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Vectors, SPEC1 2022 VCAA 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)

Show Answers Only

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.}`