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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 30

Use the vector form of the linear equations

`3x-2y = 4`  and  `3y + 2x-6 = 0`

to show they are perpendicular.   (3 marks)

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`text(Proof)\ text{(See Worked Solutions)}`

Show Worked Solution
`3x-2y` `= 4`
`3x` `= 2y + 4`
`3/2 x` `=y+2`
`x/(2/3)` `= y + 2`

 
`underset ~(v_1) = ((0), (-2)) + lambda ((2/3), (1))`

`3y + 2x-6` `= 0`
`2x` `= -3y + 6`
`-2/3 x` `= y-2`
`x/(-3/2)` `= y-2`

 
`underset ~(v_2) = ((0), (2)) + lambda ((-3/2), (1))`

`((2/3), (1)) ((-3/2), (1)) = -1 + 1 = 0`

`:. underset ~(v_1) _|_ underset ~(v_2)`