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1.10 Vectors

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Question 18

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

1.10 Vectors Questions

  1. A Level
  2. /Maths
  3. /1.10 Vectors

197 exam-style questions on OCR A Level Maths 1.10 Vectors, covering 1.10.1 Vectors in two dimensions, 1.10.2 Vectors in three dimensions (A-level only), 1.10.3 Magnitude and direction of vectors, 1.10.4 Basic operations on vectors, 1.10.5 Position vectors, 1.10.6 Distance between points, 1.10.7 Problem solving using vectors, 1.10.8 Vectors in kinematics, and 1.10 Vectors. Each one has a worked solution and a mark scheme showing where the marks go.

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