Further Mathematics · AQA 7367 · Topic F

Further vectors

Lines and planes in three dimensions, the scalar and vector products, angles, intersections and shortest distances.

  • 6 sub-topics
  • 2 in Year 12, 2 in Year 13
  • 109 practice questions
  • 6 animated lessons
Practice questions (109) Lessons Worksheets All AQA Further Mathematics topics

What the specification says

Every sub-topic AQA can examine in this topic, with its reference number from the official specification. The Y12 and Y13 tags show when each sub-topic is usually taught.

  1. F1Vector and Cartesian forms of the equation of a straight line in 3DY12
  2. F2Vector and Cartesian forms of the equation of a planeY13
  3. F3Scalar product: angles between lines, planes, and a line and a planeY13
  4. F4Check whether vectors are perpendicular using the scalar productY12
  5. F5The vector product and areas of trianglesY13
  6. F6Intersections of lines and planes, and perpendicular distancesY13

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Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 109 questions in this topic, free, no account needed.

Further vectors

Easy

Angle between two lines · Year 12 · 3 marks

Line l1l_1 has direction (122)\begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix} and line l2l_2 has direction (2−12)\begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}.
  1. Find the acute angle between the lines, to one decimal place.
    (3)

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  1. The angle between lines is the angle between their direction vectors, found with the scalar product.a⋅b\mathbf{a}\cdot\mathbf{b}=2−2+4{}= 2 - 2 + 4=4{}= 4
  2. Both vectors have magnitude 1+4+4=3\sqrt{1 + 4 + 4} = 3.cos⁡θ\cos\theta=43×3{}= \dfrac{4}{3 \times 3}=49{}= \dfrac{4}{9}
  3. cos⁡θ\cos\theta is positive, so this angle is already acute.θ=63.6∘\theta = 63.6^\circ

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Further vectors

Medium

Angle between a line and a plane · Year 12 · 4 marks

  1. Find the acute angle between the line with direction (122)\begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix} and the plane 2x−y+2z=72x - y + 2z = 7.
    (4)
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  1. First find the angle with the plane's normal (2,−1,2)(2, -1, 2).cos⁡θ\cos\theta=(1)(2)+(2)(−1)+(2)(2)3×3{}= \dfrac{(1)(2) + (2)(-1) + (2)(2)}{3 \times 3}=49{}= \dfrac{4}{9}
  2. θ=63.6∘\theta = 63.6^\circ
  3. The normal is at right angles to the plane, so subtract from 90∘90^\circ.Angle with the plane =90∘−63.6∘= 90^\circ - 63.6^\circ=26.4∘{}= 26.4^\circ

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Further vectors

Hard

Foot of a perpendicular · Year 12 · 5 marks

  1. Find the foot of the perpendicular from P(2,3,4)P(2, 3, 4) to the plane x+2y+2z=3x + 2y + 2z = 3.
    (5)
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  1. The perpendicular from PP goes in the direction of the plane's normal.r\mathbf{r}=(234){}= \begin{pmatrix} 2 \\ 3 \\ 4 \end{pmatrix}+t(122){}+ t\begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}
  2. Substitute into the plane to find where it lands.(2+t)(2 + t)+2(3+2t){}+ 2(3 + 2t)+2(4+2t){}+ 2(4 + 2t)=3{}= 3
  3. 16+9t16 + 9t=3{}= 3⇒t{}\Rightarrow t=−139{}= -\tfrac{13}{9}
  4. Put tt back into the line.Foot =(59,19,109)= \left(\tfrac{5}{9}, \tfrac{1}{9}, \tfrac{10}{9}\right)

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