Pure: Vectors and Numerical Methods Depth
Vectors and Numerical Methods Depth consolidates two Pure skills that both get harder in the same way at A Level: deciding which numerical method to use and justifying whether it will actually work, and using vectors in two and three dimensions beyond the basics of magnitude and position. For numerical methods it covers checking whether a fixed-point iteration converges, why Newton-Raphson can fail for a badly chosen starting value, and reading a trapezium rule estimate against the true value. For vectors it covers the vector equation of a straight line, deciding whether two lines in 3D intersect or are skew, and finding the angle between two vectors with the dot product. The examiner is testing judgement, choosing the right method and confirming it applies, as much as calculation.
Before you start
Make sure you're comfortable with these topics first:
Method
- Decide the numerical technique from what the question gives you. Asked to locate or verify a root: use the change of sign test on f(x) at the interval endpoints. Given a rearrangement x = g(x) and a starting value: use fixed-point iteration x_(n+1) = g(x_n). Given f(x) (and able to differentiate it) with a starting value: use Newton-Raphson, x_(n+1) = x_n - f(x_n)/f'(x_n). Given a table of y-values at equal x-intervals: use the trapezium rule.
- To check whether a fixed-point iteration converges near a root at x = alpha, find g'(x) and evaluate it near alpha: the iteration converges if |g'(alpha)| < 1, and diverges (moves away from the root) if |g'(alpha)| > 1.
- The Newton-Raphson method fails, converges to the wrong root, or converges very slowly if the starting value x0 is chosen where f'(x0) = 0 (the tangent is horizontal and never meets the x-axis), or where a turning point of the curve lies between x0 and the root and sends the tangent in the wrong direction.
- The vector equation of a straight line through the point with position vector a, in the direction of vector d, is r = a + t*d, where t is a scalar parameter; write the i, j and k components separately to get three linked equations.
- To decide whether two lines given in vector form intersect or are skew, set their general points equal component by component to get three equations in the two parameters. Solve any two of the equations simultaneously, then substitute into the third: if it is satisfied, the lines intersect there; if not, and the direction vectors are not parallel, the lines are skew.
- Find the angle between two vectors (or between the direction vectors of two lines) using the dot product: cos(theta) = (a.b) / (|a||b|). Two vectors are perpendicular exactly when a.b = 0.
- Before trusting any numerical answer, sanity-check it: does the sign of f change across the claimed root, does the iteration actually move towards a stable value rather than away from it, and for a trapezium estimate, does convexity or concavity of the curve tell you whether the true integral is a little more or a little less than the estimate.
Worked example
Line l1 has vector equation r = (1 + 2s)i + (2 - s)j + (3 + s)k. Line l2 has vector equation r = (4 + t)i + (-2 + 2t)j + (6 - t)k. Show that l1 and l2 intersect, and find the position vector of their point of intersection.
- Set the i, j and k components of l1 and l2 equal to each other: i: 1 + 2s = 4 + t, so 2s - t = 3. j: 2 - s = -2 + 2t, so s + 2t = 4. k: 3 + s = 6 - t, so s + t = 3.
- Solve two of the equations simultaneously. From s + t = 3, s = 3 - t. Substituting into 2s - t = 3: 2(3-t) - t = 3, so 6 - 3t = 3, giving t = 1, and hence s = 2.
- Check these values in the third equation: s + 2t = 2 + 2(1) = 4, which matches the j-equation exactly, confirming the lines do intersect rather than being skew.
- Substitute s = 2 into l1 (or t = 1 into l2) to find the point of intersection: r = (1 + 2(2))i + (2 - 2)j + (3 + 2)k = 5i + 0j + 5k.
- Final answer: the lines intersect at the point with position vector 5i + 5k, i.e. the point (5, 0, 5).
Practice questions
Try each question, then tap to reveal the answer.
Q1The iteration x_(n+1) = g(x_n), where g(x) = (x^2+6)/5, is used to find a root near x = 2. By finding g'(x), determine whether the iteration converges near this root.Show answer
Answer: g'(x) = 2x/5, so g'(2) = 4/5 = 0.8. Since |0.8| < 1, the iteration converges near x = 2.
Q2Let f(x) = x^3 - 3x + 1. Explain why applying the Newton-Raphson method with x0 = 1 would fail.Show answer
Answer: f'(x) = 3x^2 - 3, so f'(1) = 0. The tangent to the curve at x0 = 1 is horizontal and never meets the x-axis, so the formula x1 = x0 - f(x0)/f'(x0) involves division by zero and cannot be used.
Q3Show that x^3 - 5x - 3 = 0 has a root in the interval (2, 3), then show that this root also lies in the interval (2.4, 2.5).Show answer
Answer: f(2) = -5 and f(3) = 9; the change of sign confirms a root in (2,3). f(2.4) = -1.176 and f(2.5) = 0.125; the change of sign confirms the root lies in (2.4, 2.5).
Q4The curve y = ln(x) is used with the trapezium rule to estimate the integral from x = 1 to x = 5. State, with a reason, whether this will over-estimate or under-estimate the true area.Show answer
Answer: The second derivative of ln(x) is -1/x^2, which is negative for x > 0, so the curve is concave (concave down) on this interval. Its chords lie below the curve, so the trapezium rule under-estimates the true area.
Q5Line l has vector equation r = (3i - j + 2k) + t(i + 2j - k). Determine, showing your working, whether the point (7, 6, -2) lies on l.Show answer
Answer: From the i-component, 3+t=7 gives t=4. At t=4 the line gives y = -1+2(4) = 7, but the point has y = 6, so the point does NOT lie on l (the x- and z-components match at t=4, but the y-component does not).
Q6Find the angle between the vectors a = 2i + 3j - k and b = i - 2j + 2k, giving your answer to 1 decimal place.Show answer
Answer: a.b = 2(1)+3(-2)+(-1)(2) = -6. |a| = sqrt(14), |b| = 3. cos(theta) = -6/(3sqrt(14)) = -0.5345 (4dp), so theta = 122.3 degrees (1dp).
Q7Given that a = 4i - j + 2k and b = 3i + lambda*j - 5k are perpendicular, find the value of lambda.Show answer
Answer: a.b = 4(3) + (-1)(lambda) + 2(-5) = 2 - lambda. Setting this to 0 for perpendicular vectors gives lambda = 2.
Exam-style questions
Written in the style of a A Level Maths exam paper, with a full mark scheme.
Let f(x) = x^3 - 2x - 5. (a) Show that f(x) = 0 has a root between x = 2 and x = 3. (2) (b) Using the Newton-Raphson method with x0 = 2, find x1 and x2, giving each to 4 decimal places. (3) (c) By evaluating f(x) at appropriate values, verify that the root is x = 2.0946 correct to 4 decimal places. (2)
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Line l1 has vector equation r = (2i - j + 3k) + s(i + j - k). Line l2 has vector equation r = (i + 2j + k) + t(2i - j + k). (a) Show that l1 and l2 do not intersect (i.e. they are skew). (4) (b) Show that the direction vectors of l1 and l2 are perpendicular. (2)
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The table gives values of y = 1/(x^2+1) for x = 0, 0.5, 1, 1.5 and 2: y = 1, 0.8, 0.5, 0.3077, 0.2 (4dp where needed). (a) Use the trapezium rule with all four strips to estimate the integral of 1/(x^2+1) with respect to x, from x=0 to x=2, giving your answer to 3 significant figures. (3) (b) Given that the exact value of the integral is arctan(2) = 1.1071 (4dp), calculate the percentage error in the trapezium rule estimate, giving your answer to 1 decimal place. (3)
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