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Vectors OQ = q and OR = r. X divides OQ in ratio 1:2, T divides OR in ratio 3:4. Lines XR and TQ intersect at E. Express OE in terms of q, r, and m if XE = mXR and TE = nTQ.

Mathematics

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4 parts
a(i)

r−13q\mathbf{r} - \frac{1}{3}\mathbf{q}

a(ii)

q−37r\mathbf{q} - \frac{3}{7}\mathbf{r}

b(i)

1−m3q+mr\frac{1-m}{3}\mathbf{q} + m\mathbf{r}

b(ii)

3(1−n)7r+nq\frac{3(1-n)}{7}\mathbf{r} + n\mathbf{q}

Step-by-step solution

8 steps

Here's the solution to the vector problem:

Given:
OQ⃗=q\vec{OQ} = \mathbf{q}
OR⃗=r\vec{OR} = \mathbf{r}
Point X divides OQ in the ratio 1:21:2, so OX⃗=13OQ⃗=13q\vec{OX} = \frac{1}{3}\vec{OQ} = \frac{1}{3}\mathbf{q}.
Point T divides OR in the ratio 3:43:4, so OT⃗=37OR⃗=37r\vec{OT} = \frac{3}{7}\vec{OR} = \frac{3}{7}\mathbf{r}.
Lines XR and TQ intersect at E.

Part a(i)· 1 step

XR⃗\vec{XR}

Express in terms of q\mathbf{q} and r\mathbf{r}

  1. Use the vector subtraction rule XR⃗=OR⃗−OX⃗\vec{XR} = \vec{OR} - \vec{OX}.

    XR⃗=r−13q\vec{XR} = \mathbf{r} - \frac{1}{3}\mathbf{q} XR⃗=r−13q\vec{XR} = \boxed{\mathbf{r} - \frac{1}{3}\mathbf{q}}

Part a(ii)

q−37r\mathbf{q} - \frac{3}{7}\mathbf{r}

TQ⃗\vec{TQ}

  1. Use the vector subtraction rule TQ⃗=OQ⃗−OT⃗\vec{TQ} = \vec{OQ} - \vec{OT}.

    TQ⃗=q−37r\vec{TQ} = \mathbf{q} - \frac{3}{7}\mathbf{r} TQ⃗=q−37r\vec{TQ} = \boxed{\mathbf{q} - \frac{3}{7}\mathbf{r}}

Part b(i)

1−m3q+mr\frac{1-m}{3}\mathbf{q} + m\mathbf{r}

r\mathbf{r}, q\mathbf{q} and mm.

If XE⃗=mXR⃗\vec{XE} = m\vec{XR} and TE⃗=nTQ⃗\vec{TE} = n\vec{TQ}, express OE⃗\vec{OE} in terms of

  1. Use the given relationship XE⃗=mXR⃗\vec{XE} = m\vec{XR}.

    OE⃗−OX⃗=mXR⃗\vec{OE} - \vec{OX} = m\vec{XR}

  2. Substitute OX⃗=13q\vec{OX} = \frac{1}{3}\mathbf{q} and the expression for XR⃗\vec{XR} from part (a)(i).

    OE⃗−13q=m(r−13q)\vec{OE} - \frac{1}{3}\mathbf{q} = m\left(\mathbf{r} - \frac{1}{3}\mathbf{q}\right)

  3. Rearrange to solve for OE⃗\vec{OE}.

    OE⃗=13q+mr−m3q\vec{OE} = \frac{1}{3}\mathbf{q} + m\mathbf{r} - \frac{m}{3}\mathbf{q} OE⃗=(13−m3)q+mr\vec{OE} = \left(\frac{1}{3} - \frac{m}{3}\right)\mathbf{q} + m\mathbf{r} OE⃗=1−m3q+mr\vec{OE} = \boxed{\frac{1-m}{3}\mathbf{q} + m\mathbf{r}}

Part b(ii)

3(1−n)7r+nq\frac{3(1-n)}{7}\mathbf{r} + n\mathbf{q}

r\mathbf{r}, q\mathbf{q} and nn.

  1. Use the given relationship TE⃗=nTQ⃗\vec{TE} = n\vec{TQ}.

    OE⃗−OT⃗=nTQ⃗\vec{OE} - \vec{OT} = n\vec{TQ}

  2. Substitute OT⃗=37r\vec{OT} = \frac{3}{7}\mathbf{r} and the expression for TQ⃗\vec{TQ} from part (a)(ii).

    OE⃗−37r=n(q−37r)\vec{OE} - \frac{3}{7}\mathbf{r} = n\left(\mathbf{q} - \frac{3}{7}\mathbf{r}\right)

  3. Rearrange to solve for OE⃗\vec{OE}.

    OE⃗=37r+nq−3n7r\vec{OE} = \frac{3}{7}\mathbf{r} + n\mathbf{q} - \frac{3n}{7}\mathbf{r} OE⃗=(37−3n7)r+nq\vec{OE} = \left(\frac{3}{7} - \frac{3n}{7}\right)\mathbf{r} + n\mathbf{q} OE⃗=3(1−n)7r+nq\vec{OE} = \boxed{\frac{3(1-n)}{7}\mathbf{r} + n\mathbf{q}}
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