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Vectors, EXT2 V1 2022 HSC 11e

Let `ℓ_(1)` be the line with equation `([x],[y])=([-1],[7])+lambda([3],[2]),lambda inRR`.

The line `ℓ_(2)` passes through the point  `A(-6,5)`  and is parallel to `ℓ_(1)`.

Find the equation of the line `ℓ_(2)` in the form  `y=mx+c`.  (2 marks)

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`y=2/3x+9`

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`m_(ℓ_(1))=2/3`

`text{Equation of}\ ℓ_(2)\ text{has}\ m=2/3\ text{and passes through}\ (-6,5):`

`y-5` `=2/3(x+6)`  
`y` `=2/3x+9`  

Vectors, EXT2 V1 2020 HSC 3 MC

 What is the Cartesian equation of the line  `underset~r = ((1),(3)) + lambda ((-2),(4))`?

  1. `2y + x = 7`
  2. `y - 2x = -5`
  3. `y + 2x = 5`
  4. `2y - x = -1`
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`C`

Show Worked Solution

`((x),(y)) = ((1),(3)) + λ ((-2),(4))`
 

`x = 1 – 2λ \ => \ λ = frac{1- x}{2}`

`y = 3 + 4λ \ => \ λ = frac{y- 3}{4}`
 

`frac{y – 3}{4}` `= frac{1 – x}{2}`
`y – 3` `= 2 – 2x`
`y + 2x` `= 5`

  
`=> \ C`

Vectors, EXT2 V1 EQ-Bank 19

  1. What vector line equation, `underset~r`, corresponds to the Cartesian equation
  2. `qquad (x + 2)/5 = (y-5)/4`   (1 mark)

  3. Express   `underset~v`  in Cartesian form where,
  4. `qquad underset~v = ((1),(−4)) + lambda((3),(1))`   (1 mark)

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 a.    `underset~r = ((−2),(5)) + lambda((5),(4))`

 b.    `y = (x-13)/3`

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a.    `underset~r = ((−2),(5)) + lambda((5),(4))`

COMMENT: Ensure you know the format for conversion from Cartesian to vector form (part i)!

 

b.    `((x),(y)) = ((1),(−4)) + lambda((3),(1))`

`x = 1 + 3lambda \ \ => \ lambda = (x-1)/3`
`y = −4 + lambda\ \ => \ lambda = y + 4`

 
`y + 4 = (x-1)/3`

`:. y = (x-13)/3`

Vectors, EXT2 V1 EQ-Bank 23

  1. Find values of `a, b, c` and `d` such that  `underset~v = ((a),(b)) + 2((c),(d))`  is a vector equation of a line that passes through `((3),(1))` and `((−3),(−3))`.   (2 marks)

  2. Determine whether  `underset~u = ((4),(6)) + lambda((−2),(3))`  is perpendicular to `underset~v`.   (1 mark)

  3. Express `underset~u` in Cartessian form.   (1 mark)

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a.    `a = 3, b = 1, c = −3, d = −2,\ text{or}\ a = −3, b = −3, c = 3, d = 2`

b.    `text(See worked solutions)`

c.    `y = −3/2x + 12`

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a.    `text(Method 1)`

`overset(->)(OA) = underset~a = ((3),(1)),\ \ overset(->)(OB) = underset~b = ((−3),(−3))`

`overset(->)(AB)` `= overset(->)(OB)-overset(->)(OA)`
  `= ((−3),(−3))-((3),(1))`
  `= ((−6),(−4))`

 

`underset~v` `= underset~a + lambdaunderset~b`
  `= ((3),(1)) + lambda((−6),(−4))`
  `= ((3),(1)) + 2((−3),(−2))`

 
`:. a = 3, b = 1, c = −3, d = −2`
 

`text(Method 2)`

`overset(->)(BA)` `= overset(->)(OA)-overset(->)(OB)`
  `= ((3),(1))-((−3),(−3))`
  `= ((6),(4))`

 
`underset~v = ((−3),(−3)) + 2((3),(2))`

`:. a = −3, b = −3, c = 3, d = 2`
 

b.   `underset~u = ((4),(6)) + lambda((−2),(3))`

`underset~v = ((3),(1)) + 2((−3),(−2))`

`((−2),(3)) · ((−3),(−2)) = 6-6 = 0`

`:. underset~u ⊥ underset~v`
 

c.   `((x),(y))= ((4),(6)) + lambda((−2),(3))`

`x = 4-2lambda\ \ \ …\ (1)`

`y = 6 + 3lambda\ \ \ …\ (2)`

`text(Substitute)\ \ lambda = (4-x)/2\ \ text{from (1) into (2):}`

`y` `= 6 + 3((4-x)/2)`
`y` `= 6 + 6-(3x)/2`
`y` `= −3/2x + 12`