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So I really need help with figuring out how to do this problem. I answered two of the questions (not sure if I am right) and I am just stuck on the 2nd part.

Questions: Two planes are given by the equations Q:2x+y−3z=2 and R:−x+2y−z=1.

Question 1: Find a vector normal to the plane Q. Find an equation of the plane which is parallel to Q and passes through the point (1,2,−1).

Attempted answer: :

A vector normal to plane: <2,1,-3>

A normal vector for new plane is <2,1,-3>. A pt on plane is <1,2,-3>

T/f equation is

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

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

Question 2: Find a vector normal to the plane R. Find a vector equation of the line that is perpendicular to the plane R and passes through the point (0,0,−1).

Attempted Answer: Don't know how to start this one and help on how to start would be appreciated.

Question 3: Find a vector equation of the line in the intersection of Q and R.

Attempted answer:

  • i j k
  • 2 1 -3
  • -1 2 -1

6i+5j+5k = 9.27

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  • $\begingroup$ For question 2, think about what “normal” means. $\endgroup$
    – amd
    Commented Sep 14, 2018 at 5:28

1 Answer 1

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You have done question 1 correctly.

Question 2 asks for a line normal to $R$ and passing through $(0,0,-1)$.

Thus your equation is $$ r =(0,0,-1)+t(-1,2,-1) $$

Question 3 requires either two points on the intersection or one point on the intersection and the direction vector which is found by cross product of normals to the planes.

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