2011_Solution_6 - MA 1505 Mathematics I Tutorial 6...

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MA 1505 Mathematics I Tutorial 6 Solutions 1. Let a = 3 i + 2 j + k . Then u = proj w a is parallel to w and v = a - proj w a is perpendicular to w and a = u + v . We compute u = a · w k w k 2 w = 3 + 6 + 4 1 + 9 + 16 ( i + 3 j + 4 k ) = 1 2 ( i + 3 j + 4 k ) and v = a - u = (3 i + 2 j + k ) - 1 2 ( i + 3 j + 4 k ) = 1 2 (5 i + j - 2 k ) . 2. At the point of intersection we have 2 t + 2 = 4 s - 12 = 2 s - t = 7; t + 2 = 2 s - 5 = 2 s - t = 7; 3 t + 3 = s - 3 = s - 3 t = 6 . It follows that t = - 1 and s = 3 , and the point of intersection is (0 , 1 , 0) . A normal to the plane is given by (2 i + j + 3 k ) × (4 i + 2 j + k ) = - 5 i + 10 j + 0 k . An equation of the plane takes the form - 5 x + 10 y = - 5 · 0 + 10 · 1 + 0 · 0 = ⇒ - 5 x + 10 y = 10 = ⇒ - x + 2 y = 2 . 3. (a) --→ AB = (3 i + 0 j + k ) - (3 i + 3 j + 0 k ) = 0 i - 3 j + k , and -→ AC = (0 i + 2 j + k ) - (3 i + 3 j + 0 k ) = - 3 i - j + 1 k . A vector normal to the plane is
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2011_Solution_6 - MA 1505 Mathematics I Tutorial 6...

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