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Vectors, EXT1 V1 EQ-Bank 8

In the diagram, \(OABC\) is a parallelogram where  \(\overrightarrow{O A}=\underset{\sim}{a}\)  and  \(\overrightarrow{O C}=\underset{\sim}{c}\).

Point \(X\) divides \(O C\) in ratio \(1: 3\) and point \(Y\) divides \(A C\) in the ratio \(4: 3\), as shown.
 

Prove that points \(X, Y\) and \(B\) are collinear.   (4 marks)

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\(\text{Proof (See worked solution)}\)

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\(X, Y, B \ \ \text{are collinear if:}\)

\(m_{XY}=m_{YB} \ \ \text {or} \ \ \overrightarrow{XY}=\lambda \overrightarrow{YB}\ \ …\ (1)\)

\(\overrightarrow{XY}=\overrightarrow{XC}+\overrightarrow{CY}\)

  \(\overrightarrow{XC}=\dfrac{3}{4} \overrightarrow{OC}=\dfrac{3}{4} \underset{\sim}{c}\)

  \(\overrightarrow{CY}=\dfrac{3}{7} \overrightarrow{CA}=\dfrac{3}{7}(\underset{\sim}{\alpha}-\underset{\sim}{c})\)

\(\overrightarrow{XY}=\dfrac{3}{4} \underset{\sim}{c}+\dfrac{3}{7} \underset{\sim}{a}-\dfrac{3}{7} \underset{\sim}{c}=\dfrac{9}{28} \underset{\sim}{c}+\dfrac{3}{7} \underset{\sim}{a}\)
 

\(\overrightarrow{YB}=\overrightarrow{Y A}+\overrightarrow{A B}\)

  \(\overrightarrow{YA}=\dfrac{4}{7} \overrightarrow{C A}=\dfrac{4}{7}( \underset{\sim}{a}- \underset{\sim}{c})\)

  \(\overrightarrow{AB}=\overrightarrow{OC}=\underset{\sim}{c} \quad \text{(opposite sides of parallelogram)}\)

\(\overrightarrow{Y B}=\dfrac{4}{7}(\underset{\sim}{a}-\underset{\sim}{c})+\underset{\sim}{c}=\dfrac{4}{7} \underset{\sim}{a}+\dfrac{3}{7}\underset{\sim}{c}\)
 

\(\overrightarrow{XY}\) \(=\dfrac{9}{28} \underset{\sim}{c}+\dfrac{3}{7} \underset{\sim}{a}\)
  \(=\dfrac{3}{4}\left(\dfrac{3}{7} \underset{\sim}{c}+\dfrac{4}{7} \underset{\sim}{a}\right)\)
  \(=\dfrac{3}{4} \overrightarrow{YB}\)

 
\(\therefore X, Y \ \text{and} \ B  \ \text{are collinear (see (1) above)}\)

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)`
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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`

Vectors, EXT1 V1 EQ-Bank 6

Point  `C`  lies on  `AB`  such that  `overset(->)(AC) = lambdaoverset(->)(AB)`.

  1.  Express  `underset~c`  in terms of  `underset~a`  and  `underset~b`.   (2 marks)

  2.  Hence or otherwise, show that  `overset(->)(BC) = (1-lambda)(underset~a-underset~b)`.   (1 mark)

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  1. `lambdaunderset~b + (1-lambda)underset~a`
  2. `text{Proof (See Worked Solutions)}`
Show Worked Solution
i.      

`underset~c = underset~a + overset(->)(AC)`

`overset(->)(AB)` `= underset~b-underset~a`
`overset(->)(AC)` `= lambdaoverset(->)(AB)`
  `= lamda(underset~b-underset~a)`

 

`:.underset~c` `= underset~a + lambda(underset~b-underset~a)`
  `= lambdaunderset~b + (1-lambda)underset~a`

 

ii.    `overset(->)(BC)` `= underset~c-underset~b`
    `= lambdaunderset~b + (1-lambda)underset~a-underset~b`
    `=(1-lambda)underset~a +(lambda-1)underset~b`
    `=(1-lambda)underset~a -(1-lambda)underset~b`
    `= (1-lambda)(underset~a-underset~b)`

Vectors, EXT1 V1 SM-Bank 10

In the quadrilateral  `PQRS`, `T` lies on  `SR`  such that  `ST : TR = 3 : 1`.
 


 

  1. Find  `overset(->)(TS)`  in terms of  `underset~u`,  `underset~v`  and  `underset~w`.   (1 mark)

  2. Hence, find  `overset(->)(TP)`  in terms of  `underset~u`,  `underset~v`  and  `underset~w`.   (1 mark)

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  1. `3/4(underset~w-underset~u-underset~v)`
  2. `1/4 underset~u + 3/4 underset~w-3/4 underset~v`
Show Worked Solution

i.   `overset(->)(TS) = 3/4 xx-overset(->)(SR)`

`overset(->)(SR) = underset~u + underset~v-underset~w`

`:. overset(->)(TS)` `= 3/4 xx − (underset~u + underset~v-underset~w)`
  `= 3/4(underset~w-underset~u-underset~v)`

 

ii.    `overset(->)(TP)` `= overset(->)(SP) + overset(->)(TS)`
    `= underset~u + 3/4(underset~w-underset~u-underset~v)`
    `= 1/4 underset~u + 3/4 underset~w-3/4 underset~v`