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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 3 - 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 3 - 2020 - Paper 2

Step 1

P, Q and R lie on a straight line

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Answer

To express the relationship between the position vectors p, q, and r, we start by recognizing that since points P, Q, and R are collinear, we can represent the position vector of Q as a linear combination of the position vectors of P and R. Let us introduce a parameter t such that:

Q=(1t)P+tRQ = (1 - t)P + tR

Substituting the specific condition that Q lies one third of the way from P to R, we have:

t=13t = \frac{1}{3}

Thus, we write:

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

Now, we manipulate our equation:

Q=23P+13RQ = \frac{2}{3}P + \frac{1}{3}R

To express Q purely in terms of r and p, rearranging yields:

Q=13R+23PQ = \frac{1}{3}R + \frac{2}{3}P

Multiplying through by 3 gives:

3Q=R+2P3Q = R + 2P

From which we can obtain:

R=3Q2PR = 3Q - 2P

By substituting back:

Q=13(R+2P)Q = \frac{1}{3}(R + 2P)

This demonstrates what we set out to show.

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