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    Hi, I have two questions here, i have been trying to draw pictures to represent these, but still i have no idea how to proceed to working out what the question wants me to.

    1) Show that the vector from the origin to the point with coordinates (1,1,1) is perpendicular to the plane that cuts the x,y and z axes at the points (1,0,0), (0,1,0) and (0,0,1) respectively.

    2) Call a' the projection of the vector a onto a plane. If n(hat) is the unit vector perpendicular to this plane show that
    a' = a - (a.n)n

    Thanks, if you could point me in the right way please.
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    1. Find two non-parallel vectors in the plane (e.g. (1, -1, 0) - do you see where I got that from?), and show that (1, 1, 1) is perpendicular to both by using the dot (scalar) product.

    2. Well, a.n is (the length of) the projection of a onto n, and (a.n)n is this projection in the direction of n. What do you get if you then subtract this from a? Draw a vector triangle.
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    2. Well, a.n is (the length of) the projection of a onto n, and (a.n)n is this projection in the direction of n. What do you get if you then subtract this from a? Draw a vector triangle.[/QUOTE]

    Yes, i tried to draw a triangle, but the labelling is hell as I couldn't even understand what exactly each unit means(not all of them, at least). I have a right angled triangle with the hypotenuse being the vector a. then the vector along the bottom of the hypotenuse is a.n; does that mean that a' is the vertical then?
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    (Original post by shengoc)
    2. Well, a.n is (the length of) the projection of a onto n, and (a.n)n is this projection in the direction of n. What do you get if you then subtract this from a? Draw a vector triangle.
    Yes, i tried to draw a triangle, but the labelling is hell as I couldn't even understand what exactly each unit means(not all of them, at least). I have a right angled triangle with the hypotenuse being the vector a. then the vector along the bottom of the hypotenuse is a.n; does that mean that a' is the vertical then?[/QUOTE]
    The hypotenuse is a, and the side along the vector n is (a.n)n. We know that a' = a - (a.n)n. So what's a'?
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    Cool, i think i get what you are trying to say. Thanks.
 
 
 
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