InterviewSolution
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Find the equation of the plane through the points A( 3, 4, 2) and B(7, 0, 6) and perpendicular to the plane 2x – 5y = 15. HINT: The given plane is 2x – 5y + 0z = 15 |
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Answer» Plane passes through (3,4,2) and (7,0,6), A(x - 3) + B(y - 4) + C(z - 2) = 0 (1) A(x - 7) + B(y - 0) + C(z - 6) = 0 (2) Subtracting (1) from (2), A(x - 7 - x + 3) + B(y - y + 4) + C(z - 6 - z + 2) = 0 -4A + 4B - 4C = 0 A - B + C = 0 B = A + C (3) Now plane is perpendicular to 2x - 5y =15 2A - 5B = 0 (4) Using (3) in (4) 2A-5(A + C) = 0 2A - 5A - 5C = 0 -3A - 5C = 0 C = \(\frac{-3}5\)A B = A + \(\frac{-3}5\)A ⇒ \(\frac{2}5\)A Putting values in equation (1) A(x - 3) + \(\frac{2}5\)A (y - 4) + \(\frac{-3}5\)A(z - 2) = 0 A(5(x - 3) + 2(y - 4) - 3(z -2) = 0 5x + 2y - 3z -15-8 + 6 = 0 5x + 2y - 3z -17 = 0 So, required equation of plane is 5x + 2y - 3z -17 = 0. |
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