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2 t, 3- 3 and (-2r, 3-, ) 11.) Determine the point of intersection of the lines Note:4,1,2, with (1,2/3,-1)^k(-2.-1,1)=kv;, for any kER. So .1,. Change one of the parameters to s, then equate the corresponding coordinates of the lines and solve for t, and s. substitute the values of t, and s in their respective lines to get the required point. Locate the point of intersection of the plane 2x+ y-z-0 and the line through (3,1,0) 12.) that is perpendicular to the plane. ( Be thorough in your explanation or work) Note: let (2,1,-1)1 plane, and n | line, line, 1: (3+2t,1+t,-t). Substitute the coordinates of into the equation of the plane and solve for t. Use this value in I to get the required point Find an equation of the plane that passes through the line of intersection of the planes 13.) x-z 1 and y + 2z 3 and is perpendicular to the plane (Be thorough in your explanation or work) Note: Let n, (1,0,-1), n, = (0,1,2), n, = (1,1.-2), n, I plane i. Let P(1,3,0) Eline. Let , x Let -x,, P(xy,z). Then the equation of the plane is given by: PP-0 x+y-2z 1

Answer 1

Lines: - (1 + t, 3 + 2/3 t, 1 - t): - (- 2 t, 3 - t, t) The point intersection will pass through both the Lines. Let this point be (x, y, z) Thus we have: - x = 1 + t = -2 S 1 y = 3 + 2 t /3 = 3 - S 2 z = 1 - t = s 3 Solving these we have S = -2 t = 3 using this we have: - x = 4, y = 5, z = -2 Point of intersection = (4,5, -2) \r\n . Place: - 2 x + y - z = 0 Point: - (3, 1, 0) = P Let the point of intersection be (x, y, z) Direction cosine of P & (x, y, z) is given as