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Vectors, EXT1 V1 2025 HSC 14c

The hands of an analogue clock are \(OA\) and \(OB\),

where \(A\) is \(\left(\sin \left(\dfrac{\pi t}{360}\right), \cos \left(\dfrac{\pi t}{360}\right)\right), B\) is \(\left(2 \sin \left(\dfrac{\pi t}{30}\right), 2 \cos \left(\dfrac{\pi t}{30}\right)\right)\),

\(O\) is the origin, and  \(t \geq 0\)  is the number of minutes past midnight.

Find the values of \(t\) when the hands are perpendicular for the first and second time after midnight. Give your answers to 3 decimal places.   (3 marks)

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\(t=16.364, 49.091 \ \text{mins}\).

Show Worked Solution

\(\text{Express \(OA\) and \(OB\) as vectors:}\)

\(\overrightarrow{O A}=\displaystyle \binom{\sin \left(\frac{\pi t}{360}\right)}{\cos \left(\frac{\pi t}{360}\right)}, \quad \overrightarrow{O B}=\displaystyle \binom{2\, \sin \left(\frac{\pi t}{30}\right)}{2\, \cos \left(\frac{\pi t}{30}\right)}\)

\(\text{When hands are parallel,} \ \ \overrightarrow{OA} \cdot \overrightarrow{OB}=0:\)

\(\sin \left(\dfrac{\pi t}{360}\right) \times 2\, \sin \left(\dfrac{\pi t}{30}\right)+\cos \left(\dfrac{\pi t}{360}\right) \times 2\, \cos \left(\dfrac{\pi t}{30}\right)=0\)

\(\cos \left(\dfrac{\pi t}{30}-\dfrac{\pi t}{360}\right)\) \(=0\)
\(\cos \left(\dfrac{11 \pi t}{360}\right)\) \(=0\)

 

\(\dfrac{11 \pi t}{360}=\dfrac{\pi}{2}, \dfrac{3 \pi}{2}\)

\(t=\dfrac{\pi}{2} \times \dfrac{360}{11 \pi}=16.364 \ \text{mins}\)

\(t=\dfrac{3 \pi}{2} \times \dfrac{360}{11 \pi}=49.091 \ \text{mins}\).

Vectors, EXT1 V1 EQ-Bank 3 MC

Which of the following vectors is perpendicular to \(\displaystyle \binom{3}{-2}\) and has a magnitude of 3?

  1. \(\displaystyle 3\binom{-3}{2}\)
  2. \(\displaystyle \frac{3}{\sqrt{13}}\binom{-3}{2}\)
  3. \(\displaystyle \frac{\sqrt{10}}{\sqrt{13}}\binom{2}{3}\)
  4. \(\displaystyle \frac{3}{\sqrt{13}}\left(\frac{2}{3}\right)\)
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\(\Rightarrow D\)

Show Worked Solution

\(\text{If} \ \perp \ \Rightarrow \text {dot product}=0:\)

\(\displaystyle \binom{3}{-2}\binom{-3}{2}=-9-4=-13 \neq 0  \quad \text{(Eliminate A and B)}\)
  

\(\text{Consider Option D:}\)

\(\displaystyle \frac{3}{\sqrt{13}}\left(\frac{2}{3}\right)=\binom{\frac{6}{\sqrt{13}}}{\frac{9}{\sqrt{13}}} \)

\(\text{Magnitude }=\sqrt{\left(\frac{6}{\sqrt{13}}\right)^2+\left(\frac{9}{\sqrt{13}}\right)^2}=\sqrt{\dfrac{36+81}{13}}=3\)

\(\Rightarrow D\)

Vectors, EXT1 V1 2024 HSC 12a

The vectors \(\displaystyle \binom{a^2}{2}\) and \(\displaystyle  \binom{a+5}{a-4}\) are perpendicular.

Find the possible values of \(a\).   (3 marks)

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\(x=1,-4 \text { or }-2\)

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\(\text{If vectors are }\perp:\)

\(\displaystyle\binom{a^2}{2} \cdot\binom{a+5}{a-4}=0\)

\(a^3+5 a^2+2 a-8=0\)
 

\(\text{Test for roots:}\)

\(1^3+5 \times 1^2+2\times 1-8=0 \, \checkmark\)

\((a-1) \text{ is a factor.}\)

\(\text{By polynomial long division:}\)

\((a-1)\left(a^2+6 a+8\right)=0\)

\((a-1)(a+4)(a+2)=0\)

\(\therefore x=1,-4 \text { or }-2\)

Vectors, EXT1 V1 2022 HSC 11d

The vectors  `underset~u=([a],[2])`  and  `underset~v=([a-7],[4a-1])`  are perpendicular.

