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    fraser james's Avatar
    fraser james Posts: 2, Reputation: 1
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    #1

    May 25, 2012, 10:01 PM
    A vector equation for a given straight line is r = (i + 3j) + lambda (-i - j)
    a) Show that the point (1; 2) does not lie on this line.
    b) Construct a vector equation for the line that does go through the point (1; 2), and is perpendicular
    to r.
    c) Determine the point of intersection of the two lines.
    jo_king's Avatar
    jo_king Posts: 1, Reputation: 1
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    #2

    May 28, 2012, 12:51 AM
    Not an answer exactly, but here are some tips:
    Try drawing the picture with as much info as you can. First part is the position vector (1+3j) so point (1,3) exists on the line - call it A. Second part is the direction vector (-i-j). You should be able to work out a 2nd point on the line from these two vectors - call it B. The direction vector is calculated from the difference in 2 points (A and B).

    You can also plot a third point (1,2) to create a triangle. Which should conveniently be a std triangle, and will tell you the magnitude of vector AB.

    From there use the theory of projecting one vector onto another to work out the various components of your triangle. Remember that the hypotenuse vector = the sum of the two side vectors. This can also be rearranged and will give you the perpendicular vector. Then you just need to convert it into a vector equation line - meaning you will have to have a position vector (use point 1,2 to find this) and a direction vector (which you'll find above).

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