Subject: Mathematics | Level: A-Level | Exam Board: OCR
Master the geometry and algebra of 3D space with this guide to A-Level Vectors. We will break down complex proofs and mechanics problems into manageable steps, ensuring you can secure every mark available for this crucial topic."
Revision Notes & Key Concepts
Worked Examples
Worked Example
Question: Points A, B and C have position vectors `a = 2i + 4j - k`, `b = 3i + j + k` and `c = 6i - 8j + 7k` respectively. (i) Find the displacement vector AB. (ii) Find the magnitude of vector AC.
Solution: Step 1: (i) To find the displacement vector AB, we calculate `b - a`. `AB = (3i + j + k) - (2i + 4j - k) = (3-2)i + (1-4)j + (1 - (-1))k = i - 3j + 2k`. Step 2: (ii) First find the displacement vector AC. `AC = c - a = (6i - 8j + 7k) - (2i + 4j - k) = 4i - 12j + 8k`. Step 3: Calculate the magnitude of AC using the 3D Pythagoras formula. `|AC| = sqrt(4^2 + (-12)^2 + 8^2) = sqrt(16 + 144 + 64) = sqrt(224)`. Final answer: `sqrt(224)` or `4 * sqrt(14)`.
Worked Example
Question: Show that the points P, Q and R with position vectors `p = i + 5j`, `q = 3i + j` and `r = 4i - j` are collinear.
Solution: Step 1: State the condition for collinearity: we need to show that the vectors between the points are parallel and that they share a common point. Step 2: Calculate two displacement vectors, for example PQ and QR. `PQ = q - p = (3i + j) - (i + 5j) = 2i - 4j`. `QR = r - q = (4i - j) - (3i + j) = i - 2j`. Step 3: Show that one vector is a scalar multiple of the other. By inspection, `2i - 4j = 2(i - 2j)`. Therefore, `PQ = 2 * QR`. Step 4: Write a concluding statement. Since `PQ` is a scalar multiple of `QR`, the vectors are parallel. They also share a common point, Q. Therefore, the points P, Q, and R are collinear.
Worked Example
Question: A particle of mass 0.5 kg is acted on by two forces, `F1 = (2i + 3j - k) N` and `F2 = (i - 2j + 4k) N`. Find the magnitude of the acceleration of the particle.
Solution: Step 1: Find the resultant force, F, by adding the two force vectors. `F = F1 + F2 = (2i + 3j - k) + (i - 2j + 4k) = 3i + j + 3k` N. Step 2: Apply Newton's Second Law in vector form, `F = ma`. `(3i + j + 3k) = 0.5 * a`. Step 3: Solve for the acceleration vector, `a`. `a = (3i + j + 3k) / 0.5 = 2 * (3i + j + 3k) = 6i + 2j + 6k` m/s^2. Step 4: The question asks for the magnitude of the acceleration. Calculate `|a|`. `|a| = sqrt(6^2 + 2^2 + 6^2) = sqrt(36 + 4 + 36) = sqrt(76)`. Final answer: `sqrt(76)` or `2 * sqrt(19)` m/s^2.
Practice Questions
Question: Given the vector `p = 2i - 6j + 3k`, calculate its magnitude.
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Question: The vectors `a = 3i + mj` and `b = 6i - 8j` are parallel. Find the value of m.
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Question: OABC is a parallelogram. The position vector of A is `a` and the position vector of C is `c`. Find the position vector of the midpoint of the diagonal OB.
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Question: The vector equation of two lines are `r1 = (2i + j) + t(i - 3j)` and `r2 = (i - 2j) + s(2i + j)`. Find the position vector of their point of intersection.
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Question: A particle starts from rest. It moves with constant acceleration `a = (2i - 3j) m/s^2`. Find its speed after 4 seconds.
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