What are the possible values of `a`?  (2 marks)

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`a=1, -2`

Show Worked Solution

`text{If}\ \ underset~u ⊥ underset~v:`

`([a],[2])*([a-7],[4a-1])` `=0`  
`a(a-7)+2(4a-1)` `=0`  
`a^2-7a+8a-2` `=0`  
`a^2+a-2` `=0`  
`(a+2)(a-1)` `=0`  

 
`:.a=1\ \ text{or}\ \ -2`

Vectors, EXT1 V1 SM-Bank 17

Two vectors are given by `underset~a = 2underset~i + m underset~j`  and  `underset~b = −5underset~i + n underset~j`  where  `m, n > 0`.

If  `|underset~a| = 3`  and  `underset~a`  is perpendicular to  `underset~b`, find the values of  `m`  and  `n`.  (2 marks)

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`2sqrt5`

Show Worked Solution

`underset~a = [(2),(m)],\ \ underset~b = [(−5),(n)]`

 
`text(Using)\ |underset~a| = 3:`

`3` `= sqrt(2^2 + m^2)`
`m^2` `= 5`
`:.m` `= sqrt5,\ \ \ (m > 0)`

 
`text(S)text(ince)\ underset~a ⊥ underset~b:`

`a · b` `= 0`
`2xx −5 + mn` `= 0`
`sqrt5 n` `= 10`
`n` `= 10/sqrt5`
  `= 2sqrt5`

Vectors, EXT1 V1 SM-Bank 20

Consider the vector  `underset~a = underset~i + sqrt3underset~j`, where  `underset~i`  and  `underset~j`  are unit vectors in the positive direction of the `x` and `y` 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.   (1 mark)

  3. The vector  `underset~b = m underset~i - 2underset~j`.

     

    Given that  `underset~b`  is perpendicular to  `underset~a`, find the value of  `underset~m`.   (1 mark)

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  1. `1/2(underset~i + sqrt3underset~j)`
  2. `60°`
  3. `2sqrt3`
Show Worked Solution

i.   `underset~a = underset~i + sqrt3underset~j`

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

`overset^a = (underset~a)/(|underset~a|) = 1/2(underset~i + sqrt3underset~j)`

 

ii.   `text(Solution 1)`

`underset~a\ =>\ text(Position vector from)\ \ O\ \ text{to}\ \ (1, sqrt3)`

`tan theta` `=sqrt3`  
`:. theta` `=60°`  
     

`text(Solution 2)`

`text(Angle with)\ xtext(-axis = angle with)\ \ underset~b = underset~i`

`underset~a · underset~i = 1 xx 1 = 1`

`underset~a · underset~i` `= |underset~a||underset~i|costheta`
`1` `= 2 xx 1 xx costheta`
`costheta` `= 1/2`
`:. theta` `= 60°`

 

iii.   `underset~b = m underset~i – 2underset~j`

`underset~a · underset~b = [(1),(sqrt3)] · [(m),(−2)] = m – 2sqrt3`

`text(S)text(ince)\ underset~a ⊥ underset~b:`

`m – 2sqrt3` `= 0`
`m` `= 2sqrt3`

Vectors, EXT1 V1 SM-Bank 19

Consider the following vectors

`overset(->)(OA) = 2underset~i + 2underset~j,\ \  overset(->)(OB) = 3underset~i - underset~j,\ \ overset(->)(OC) = 5underset~i + 3underset~j`

  1. Find  `overset(->)(AB)`.  (1 mark)

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

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

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  1. `underset~i – 3underset~j`
  2. `text(See Worked Solutions)`
  3. `2sqrt5`
Show Worked Solution

i.  `text(Find)\ overset(->)(AB):`

COMMENT: Many teachers recommend column vector notation to simplify calculations and minimise errors – we agree!

