[In this question i and j are unit vectors due east and due north respectively: Position vectors are given relative to a fixed origin O.] Two ships P and Q are moving with constant velocities. Ship P moves with velocity $(2i - 3j)$ km h$^{-1}$ and ship Q moves with velocity $(3i + 4j)$ km h$^{-1}$. At 2 pm, ship P is at the point with position vector $(i + j)$ km and ship Q is at the point with position vector $(-2j)$ km. At time $t$ hours after 2 pm, the position vector of P is $p$ km and the position vector of Q is $q$ km. (b) Write down expressions, in terms of $t$, for (i) $p$. (ii) $q$. (iii) $PQ$. (c) Find the time when (i) Q is due north of P, (ii) Q is north-west of P.

A-Level Edexcel Maths Mechanics Working with Vectors: [In this question i and j are un

[In this question i and j are unit vectors due east and due north respectively: Position vectors are given relative to a fixed origin O.] Two ships P and Q are moving with constant velocities - Edexcel - A-Level Maths Mechanics - Question 7 - 2011 - Paper 1

Question 7

[In this question i and j are unit vectors due east and due north respectively: Position vectors are given relative to a fixed origin O.] Two ships P and Q are movi... show full transcript

Worked Solution & Example Answer:[In this question i and j are unit vectors due east and due north respectively: Position vectors are given relative to a fixed origin O.] Two ships P and Q are moving with constant velocities - Edexcel - A-Level Maths Mechanics - Question 7 - 2011 - Paper 1

Step 1

Find, to the nearest degree, the bearing on which Q is moving.

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Answer

To find the bearing of Q, we first determine its velocity vector, which is given as $(3i + 4j)$ km h $^{-1}$ . The angle $heta$ can be calculated as follows:

Calculate the angle using the tangent function: $\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{4}{3}$
Solving for $heta$ gives: $\theta = \arctan\left(\frac{4}{3}\right)$
Finally, converting to bearing, we find: $\theta = 37^{\circ}$ Thus, the bearing to the nearest degree is 37°.

Step 2

Write down expressions, in terms of t, for

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Answer

(i) p. The position vector of ship P at time $t$ after 2 pm is given by: $p = (i + j) + (2i - 3j)t = (1 + 2t)i + (1 - 3t)j$

(ii) q. The position vector of ship Q at time $t$ after 2 pm is given by: $q = (-2j) + (3i + 4j)t = (3t)i + (-2 + 4t)j$

(iii) PQ. The vector from P to Q is given by: $PQ = q - p = [(3t - (1 + 2t))i + ((-2 + 4t) - (1 - 3t))j] = (3t - 1 - 2t)i + (-2 + 4t - 1 + 3t)j = (t - 1)i + (7t - 3)j$

Step 3

Find the time when

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Answer

(i) Q is due north of P. For Q to be due north of P, the x-coordinates of P and Q must be equal. Thus: $1 + 2t = 3t$ Solving for $t$ gives: $-1 + t = 0 \Rightarrow t = 1 ext{ hour or } 3 ext{ pm}$

(ii) Q is north-west of P. For Q to be north-west of P, Q must lie on the same vertical line as P while being westward. The condition is given by the vector's direction, which results in: $-1 + t = -(3 - 7t)$ Simplifying gives: $6t = 4 \Rightarrow t = \frac{2}{3} ext{ hour or } 2:30 ext{ pm}$

[In this question i and j are unit vectors due east and due north respectively: Position vectors are given relative to a fixed origin O.] Two ships P and Q are moving with constant velocities - Edexcel - A-Level Maths Mechanics - Question 7 - 2011 - Paper 1

Worked Solution & Example Answer:[In this question i and j are unit vectors due east and due north respectively: Position vectors are given relative to a fixed origin O.] Two ships P and Q are moving with constant velocities - Edexcel - A-Level Maths Mechanics - Question 7 - 2011 - Paper 1

Find, to the nearest degree, the bearing on which Q is moving.

Write down expressions, in terms of t, for

Find the time when

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