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# Vectors P6 watch

1. 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
2. (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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