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Vectors, EXT1 V1 2022 SPEC1 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)

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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, EXT1 V1 2022 HSC 13a

Three different points `A, B` and `C` are chosen on a circle centred at `O`.

Let  `underset~a= vec(OA),underset~b= vec(OB)`  and  `underset~c= vec(OC)`. Let  `underset~h=underset~a + underset~b+ underset~c`  and let `H` be the point such that  `vec(OH)= underset~h`, as shown in the diagram.
 

     
 

Show that  `vec(BH)`  and  `vec(CA)`  are perpendicular.  (3 marks)

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`text{Proof (See Worked Solutions)}`

Show Worked Solution

`text{Prove}\ \ vec(BH) ⊥ vec(CA):`

`vec(BH)*vec(CA)=(-underset~b+underset~h)*(-underset~c+underset~a)`
 

`text{Given}\ \ underset~h=underset~a + underset~b+ underset~c\ \ =>\ \ underset~h-underset~b=underset~a + underset~c`

`vec(BH)*vec(CA)` `=(underset~a+underset~c)*(underset~a-underset~c)`  
  `=|underset~a|^2-|underset~c|^2`  

 
`underset~a and underset~c\ \ text{are radii}\ \ => \ \ |underset~a|=|underset~c|`

 
`:.\ text{S}text{ince}\ \ vec(BH)*vec(CA)=0\ \ =>\ \ vec(BH)⊥vec(CA)`

Vectors, EXT1 V1 EQ-Bank 2

In the diagram below, `ROT` is a diameter of the circle with centre `O`.

`S`  is a point on the circumference.
 


 

Using the properties of vectors  `underset~r`, `underset~s`  and  `underset~t`, show that  `angleRST`  is a right angle.  (2 marks)

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`text{Proof (See Worked Solutions)}`

Show Worked Solution
`|underset~r|` `= |underset~s| = |underset~t|\ \ \ text{(radii)}`
`underset~r` `= −underset~t`
`overset(->)(RS)` `= underset~s – underset~r`
`overset(->)(TS)` `= underset~s – underset~t`
`overset(->)(RS) · overset(->)(TS)` `= (underset~s – underset~r) · (underset~s – underset~t)`
  `= (underset~s – underset~r) · (underset~s + underset~r)\ \ \ \ (text{using}\ \ underset~r = – underset~t)`
  `= underset~s · (underset~s + underset~r) – underset~r · (underset~s + underset~r)`
  `= underset~s · underset~s + underset~s · underset~r – underset~r · underset~s – underset~r · underset~r`
  `= |underset~s|^2 – |underset~r|^2`
  `= 0`

 
`:. overset(->)(RS) ⊥ overset(->)(TS)`

`:. angleRST\ \ text(is a right angle.)`

Vectors, EXT1 V1 SM-Bank 2


 

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)`, express  `vec(LN)`  in terms of  `underset~r `  and  `underset~s`.   (2 marks)

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`vec(LN)= 2 underset~r –  underset~s`

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`