Find a Cartesian equation for the plane containing the line "l" and the point c?

Sonsider the points A(1,1,13), B(2,0,8) and C(1,-1,7).


Find "l" the parametric vector equation of line through A and B.


Find a Cartesian equation for the plane containing the line "l" and the point C

Find a Cartesian equation for the plane containing the line "l" and the point c?
1) Find the parametric vector equation of the line thru the points A(1,1,13) and B(2,0,8).





Define vector u.





u = AB = %26lt;B - A%26gt; = %26lt;2-1, 0-1, 8-13%26gt; = %26lt;1, -1, -5%26gt;





The equation of the line thru points A and B is:





L(t) = A + tu


L(t) = %26lt;1, 1, 13%26gt; + t%26lt;1, -1, -5%26gt;


where parameter t ranges over the real numbers


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2) Find a Cartesian equation for the plane containing the line L(t) and the point C(1, -1, 17).





Define a second directional vector v, of the plane.





v = AC = %26lt;C - A%26gt; = %26lt;1-1, -1-1, 7-13%26gt; = %26lt;0, -2, -6%26gt;





Any non-zero multiple of v is also a directional vector of the plane. Divide by -2.





v = %26lt;0, 1, 3%26gt;





The normal vector n, of the plane is orthogonal to any vector that lies in the plane. Take the cross product.





n = u X v = %26lt;1, -1, -5%26gt; X %26lt;0, 1, 3%26gt; = %26lt;2, -3, 1%26gt;





With a point in the plane and a normal vector to the plane we can write the equation of the plane. Remember, the normal vector is orthogonal to any vector that lies in the plane. And the dot product of orthogonal vectors is zero. Define R(x,y,z) as an arbitrary point in the plane. Then vector AR lies in the plane.





n • AR = 0


n • %26lt;R - A%26gt; = 0


%26lt;2, -3, 1%26gt; • %26lt;x - 1, y - 1, z - 13%26gt; = 0


2(x - 1) - 3(y - 1) + 1(z - 13) = 0


2x - 2 - 3y + 3 + z - 13 = 0


2x - 3y + z - 12 = 0
Reply:A*2.36842*b/c*0.66(c*a)


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1/c*(c/b/a) =l





This works.

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