Further Mathematics · OCR H245 · Topic 4.04

Further vectors

Equation of a straight line, equation of a plane, scalar product, intersections, vector product, shortest distances.

  • 6 sub-topics
  • 5 in Year 12, 1 in Year 13
  • 114 practice questions
  • 6 animated lessons
Practice questions (114) Lessons Worksheets coming soon All OCR Further Mathematics topics

What the specification says

Every sub-topic OCR 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. 4.04aEquation of a straight lineY12
  2. 4.04bEquation of a planeY12
  3. 4.04c-dScalar productY12
  4. 4.04e-fIntersectionsY12
  5. 4.04gVector productY12
  6. 4.04h-jShortest distancesY13

Try 3 questions

Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 114 questions in this topic, free, no account needed.

Further vectors

Easy

Equation of a line · Year 12 · 2 marks

A line passes through (-1, 0, -1) with direction (03−1)\begin{pmatrix} 0 \\ 3 \\ -1 \end{pmatrix}.
  1. Write its vector equation and its Cartesian equation.
    (2)
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  1. r\mathbf{r}=(−10−1){}= \begin{pmatrix} -1 \\ 0 \\ -1 \end{pmatrix}+λ(03−1){}+ \lambda\begin{pmatrix} 0 \\ 3 \\ -1 \end{pmatrix}
  2. Each coordinate solved for λ\lambda.y3=z+1−1\frac{y}{3} = \frac{z + 1}{-1}

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Further vectors

Medium

Equation of a plane · Year 12 · 2 marks

A plane passes through (-1, -2, -2) and has normal (231)\begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix}.
  1. Find its Cartesian equation.
    (2)
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  1. r⋅n=a⋅nr\cdot n = a\cdot n.r⋅(231)\mathbf{r}\cdot\begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix}=(−1−2−2)⋅(231){}= \begin{pmatrix} -1 \\ -2 \\ -2 \end{pmatrix}\cdot\begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix}
  2. 2x+3y+z=−102x + 3y + z = -10

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Further vectors

Hard

Perpendicular distance to a plane · Year 12 · 2 marks

Plane: x−2y+2z=8x - 2y + 2z = 8; point (1, -2, 1).
  1. Find the perpendicular distance from the point to the plane.
    (2)

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  1. |ax₀ + by₀ + cz₀ − d| / |n|.∣7−8∣9\frac{|7 - 8|}{\sqrt{9}}
  2. 13\tfrac{1}{3}

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