Show that the line of intersection of the planes 2x+y-3z=1 and 2x-3y-2z=-4 is perpendicular to the line r = (1,3,2) + k(2,-3,-2)
Added by Cassandra J.
Step 1
- The normal vector of the plane 2x + y - 3z = 1 is \(\mathbf{n_1} = (2, 1, -3)\). - The normal vector of the plane 2x - 3y - 2z = -4 is \(\mathbf{n_2} = (2, -3, -2)\). Show more…
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