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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`

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`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`

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`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`