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Relative to a fixed origin, points P, Q and R have position vectors p, q and r respectively - Edexcel - A-Level Maths Pure - Question 4 - 2020 - Paper 2

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Relative to a fixed origin, points P, Q and R have position vectors p, q and r respectively. Given that - P, Q and R lie on a straight line - Q lies one third of t... show full transcript

Worked Solution & Example Answer:Relative to a fixed origin, points P, Q and R have position vectors p, q and r respectively - Edexcel - A-Level Maths Pure - Question 4 - 2020 - Paper 2

Step 1

P, Q and R lie on a straight line

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Answer

Since points P, Q, and R are collinear, we can express the position vector of Q in relation to P and R. This means that Q can be represented as a linear combination of the vectors corresponding to P and R.

We can express this as: Q=(1t)P+tRQ = (1 - t)P + tR where tt is a scalar representing the position of Q on the line segment connecting P and R. Given that Q lies one third of the way from P to R, we have: t=13t = \frac{1}{3}. Thus, q=(113)p+13r=23p+13rq = \left(1 - \frac{1}{3}\right)p + \frac{1}{3}r = \frac{2}{3}p + \frac{1}{3}r

Step 2

Q lies one third of the way from P to R

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Answer

Given our previous expression for q, we will express it in a more detailed form:

Starting from our equation, q=23p+13rq = \frac{2}{3}p + \frac{1}{3}r we can manipulate it to isolate q on one side to compare with what we need to show:

Multiplying through by 3 to eliminate the fraction gives: 3q=2p+r3q = 2p + r Rearranging this gives: r2p=3qr - 2p = 3q Now if we write r in terms of p and q, we have: q=13(r+2p)q = \frac{1}{3}(r + 2p) Thus, we have shown the required result.

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