Vector Pathways0%

Vectors · Topic 3 of 12

Vector Pathways

Video lesson · from 1:00:083 worked examples

One lesson video covers all of Vectors, so it opens at 1:00:08 for this topic — not from the beginning.

Theory

A pathway writes a vector as a route through vectors you already know. Any route from the start point to the end point gives the same vector, so choose one along edges you can describe.

AC⃗=AB⃗+BC⃗\vec{AC} = \vec{AB} + \vec{BC}

Each step starts where the last one ended. To turn each step into a vector you know, use these three facts:

  • Equal vectors: edges with the same length and direction are the same vector, wherever they are. In a cuboid, every edge parallel to AB⃗\vec{AB} and pointing the same way is equal to it.
  • Reversing: going against a vector changes its sign: BA⃗=−AB⃗.\vec{BA} = -\vec{AB}.
  • Part of an edge: if M is the midpoint of HG, then HM⃗=12HG⃗.\vec{HM} = \tfrac{1}{2}\vec{HG}. If T divides EF in the ratio 1:3,1:3, then ET⃗=14EF⃗.\vec{ET} = \tfrac{1}{4}\vec{EF}.

⚠️ Common Examiner Traps

  • Write the route first: name each step, such as BH⃗=BA⃗+AD⃗+DH⃗,\vec{BH} = \vec{BA} + \vec{AD} + \vec{DH}, before you put anything in. A wrong route is a common reason for losing marks here.
  • Check the letters join up: each step must start where the one before it ended.
  • Mind the direction: a step against a given vector needs a minus sign, and in i,j,ki, j, k form that changes all three signs.
  • Ratio parts, not the ratio: a ratio of 1:31:3 means 14\tfrac{1}{4} of the edge, not 13.\tfrac{1}{3}.
  • Finish the sum: collect like terms, and add the i,i, jj and kk parts separately. A route on its own is not the answer.

Worked examples

Example 1

ABCD, EFGH is a cuboid. AB⃗=p,\vec{AB} = \mathbf{p}, AD⃗=q\vec{AD} = \mathbf{q} and AE⃗=r.\vec{AE} = \mathbf{r}. M is the midpoint of HG.

ABCDEFGHMpqr

Express each vector in terms of p,\mathbf{p}, q\mathbf{q} and r.\mathbf{r}.

  • AG⃗\vec{AG}
  • BH⃗\vec{BH}
  • AM⃗\vec{AM}

1. Go along AB, then BC, then up CG. BC is parallel to AD and CG is parallel to AE:

AG⃗=AB⃗+BC⃗+CG⃗=p+q+r\begin{aligned} \vec{AG} &= \vec{AB} + \vec{BC} + \vec{CG} \\ &= \mathbf{p} + \mathbf{q} + \mathbf{r} \end{aligned}

2. Go back along BA, then AD, then up DH. BA is against p:\mathbf{p}:

BH⃗=BA⃗+AD⃗+DH⃗=−p+q+r\begin{aligned} \vec{BH} &= \vec{BA} + \vec{AD} + \vec{DH} \\ &= -\mathbf{p} + \mathbf{q} + \mathbf{r} \end{aligned}

3. Go along AD, up DH, then half of HG. HG is parallel to AB:

AM⃗=AD⃗+DH⃗+12HG⃗=12p+q+r\begin{aligned} \vec{AM} &= \vec{AD} + \vec{DH} + \tfrac{1}{2}\vec{HG} \\ &= \tfrac{1}{2}\mathbf{p} + \mathbf{q} + \mathbf{r} \end{aligned}

Example 2

V, PQRS is a rectangular-based pyramid, where

PQ⃗=4i+2jPS⃗=−i+2j+3kSV⃗=3i−j+2k\begin{aligned} \vec{PQ} &= 4\mathbf{i} + 2\mathbf{j} \\ \vec{PS} &= -\mathbf{i} + 2\mathbf{j} + 3\mathbf{k} \\ \vec{SV} &= 3\mathbf{i} - \mathbf{j} + 2\mathbf{k} \end{aligned}
PQRSV

Express QV⃗\vec{QV} in terms of i,\mathbf{i}, j\mathbf{j} and k.\mathbf{k}.

Go back from Q to P, across to S, then up to V. QP is against PQ⃗,\vec{PQ}, so all its signs change:

QV⃗=QP⃗+PS⃗+SV⃗=(−4i−2j)+(−i+2j+3k)+(3i−j+2k)=−2i−j+5k\begin{aligned} \vec{QV} &= \vec{QP} + \vec{PS} + \vec{SV} \\ &= (-4\mathbf{i} - 2\mathbf{j}) \\ &\quad + (-\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}) \\ &\quad + (3\mathbf{i} - \mathbf{j} + 2\mathbf{k}) \\ &= -2\mathbf{i} - \mathbf{j} + 5\mathbf{k} \end{aligned}

Example 3

ABC, DEF is a triangular prism. AB⃗=u,\vec{AB} = \mathbf{u}, AC⃗=v\vec{AC} = \mathbf{v} and AD⃗=w.\vec{AD} = \mathbf{w}. The point T divides EF in the ratio 1:3.1:3.

ABCDEFTuvw

Express AT⃗\vec{AT} in terms of u,\mathbf{u}, v\mathbf{v} and w.\mathbf{w}.

EF is parallel to BC, and BC goes back along AB and then along AC:

EF⃗=BC⃗=−u+v\vec{EF} = \vec{BC} = -\mathbf{u} + \mathbf{v}

The ratio 1:31:3 has 4 parts, so ET⃗=14EF⃗.\vec{ET} = \tfrac{1}{4}\vec{EF}. Go along AD, then DE, then ET. DE is parallel to AB:

AT⃗=AD⃗+DE⃗+14EF⃗=w+u+14(−u+v)=34u+14v+w\begin{aligned} \vec{AT} &= \vec{AD} + \vec{DE} + \tfrac{1}{4}\vec{EF} \\ &= \mathbf{w} + \mathbf{u} + \tfrac{1}{4}(-\mathbf{u} + \mathbf{v}) \\ &= \tfrac{3}{4}\mathbf{u} + \tfrac{1}{4}\mathbf{v} + \mathbf{w} \end{aligned}