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1.11 Vectors

1.11 Vectors

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Question 23

The point A A\,A has position vector

a=5i−3j+4k\mathbf{a} = 5\mathbf{i} - 3\mathbf{j} + 4\mathbf{k}a=5i−3j+4k.

The point P P\,P has position vector p=λi+2j+k\mathbf{p} = \lambda\mathbf{i} + 2\mathbf{j} + \mathbf{k}p=λi+2j+k, where λ \lambda\,λ is a constant.

a.

Show that ∣AP⃗∣2=λ2−10λ+59\left|\vec{AP}\right|^2 = \lambda^2 - 10\lambda + 59​AP​2=λ2−10λ+59.

[3]
b.

Find the value of λ \lambda\,λ for which ∣AP⃗∣\left|\vec{AP}\right|​AP​ is least, and state that least value in exact form.

[3]
c.

Find the set of values of λ \lambda\,λ for which ∣AP⃗∣>7\left|\vec{AP}\right| > 7​AP​>7.

[2]
Markscheme

1.11 Vectors Questions

  1. A Level
  2. /Maths
  3. /1.11 Vectors

168 exam-style questions on OCR (MEI) A Level Maths 1.11 Vectors, covering 1.11.1 Language of vectors in two dimensions, 1.11.2 Add, subtract and scale vectors, 1.11.3 Magnitude and direction of a vector, 1.11.4 Position vectors, 1.11.5 Distance between points by position vectors, 1.11.6 Vectors to solve problems, 1.11.7 Language of vectors in three dimensions (A-level only), and 1.11 Vectors. Each one has a worked solution and a mark scheme showing where the marks go.

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