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Math Help - how to convert vector equation to scalar equation

  1. #1
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    how to convert vector equation to scalar equation

    please help turn this vector equation [x,y,z]=[3,7,-5]+s[1,2,-1]+t[1,-2,3] into a scalar equation
    in the form of x+y+z+D=0

    i understand in the case of [x,y]=[3,7,-5]+s[1,2,-1]
    the direction vector is [1,2,-1] and the normal vector is [3,7,-5]
    all i need to do in this case was to do
    as the following
    [x-1](3)+[y-2](7)+[z+5](-5)
    which would give me the scalar equation 3x+7y-5z-42=0

    but how would you find the scalar equation if you have 2 direction vectors?

    i have a quiz tommorwT_T but still dont get this question... OTL
    hugs and kisses to anyone who can help/solve it for me....thanks in advance
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by antikv View Post
    please help turn this vector equation [x,y,z]=[3,7,-5]+s[1,2,-1]+t[1,-2,3] into a scalar equation
    in the form of x+y+z+D=0

    i understand in the case of [x,y]=[3,7,-5]+s[1,2,-1]
    the direction vector is [1,2,-1] and the normal vector is [3,7,-5]
    all i need to do in this case was to do
    as the following
    [x-1](3)+[y-2](7)+[z+5](-5)
    which would give me the scalar equation 3x+7y-5z-42=0

    but how would you find the scalar equation if you have 2 direction vectors?

    i have a quiz tommorwT_T but still dont get this question... OTL
    hugs and kisses to anyone who can help/solve it for me....thanks in advance
    To find the scalar equation of a plane you need its normal vector. So cross the two direction vectors

    (\mathbf{i}+2\mathbf{j}-\mathbf{k})\times (\mathbf{i}-2\mathbf{j}+3\mathbf{k})=4\mathbf{i}-4\mathbf{j}-4\mathbf{k}

    So the equation has the form

    4x-4y-4z=c

    Now just use the point (3,7,-5) to determine c.
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  3. #3
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    yay thank you so much..i love u
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  4. #4
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    A slightly different way: From the vector equation, x= 3+ s+ t, y= 7+ 2s- 2t, z= -5- s+ 3t. We can eliminate s by adding the first and third equations: x+ z= -2+ 4t. If we add two times the third equation to the second equation, we also eliminate s: y+ 2z= -3+ 4t. Finally, subtract y+ 2z=-3+ 4t from x+z= -2+ 4t to eliminate t: x+ z- y- 2z= x- y- z= 1. So the equation of the plane is x- y- z= 1, the same equation as before.
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