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# Vectors

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You are given the vectors
X = (1,1,1), y = (2,1,1) and z = (6,2,2).

(i) Find the Cartesian equation of the plane &#928; normal to the vector x containing the point (2,1,1).

(ii) Find the parametric equation of the line l through the points (2,1,1) and (6,2,2).

(iii) If l' is given parametrically by l' = x + ty (with x and y defined above), find the point of intersection of l' with the plane &#928;.

(iv) Find x x y, y x z and [x, y, z].

Give an explanation in geometric terms of what you have found.

##### Solution Summary

This shows how to find Cartesian and parametric equations, and describe findings geometrically.

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You are given the vectors
x = (1,1,1), y = (2,1,1) and z = (6,2,2).

(i) Find the Cartesian equation of the plane _ normal to the vector x containing the point (2,1,1).

So what can we say about the plane? Well, we know that if it's normal to x, any vector within the plane is going to have a zero scalar (dot) product with x. Therefore, we can use this knowledge to completely define the orientation of the plane. Trouble is, it could be any one of an infinite number of parallel planes! But we're told that it contains the point (2,1,1), so this makes the answer unique.

Let's pick a vector in the plane that we're trying to find. We don't know anything about it, so we'll call it (x,y,z). Now if this vector (a,b,c) is normal to x, then their scalar product will be zero.

x . (x, y, z) = (1*x) +(1*y) + (1*z) = 0

so x+y+z=0 is a plane with the correct orientation.

But here's something interesting. How about the plane with the ...

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