`overset(->)(OA) = [(2),(2)],\ \ overset(->)(OB)[(3),(−1)]`

`overset(->)(AB)` `= overset(->)(OB) – overset(->)(OA)`
  `= [(3),(−1)] – [(2),(2)]`
  `= [(1),(−3)]`
  `= underset~i – 3underset~j`

 

ii.    `overset(->)(AC)` `= overset(->)(OC) – overset(->)(OA)`
    `= [(5),(3)] – [(2),(2)]`
    `= [(3),(1)]`
    `= 3underset~i + underset~j`

 

`overset(->)(AB) · overset(->)(AC)` `= 1 xx 3 + −3 xx 1=0`

`=> AB ⊥ AC`

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

 

iii.   `overset(->)(BC)\ text(is the hypotenuse)`

`overset(->)(BC)` `= overset(->)(OC) – overset(->)(OB)`
  `= [(5),(3)] – [(3),(−1)]`
  `= [(2),(4)]`
`|overset(->)(BC)|` `=\ text(length of hypotenuse)`
  `= sqrt(2^2 + 4^2)`
  `= sqrt(20)`
  `= 2sqrt5`

Vectors, EXT1 V1 SM-Bank 16 MC

The vectors  `underset~a = 2underset~i + m underset~j`  and  `underset~b = m^2underset~i - underset~j`  are perpendicular for

A.   `m = −2`  and  `m = 0`

B.   `m = 2`  and  `m = 0`

C.   `m = -1/2`  and  `m = 0`

D.   `m = 1/2`  and  `m = 0`

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`D`

Show Worked Solution

`underset ~a ⊥ underset ~b\ \ =>\ \ underset ~a ⋅ underset ~b=0`

`underset ~a ⋅ underset ~b` `= 2m^2 + m(-1)`
`0` `= 2m^2 – m`
`0` `= m(2m – 1)`

 
`:. m = 0, quad m = 1/2`

`=> D`

Vectors, EXT1 V1 SM-Bank 15

Consider the vectors

`underset~a = 6underset~i + 2underset~j,\ \ underset~b = 2underset~i - m underset~j`

  1. Calculate  `2underset~a - 3underset~b`.  (1 mark)

  2. Find the values of  `m`  for which  `|underset~b| = 3sqrt2`.  (2 marks)

  3. Find the value of  `m`  such that  `underset~a`  is perpendicular to  `underset~b`.  (1 mark)

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  1. `[(6),(4 + 3m)]`
  2. `±sqrt14`
  3. `6`
Show Worked Solution
i.    `2underset~a – 3underset~b` `= 2[(6),(2)] – 3[(2),(−m)]`
    `= [(12),(4)] – [(6),(−3m)]`
    `= [(6),(4 + 3m)]`

 

ii.   `underset~a = [(6),(2)], \ \ underset~b = [(2),(−m)]`

`|underset~b|` `= sqrt(4 + m^2)`
`3sqrt2` `= sqrt(4 + m^2)`
`18` `= 4 + m^2`
`m^2` `= 14`
`m` `= ±sqrt14`

 

iii.   `text(If)\ \ underset~a ⊥ underset~b \ => \ underset~a · underset~b = 0`

`6 xx 2 + 2 xx – m` `= 0`
`2m` `= 12`
`:. m` `= 6`

Vectors, EXT1 V1 2018 SPEC2 12 MC

If  `|underset ~a + underset ~b| = |underset ~a| + |underset ~b|`  and  `underset ~a, underset ~b != underset ~0`, which one of the following is necessarily true?

A.   `underset ~a\ text(is parallel to)\ underset ~b`

B.   `|underset ~a| = |underset ~b|`

C.   `underset ~a = underset ~b`

D.   `underset ~a\ text(is perpendicular to)\ underset ~b` 

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`A`

Show Worked Solution
`|underset ~a + underset ~b|^2` `= (|underset ~a| + |underset ~b|)^2\ \ \ text{(given)}`
  `= |underset ~a|^2 + 2|underset ~a||underset ~b|+|underset ~b|^2`
`underset ~a ⋅ underset ~b` `= |underset ~a||underset ~b| cos theta`

 
`=> 2|underset ~a||underset ~b| = (2 underset ~a ⋅ underset ~b)/(cos theta)`

♦♦♦ Mean mark 36%.

`=>|underset ~a + underset ~b|^2 = |underset ~a|^2 + (2 underset ~a ⋅ underset ~b)/(cos theta) + |b|^2`

`(underset ~a + underset ~b) * (underset ~a + underset ~b) = underset ~a ⋅ underset ~a + (2 underset ~a ⋅ underset ~b)/(cos theta) + underset ~b ⋅ underset ~b`

`underset ~a ⋅ underset ~a + 2underset ~a ⋅ underset ~b + underset ~b ⋅ underset ~b = underset ~a ⋅ underset ~a + (2 underset ~a ⋅ underset ~b)/(cos theta) + underset ~b ⋅ underset ~b`

`2 underset ~a ⋅ underset ~b = (2 underset ~a ⋅ underset ~b)/(cos theta)`

`:. cos theta = 1\ \ =>\ \  theta = 0`

`=>  A`