# 2.7 Further exercises

## Example 31

Let **p** = 2**i** − 3**j** + **k** and **q** = −**i** −2**j** −4**k** be two vectors in ^{3}. Determine **p** + **q**, **p** − **q** and 2**p** − 3**q**.

### Answer

Since **p** = 2**i** − 3**j** + **k** and **q** = −**i** −2**j** −4**k**, we have

and

## Example 32

Let **u** and **v** be the position vectors (1, 1) and (2, 1), respectively.

**(a)**Determine the position vectors**u**+ 2**v**, −**u**, −**u**+ 3**v**and**u**− 3**v**.**(b)**On a single diagram, sketch**u**,**v**and the position vectors that you found in part (a).

### Answer

**(a)**Here,and

**(b)**

## Example 33

Let **u** = (2, 6) and **v** = (4, 2).

**(a)**Determine numbers*α*and*β*such that*Hint:*Obtain simultaneous equations involving*α*and*β*by equating the first and second components of the vectors on each side of equation (2.9).**(b)**Sketch the position vectors**u**,**v**and (3, 4) on a single diagram.

### Answer

**(a)**First we writeThus

Then, following the hint, we equate the first and second components of these vectors; this gives two simultaneous equations:

These two equations can be solved to give

Thus we can write

**(b)**

## Example 34

Let *A* and *B* be the points (5, 4) and (−2, −3).

Determine the point *R* that divides *AB* in the ratio 2 : 5, and the midpoint *M* of *AB*.

### Answer

The point *R* divides *AB* in the ratio . It follows from the Section Formula, with , that the position vector **r** of *R* is

Hence the coordinates of *R* are (3, 2).

The midpoint *M* of *AB* has position vector **m**, where

Hence the coordinates of *M* are
.

## Example 35

**(a)**Determine the vector form of the equation of the line ℓ through the points (2, 3) and (5, −1).**(b)**Hence determine whether the points (7, 2) and (−1, 7) lie on ℓ.

### Answer

**(a)**Let**p**= (2, 3) and**q**= (5, −1). Then the vector form of the equation of the line ℓ isthat is,

where λ ∈ .

**(b)**The point (7, 2) lies on the line ℓ if there is a value of λ such thatEquating components in turn, we obtain

The second equation gives , but this value of λ does not satisfy the first equation. It follows that (7, 2) does not lie on the line ℓ.

The point (−1, 7) lies on the line ℓ if there is a value of λ such that

Equating components in turn, we obtain

The second equation gives λ = 2, and this value also satisfies the first equation. It follows that the point (−1, 7) lies on the line ℓ.

## Example 36

The rectangle *OABC* has vertex *O* at the origin, and the position vectors of the vertices *A*, *B* and *C* are **a**, **b** and **c**, respectively. *P* is the midpoint of *BC*, and the point *Q* is one-third of the way along *AB* from *A*. *S* is the point of intersection of the lines *OC* and *QP*.

**(a)**Determine the position vectors of*P*and*Q*, in terms of**a**and**c**.**(b)**Determine the vector form of the equation of the line that passes through*P*and*Q*, in terms of**a**and**c**.**(c)**Use the equation that you obtained in part (b), and the fact that*S*lies on the line*OC*, to find the position vector of*S*, in terms of**a**and**c**.

### Answer

**(a)***P*is the midpoint of*BC*, so its position vector**p**can be written asNow

**b**=**a**+**c**by the Parallelogram Law, since*OB*is the diagonal of the parallelogram with adjacent sides**a**and**c**. ThusNext, the position vector

**q**of the point*Q*is given by . Here ; also , since (as they have the same magnitude and direction). Hence**(b)**The vector form of the line through*P*and*Q*isMultiplying out and collecting terms, we obtain

where λ ∈ .

**(c)**The point*S*lies on*QP*, so its position vector**s**can be written in the formBut

**s**can also be written solely as a multiple of**c**, since*S*is also on*OC*and so**s**is in the same direction as**c**. This means that the coefficient of**a**in equation (S.2) for**s**must be 0.Thus , so λ = 2.

It follows that