By Aleksandrov, Pavel Sergeevich

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**Sample text**

6 20 Vectors in 2 or 3 Dimensions There are two possibilities here. As you found in Exercise 2, a point dividing the line segment in a negative ratio must lie outside that line segment. Now any line cutting one side of a triangle internally must cut one of the other two sides internally. Since it goes into the triangle it must also come out. ) It must therefore cut the third side externally. The only other possibility is that the line does not cut any side of the triangle internally. In each case there is an odd number of negative ratios, and therefore the product will be negative, so we can see that the negative sign will always be correct.

Scalar Products and Equations of Planes 35 (ii) Consider the three planes whose equations are x+ y+2z==3 2x + 2y + 4z == 4 x - y + 3z == 2. We can see by inspection that the first two planes are parallel (and not coincident) so we immediately know that there is no point which is on all three planes. Thus the intersection of the three planes is the empty set which we denote by 0. Summary 1. The scalar product of the vectors a and b is defined as a. b == alb l and b == bli + b2j + b3k, + a2b2 + a3b3.

Ii) Suppose that Then == a + 2{1 + l' Y == 2a + {3 z == 2a + {3. x (1) (2) (3) From this we see that y == Z, and so there would be no solution in a, (3, l' for any vector for which the y and Z components are different. 3 cannot be expressed as a linear combination of u, v, t, and hence {u, v, t} is not a spanning set for ]R3. EXERCISE 4 Determine whether or not the set of vectors {u, v, w} is a spanning set for ]R3 in each of the following cases: In Chapter 1 we saw that any vector in lR? can be written as a linear combination of the vectors i, j and k.