Hello, I was looking at the 2020 HSC marking criteria provided by NESA for the recent ext. 2 exam, and I noticed that for 14a ii) my solution was nothing like the two sample responses provided and the ones mentioned in the marking scheme criteria as it only seems to mention two methods. I can't remember exactly how I proved it, but I remember obtaining the result, albeit by using an overly complicated method. My question is, are there any other possible methods for this question like by using the cosine rule/method or something else? Or are the two solutions provided by NESA the only possible solutions and hence whatever method I used is wrong? Thanks.
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Question Help (2020 Paper) (1 Viewer)
- Thread starter FookHSC
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xilbur
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Hey boss,
If you still got the answer and your working out is logical, then you'll get the marks. The marking criteria provides SAMPLE answers; not necessarily ALL of the methods to obtain a result. These answers are typically the most common methods. From what my teachers have told me, the sample answers are the minimum requirement to get full marks for a question. So don't stress! As long as your answer is logical, you should still access the full marks.
There are definitely other approaches.
For example, let P be the point representing
, making OAPB a parallelogram. Since OAB is an equilateral triangle, OAPB has a pair of adjacent sides equal and thus must be a rhombus. The diagonals of a rhombus must be perpendicular (
), thus:
 \\ (z_1 + z_2)^2 &= k^2 i^2 (z_2 - z_1)^2 \\ {z_1}^2 + 2z_1 z_2 + {z_2}^2 &= -k^2 \left({z_2}^2 -2z_1 z_2 + {z_1}^2\right) \\ (1 + k^2){z_1}^2 + (1 + k^2){z_2}^2 &= 2\left(k^2 - 1\right) z_1 z_2 \\ {z_1}^2 + {z_2}^2 &= \frac{2\left(k^2 - 1\right)}{k^2 + 1} z_1 z_2 \quad \text{. . . . . (*)} \end{align*})
We now only need to find the value of
.
In equilateral triangles OAB and ABP, let the sides all be of length
, from which it follows that 
and 
. Applying the cosine rule to either of OAP or OBP gives:
^2 &= d^2 + d^2 - 2d^2\cos{\frac{2\pi}{3}} \\ k^2 d^2 &= 2d^2 - 2d^2 \times -\cos{\frac{\pi}{3}} \\ k^2 d^2 &= 2d^2\left(1 + \frac{1}{2}\right) \\ k^2 &= \frac{2d^2}{d^2} \times \frac{3}{2} \\ k^2 &= 3 \\ \\ \text{so, from (*), we have} \quad {z_1}^2 + {z_2}^2 &= \frac{2\left(k^2 - 1\right)}{k^2 + 1} z_1 z_2 \quad \text{with $k^2 = 3$} \\ &= \frac{2(3 - 1)}{3 + 1} z_1 z_2 \\ &= z_1 z_2 \quad \text{as required} \end{align*})
For example, let P be the point representing
We now only need to find the value of
In equilateral triangles OAB and ABP, let the sides all be of length
Turning it around may be quicker:
Let OP be the position vector representing, which makes OAPB a parallelogram. Since OAB is an equilateral triangle, OAPB has a pair of adjacent sides equal and thus must be a rhombus - and so its diagonals must be perpendicular. The rhombus can be divided into two congruent equilateral triangles, 
(SSS - equilateral triangles with a common side AB). Let all the sides of OAB and PAB be of length . Looking at the perpendicular vectors , we have:
.
Noting that both
and 
are in equilateral triangles, it follows that:
Applying the cosine rule then gives
^2 &= d^2 + d^2 - 2d \times d\cos{\frac{2\pi}{3}} \\ k^2 d^2 &= 2d^2 - 2d^2 \times -\cos{\frac{\pi}{3}} \\ k^2 d^2 &= 2d^2\left(1 + \frac{1}{2}\right) \\ k^2 &= 2 \times \frac{3}{2} = 3 \qquad \text{as $d^2 \neq 0$} \end{align*})
The two possible values of
depend on whether vector OP is rotated 90 degrees clockwise to vector BA or anticlockwise to vector AB. In either case, we have established that
^2 &= \left[\pm i\sqrt{3}\left(z_2 - z_1\right)\right]^2 \\ {z_1}^2 + 2z_1 z_2 + {z_2}^2 &= 3i^2 \left({z_2}^2 - 2z_1 z_2 + {z_1}^2\right) \\ &= -3{z_2}^2 + 6z_1 z_2 - 3{z_1}^2 \quad \text{as $i^2 = -1$} \\ 4{z_1}^2 + 4{z_2}^2 &= 6 z_1 z_2 - 2 z_1 z_2 \\ 4\left({z_1}^2 + {z_2}^2\right) &= 4z_1 z_2 \\ {z_1}^2 + {z_2}^2 &= z_1 z_2 \quad \text{as required} \end{align*})
Let OP be the position vector representing
Noting that both
Applying the cosine rule then gives
The two possible values of