Find parametric equations for the line in which the planes x+2y+z=1 and x−y+2z=−8 intersect.

Question

Find parametric equations for the line in which the planes  x+2y+z=1 and x−y+2z=−8 intersect.

Lauren Fuller · Accepted Answer

Step 1  To find the parametric equations for the line in which the planes  x+2y+z=1 and x−y+2z−8 intersect  Step 2  We have given the equations of the planes:  1) x+2y+z=1  and  2) x−y+2z=−8  we have to find the parametric equations for the line in which the planes intersect.  Firstly we find the point on the line by setting z=0  From (1) we get:   3) x+2y=1  and from (2)  4) x−y=−8  From (3) and (4) we get:  x=−5 and y=3  So we get: (−5, 3, 0) is a point on the line.  Now the Normal vectors of the planes are  n1→=(1, 2, 1)  n2→=(1, −1, 2)  Now the normal vector perpendicular to both is:  a→=n1→×n2→  =|i^j^k^1211−12|  =i^(5)−j^(1)+k^(−3)  =⟨5, −1, −3⟩  Therefore the line passes through (−5, 3, 0) and parallel to the vector a→  =⟨5, −1, −3⟩  As we know the parametric equation of a line passing through the point  (x0, y0, z0)  and parallel to the vector ⟨a, b, c⟩ is given by  x=x0+at, y=y0+bt, z=z0=ct  Here (x0, y0, z0)=(−5, 3, 0)  and ⟨a, b, c⟩=⟨5, −1, −3⟩  So the parametric equation of the line is  x=−5+5t, y=3−t, z=−3t  Therefore the parametric equations for the line in which planes x+2y+z=1  and x−y+2z=−8 intersect is:  x=−5+5t, y=3−t, z=−3t

Find parametric equations for the line in which the planes x+2y+z=1

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