16 determine whether the following system of

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16. Determine whether the following system of equations has a single point of intersection. If so, find the point of intersection. (7 marks)
Step #2: Calculate the cross product n 2 × n 3 . Let n 2 = u Let n 3 = w w = ( 1 , 2 , 3 ) × ( 2 , 3 , 0 ) w = u 2 w 3 w 2 u 3 ,u 3 w 1 w 3 u 1 ,u 1 w 2 w 1 u 2 w =( ( 2 ) ( 0 ) ( 3 ) ( 3 ) , ( 3 ) ( 2 ) ( 0 ) ( 1 ) , ( 1 ) ( 3 ) ( 2 ) ( 2 ) ) w = ( ( 0 −(− 9 ) ) , ( 6 0 ) , ( 3 4 ) ) w =( 9 , 6 , 7 ) Let w = n 2 × n 3 n 2 × n 3 =( 9,6, 7 ) Step #3: Calculate the dot product; n 1 ( n 2 × n 3 ) . n 1 ( n 2 × n 3 ) = ( 4,1, 9 ) ( 9,6, 7 ) n 1 ( n 2 × n 3 ) = [ ( 4 ) ( 9 ) + ( 1 ) ( 6 ) + ( 9 ) ( 7 ) ] n 1 ( n 2 × n 3 ) = 36 + 6 + 63 n 1 ( n 2 × n 3 ) = 105 Therefore, because n 1 ( n 2 × n 3 ) 0 the normal vectors are not coplanar and intersect at one point. Step #4: Use Elimination. 4 x + y 9 z = 0 [1] x + 2 y + 3 z = 0 [2] 2 x 3 y 5 = 0 [3]
Part 1: Subtract [ 1 ] × 2 by [ 2 ] . 8 x + 2 y 18 z = 0 [ 1 ] × 2 x + 2 y + 3 z = 0 [ 2 ] ¿ ¿ ¿ ¿ ¿ ¿ ¿ ¿ ¿ ¿ ¿ ¿ ¿ 7 x 21 z = 0 [ 4 ] Part 2: Add [ 1 ] × 2 by [ 3 ] . 12 x + 3 y 27 z = 0 [ 1 ] × 3 2 x 3 y 5 = 0 [ 3 ] ¿ ¿ ¿ ¿ ¿ ¿ ¿ ¿ ¿ ¿ ¿ ¿ ¿ 14 x 27 z 5 = 0 [ 5 ] Step #5: Determine z by Elimination; Subtract [ 4 ] × 2 by [ 5 ] . 14 x 42 z = 0 [ 4 ] × 2 14 x 27 z 5 = 0 [ 5 ]
¿ ¿ ¿ ¿ ¿ ¿ ¿ ¿ ¿ ¿ ¿ ¿ ¿ 15 z + 5 = 0 15 z =− 5 15 z 15 = 5 15 z = 1 3 Step #6: Substitute z into equation [4] to solve for x. 7 x 21 z = 0 [ 4 ] 7 x 21 ( 1 3 ) = 0 7 x 7 = 0 7 x = 7 7 x 7 = 7 7 x = 1
Step #7: Substitute x and z into equation one to solve for y. 4 x + y 9 z = 0 4 ( 1 ) + y 9 ( 1 3 ) = 0 4 + y 3 = 0 y + 1 = 0 y =− 1 Therefore, the point of intersection is ( 1, 1, 1 3 ) . 17. Find the shortest distance from P (–4, 2, 6) to the plane 2 x 3 y + z 8 = 0. marks) (4
Step #2: Determine another direction vector from point P to the plane by finding another point on the plane. .
Step #3: Determine the direction of PR.
Step #4: Calculate the projection of PQ on n to determine the distance between point P and the plane.
| pro j n ( PR ) | = | ( 5, 3, 3 ) ( 2, 3,1 ) ( 2, 3,1 ) ( 2, 3,1 ) | | ( 2, 3,1 ) | | pro j n ( PQ ) | = | [ ( 5 ) ( 2 ) + ( 3 ) ( 3 ) + ( 3 ) ( 1 ) ] [ ( 2 ) ( 2 ) + ( 3 ) ( 3 ) + ( 1 ) ( 1 ) ] | ( 2 ) 2 + ( 3 ) 2 + ( 1 ) 2 | pro j n ( PQ ) | = | ( 10 + 9 3 ) ( 4 + 9 + 1 ) | 4 + 9 + 1 | pro j n ( PQ ) | = | 16 14 | 14 | pro j n ( PQ ) | = 16 14 14 | pro j n ( PQ ) | = 4.28 units Therefore, the shortest distance between P and the plane is about 4.28 units.

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