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    Find a cartesian equation of a plane which passes through the origin 0 and contains the line with equations

    (x-1)/2=(y-2)/3=(z-3)/4
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    (Original post by ziya_the_king)
    Find a cartesian equation of a plane which passes through the origin 0 and contains the line with equations

    (x-1)/2=(y-2)/3=(z-3)/4
    Let (x-1)/2=(y-2)/3=(z-3)/4 = t.
    Hence x=2t+1, y=3t+2, z=4t+3.
    So the vector equation of the line is i[2t+1] + j[3t+2] + k[4t+3]=i+2j+3k+t[2i+3j+4k]
    So the plane passes through (0,0,0) and (1,2,3) and is parallel to the line with direction vector 2i+3j+4k [As the line lies in the plane].
    The plane is also parallel to i+2j+3k [The direction between the two points on the plane mentioned above.
    You can hence find 'n', a vector normal to the plane, by evaluating the determinant of i,j,k and the two vectors which the plane is parallel to.
    You can use 'a', a point on the plane as the origin.
    You can thus express the equation of the plane in the form r.n=p and after letting r=xi+yj+zk you have the cartesian equation.
 
 
 
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Updated: May 17, 2005

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