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Vectors, EXT1 V1 2021 HSC 14c

  1. For vector `underset~v`, show that  `underset~v · underset~v = |underset~v|^2`.  (1 mark)

  2. In the trapezium `ABCD`, `BC` is parallel to `AD` and  `|overset(->)(AC)| = |overset(->)(BD)|`.
     
         

  3. Let  `underset~a = overset(->)(AB), underset~b = overset(->)(BC)`  and  `overset(->)(AD) = koverset(->)(BC)`,  where  `k > 0`.
  4. Using part (i), or otherwise, show  `2underset~a · underset~b + (1 - k)|underset~b|^2 = 0`.  (3 marks)

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  1. `text(See Worked Solution)`
  2. `text(See Worked Solution)`
Show Worked Solution

i.  `text(Let)\ \ underset~v = ((x),(y))`

`|underset~v| = sqrt(x^2 + y^2)`

`underset~v · underset~v = x^2 + y^2 = (sqrt(x^2 + y^2))^2 = |underset~v|^2`

 

♦♦♦ Mean mark part (ii) 18%.

ii.   `text(Show)\ \ 2underset~a · underset~b + (1 – k) |underset~b|^2 = 0`

`|overset(->)(AC)|` `= |overset(->)(BD)|`
`|underset~a + underset~b|` `= |kunderset~b – underset~a|`
`|underset~a + underset~b|^2` `= |kunderset~b – underset~a|^2`
`(underset~a + underset~b)(underset~a + underset~b)` `= (kunderset~b – underset~a)(kunderset~b – underset~a)\ \ text{(see part a.)}`
`underset~a · underset~a + 2underset~a · underset~b + underset~b · underset~b` `= k^2 underset~b · underset~b – 2kunderset~a · underset~b + underset~a · underset~a`
`2underset~a · underset~b + 2kunderset~a · underset~b + underset~b · underset~b – k^2 underset~b · underset~b` `= 0`
`2underset~a · underset~b(1 + k) + underset~b · underset~b(1 – k^2)` `= 0`
`2underset~a · underset~b(1 + k) + underset~b · underset~b(1 + k)(1 – k)` `= 0`
`2underset~a · underset~b +  (1 – k)|underset~b|^2` `= 0`