# Maths Higher Tomorrow!

1. May 20, 2004

### KaNe

I have the biggest exam of my life tomorrow -_-

and i need to know how to do these questions :

The point Q divides the line joining P(-1, -1, 0) to R(5, 2, -3) in the ratio 2:1. Find the co-ordinates of Q

Vectors.

The answer is (3, 1, -2)

Help is appreciated.

2. May 20, 2004

### AKG

$$\mbox{Let }\vec{Q}\ =\ (a,\ b,\ c)$$
$$\vec{Q}\ -\ \vec{P}\ =\ 2(\vec{R}\ -\ \vec{Q})$$
$$(a,\ b,\ c)\ -\ (-1,\ -1,\ 0)\ =\ 2[(5,\ 2,\ -3)\ -\ (a,\ b,\ c)]$$
$$a\ +\ 1\ =\ 10\ -\ 2a$$

From here, you should be able to figure out two more equations, and should find them very easy to solve, giving you a, b, and c, the co-ordinates of Q.

3. May 24, 2004

### VLHN

Disagree with the symbol of AKG, remember POIN is a non-direction Quantity, so that u cant use the VECTOR symbol . And your answer is right. When u take Q-P u will get the vector(PQ) , and the same with R-Q :)

4. May 24, 2004

### AKG

Every point in $\mathbb R^3$ has a vector representation. The point $Q(a, b, c)$ has a representation $\vec Q = (a, b, c) = a\mathbf i + b\mathbf j + c\mathbf k$. It made more sense to work with the vectors than to work with the points.

5. May 24, 2004

### VLHN

I think
1/If we get Q(a,b,c) >> a,b,c are the co-ordinate of the POIN Q
2/If we get vector(Q(a,b,c)) >> a,b,c are the co-ordinate of vector(OQ), with O(0,0,0)

I agree with you 100% in this kind, may be i understand Q = POINT, but u want to mean Q = vector in your solving :)