Linear Algebra Solutions 89 - � 2 = √ 11134 38 = √ 293 × 38 38 = √ 293 √ 38 12 Let P be a point inside the triangle ABC Then the line through P

11. A direction vector for L is given by - BC = [ - 5 , - 2 , 3] t . Hence the plane through A perpendicular to L is given by - 5 x - 2 y + 3 z = ( - 5) × 3 + ( - 2) × ( - 1) + 3 × 2 = - 7 . The position vector P of an arbitrary point P on L is given by P = B + t - BC , or   x y z   =   2 1 4   + t   - 5 - 2 3   , or equivalently x = 2 - 5 t, y = 1 - 2 t, z = 4 + 3 t . To find the intersection of line L and the given plane, we substitute the expressions for x, y, z found in terms of t into the plane equation and solve the resulting linear equation for t : - 5(2 - 5 t ) - 2(1 - 2 t ) + 3(4 + 3 t ) = - 7 , which gives t = - 7 / 38. Hence P = ( 111 38 , 52 38 , 131 38 ) and AP = s 3 - 111 38 ¶ 2 + -

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Unformatted text preview: ¶ 2 = √ 11134 38 = √ 293 × 38 38 = √ 293 √ 38 . 12. Let P be a point inside the triangle ABC . Then the line through P and parallel to AC will meet the segments AB and BC in D and E , respectively. Then P = (1-r ) D + r E , < r < 1; D = (1-s ) B + s A , < s < 1; E = (1-t ) B + t C , < t < 1 . Hence P = (1-r ) { (1-s ) B + s A } + r { (1-t ) B + t C } = (1-r ) s A + { (1-r )(1-s ) + r (1-t ) } B + rt C = α A + β B + γ C , 92...
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