a||underset ~i| cos theta`
    `= sqrt 6 cos theta`
  `sqrt 3` `= sqrt 6 cos theta`
`cos theta` `= 1/sqrt 2`
`:. theta` `= pi/4 = 45^@`

 

c.   `underset ~a ⋅ underset ~b = sqrt 3 (2 sqrt 3) + (-1)(m) + (-sqrt 2)(-5) = 0`

`6 – m + 5 sqrt 2` `=0`  
`:. m` `=6 + 5 sqrt 2`  

Vectors, SPEC1 2017 VCAA 5

Relative to a fixed origin, the points `B`, `C` and `D` are defined respectively by the position vectors  `underset~b = underset~i - underset~j + 2underset~k, \ underset~c = 2underset~i - underset~j + underset~k`  and  `underset~d = aunderset~i - 2underset~j`  where `a` is a real constant.

Given that the magnitude of angle `BCD` is  `pi/3`, find `a`.  (4 marks)

Show Answers Only

`a = −2`

Show Worked Solution

`text(Angle between)\ overset(->)(CB)\ text(and)\ overset(->)(CD) = pi/3`

♦ Mean mark 45%.

`overset(->)(CB)` `= (1 – 2)underset~i + (−1 – −1)underset~j + (2 – 1)underset~k`
  `= −underset~i + underset~k`
`overset(->)(CD)` `= (a – 2)underset~i + (−2 – −1)underset~j + (−1 + 0)underset~k`
  `= (a – 2)underset~i – underset~j – underset~k`

 

`overset(->)(CD) · overset(->)(CB)` `= −(a – 2) + 0 – 1`
  `= 1 – a`
  `= |overset(->)(CD)||overset(->)(CB)|cos(pi/3)`

 

`1 – a` `= sqrt((a – 2)^2 + (−1)^2 + (−1)^2)sqrt((−1)^2 + (1)^2)cos(pi/3)`
`1 + a` `= sqrt(a^2 – 4a + 4 + 1 + 1) xx sqrt2 xx 1/2`
`2(1 – a)` `= sqrt(2a^2 – 8a + 12), \ \ a < 1`
`4(1 – a)^2` `= 2a^2 – 8a + 12, \ \ a < 1`
`4(1 – 2a + a^2)` `= 2a^2 – 8a + 12`
`4 – 8a + 4a^2` `= 2a^2 – 8a + 12`
`2a^2` `= 8`
`a^2` `= 4`
`:. a` `= −2\ \ \ (a<1)`

Vectors, SPEC2-NHT 2018 VCAA 11 MC

Let  `underset ~a = 2 underset ~i - 2 underset ~j + underset ~k`  and  `underset ~b = 2 underset ~i + 3 underset ~j + 6 underset ~k`.

The acute angle between  `underset ~a`  and  `underset ~b`  is closest to

  1. `11º`
  2. `75º`
  3. `79º`
  4. `86º`
  5. `88º`
Show Answers Only

`C`

Show Worked Solution
`underset ~a ⋅ underset ~b` `= 4 – 6 + 6 = 4`
`4` `=sqrt(4 + 4 + 1) sqrt(4 + 9 + 36) cos theta`
`4` `= sqrt 9 xx sqrt 49 cos theta`
`4` `= 21 cos theta`
`theta` `= cos^(-1) (4/21)`
`:. theta` `= 79.0194…`

 
`=>  C`

Vectors, SPEC2 2018 VCAA 11 MC

Consider the vectors given by  `underset ~a = m underset ~i + underset ~j`  and  `underset ~b = underset ~i + m underset ~j`, where  `m in R`.

If the acute angle between  `underset ~a`  and  `underset ~b`  is 30°, then `m` equals

  1. `sqrt 2 +- 1`
  2. `2 +- sqrt 3`
  3. `sqrt 3, 1/sqrt 3`
  4. `sqrt 3/(4 - sqrt 3)`
  5. `sqrt 39/13` 
Show Answers Only

`C`

Show Worked Solution
`underset ~a *underset ~b` `= m + m = 2m`
`underset ~a *underset ~b` `= |underset ~a||underset ~b| cos 30^@`
  `= sqrt(m^2 + 1) *sqrt(1 + m^2) *cos 30^@`
  `= {(m^2 + 1) sqrt 3}/2`

 

`{(m^2 + 1) sqrt 3}/2` `=2m`  
`m^2 sqrt 3 + sqrt 3` `=4m`  
`m^2 sqrt 3 – 4m + sqrt 3` `=0`  
`(sqrt 3 m)^2 – 4(sqrt 3 m) + 3` `=0`  
`(sqrt 3 m)^2 – 4(sqrt 3 m) + 2^2 – 1` `=0`  
`(sqrt 3 m – 2)^2 – 1` `=0`  
`sqrt 3 m – 2` `= +-1`  
`sqrt 3 m` `= 2 +- 1`  

 
`:. m = (2 +- 1)/sqrt 3 = 3/sqrt 3 or 1/sqrt 3`

`= sqrt 3, 1/sqrt 3`

`=>  C`