Section The given line has equations. x = 3 + t(13 3) = t, y = 2 + t(3 + 2) = 2 + 5t, z = 7 + t( 8 7) = 7 15t.

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1 . The given line has equations Section 8.8 x + t( ) + 0t, y + t( + ) + t, z 7 + t( 8 7) 7 t. The line meets the plane y 0 in the point (x, 0, z), where 0 + t, or t /. The corresponding values for x and z are 7 and, respectively.. E (B + C), F ( t)a + te, where t AF AE AF AF + F E AF/F E (AF/F E) +. F A + ( ) (B + C) A + (B + C) (A + B + C).. Let A (,, 4), B (,, ), C (,, 6). Then we prove AC t AB for some real t. We have AC AB, AC ( ) AB and consequently C is on the line AB. In fact A is between C and B, with AC AB. 4. The points P on the line AB which satisfy AP P B are given by P A + t AB, where t/( t) /. t/( t) ±/. The equation t/( t) / gives t /7 and hence P + 6/7 4 9/7. 7 /7 87.

2 P (6/7, 9/7, /7). The equation t/( t) / gives t / and hence P P (4/, /, /).. An equation for M is P A + t An equation for N is Q E + s 4 4/ / /. BC, which reduces to x + 6t y t z + 7t. EF, which reduces to x + 9s y z 8 + s. To find if and where M and N intersect, we set P Q and attempt to solve for s and t. We find the unique solution t, s /, proving that the lines meet in the point (x, y, z) ( + 6,, + 7) (7,, 0). 6. Let A (,, 6), B (, 7, 9), C (,, 7). Then (i) where cos ABC ( BA BC)/(BA BC), BA [,, ] t and BC [4, 6, ] t. cos ABC ABC π/ radians or

3 (ii) where cos BAC ( AB AC)/(AB AC), AB [,, ] t and AC [, 4, ] t. cos BAC (iii) ABC π/ radians or 90. where cos ACB ( CA CB)/(CA CB), CA [, 4, ] t and CB [ 4, 6, ] t. cos ACB ACB π/6 radians or By Theorem 8.., the closest point P on the line AB to the origin O is given by P A + t AB, where Now t / and A P AO AB t AB A AB AB. AB +. 6/ / / and P ( 6/, /, /). Consequently the shortest distance OP is given by ( 6 ) ( ) ( )

4 Alternatively, we can calculate the distance OP, where P is an arbitrary point on the line AB and then minimize OP : P A + t AB Consequently for all t; moreover when t /. + t + t + t + t OP ( + t) + ( + t) + ( + t) t ( 4t + 4 t 4 t + 4 ) ( { t } ) ( { t } ) + 0. OP 0 OP 0 8. We first find parametric equations for N by solving the equations x + y z x + y z 4.. The augmented matrix is [ 4 ], which reduces to [ 0 / / 0 / / x + z, y z, with z arbitrary. Taking z 0 gives a point A (,, 0), while z gives a point B (,, ). 90 ].

5 if C (, 0, ), then the closest point on N to C is given by AB, where t ( AC AB)/AB. P A + t Now so AC P t / / / / 0 and AB + ( + ) ( + ) + + / / / /. 4/ 7/ / so P (4/, 7/, /). Also the shortest distance P C is given by ( P C 4 ) ( ) ( + ),, The intersection of the planes x + y z 4 and x y + z is the line given by the equations x 9 + z, y + 7 z, where z is arbitrary. the line L has a direction vector [/, 7/, ] t or the simpler [, 7, ] t. Then any plane of the form x + 7y + z d will be perpendicualr to L. The required plane has to pass through the point (6, 0, ), so this determines d: d The length of the projection of the segment AB onto the line CD is given by the formula CD AB. CD Here CD [ 8, 4, ] t and AB [4, 4, ] t, so CD AB ( 8) ( 4) + ( ) CD ( 8) ( )

6 . A direction vector for L is given by BC [,, ] t. the plane through A perpendicular to L is given by x y + z ( ) + ( ) ( ) + 7. The position vector P of an arbitrary point P on L is given by P B+t BC, or x y + t, z 4 or equivalently x t, y t, z 4 + 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: ( t) ( t) + (4 + t) 7, which gives t 7/8. P ( 8, 8, ) 8 and ( AP ) ( ) ( 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 ( r)d + re, 0 < r < ; D ( s)b + sa, 0 < s < ; E ( t)b + tc, 0 < t <. ) P ( r) {( s)b + sa} + r {( t)b + tc} ( r)sa + {( r)( s) + r( t)} B + rtc αa + βb + γc, 9

7 where α ( r)s, β ( r)( s) + r( t), γ rt. Then 0 < α <, 0 < γ <, 0 < β < ( r) + r. Also α + β + γ ( r)s + ( r)( s) + r( t) + rt.. The line AB is given by P A + t[, 4, ] t, or x 6 + t, y + 4t, z + t. Then B is found by substituting these expressions in the plane equation x + 4y + z 0. We find t 9/0 and consequently ( B 6 77 ) 6 9,, ( 0, 86 0, ). 0 Then AB AB t 4 t Let A (, 0, ), B (6,, 4), C (,, 0). Then the area of triangle ABC is AB AC. Now AB AC AB AC. 9. Let A (,, 4), A (,, ), A (4,, ). Then the point P (x, y, z) lies on the plane A A A if and only if A P ( A A A A ) 0,

8 or x y z 4 x 7y + 6z Non parallel lines L and M in three dimensional space are given by equations P A + sx, Q B + ty. (i) Suppose P Q is orthogonal to both X and Y. Now P Q Q P (B + ty ) (A + sx) AB +ty sx. More explicitly ( AB +ty + sx) X 0 ( AB +ty + sx) Y 0. t(y X) s(x X) AB X t(y Y ) s(x Y ) AB Y. However the coefficient determinant of this system of linear equations in t and s is equal to Y X X X Y Y X Y (X Y ) + (X X)(Y Y ) X Y 0, as X 0, Y 0 and X and Y are not proportional (L and M are not parallel). (ii) P and Q can be viewed as the projections of C and D onto the line P Q, where C and D are arbitrary points on the lines L and M, respectively. by equation (8.4) of Theorem 8.., we have P Q CD. Finally we derive a useful formula for P Q. Again by Theorem 8.. P Q AB P Q AB ˆn, P Q 94

9 where ˆn P Q L z C M D P Q O y x P Q is a unit vector which is orthogonal to X and Y. where t ±/ X Y. ˆn t(x Y ), P Q AB (X Y ). X Y 7. We use the formula of the previous question. Line L has the equation P A + sx, where X AC. Line M has the equation Q B + ty, where Y BD. X Y [ 6,, ] t and X Y 6. 9

10 the shortest distance between lines AC and BD is equal to 0 6 AB (X Y ). X Y Let E be the foot of the perpendicular from A 4 to the plane A A A. Then vol A A A A 4 ( area A A A ) A 4 E. Now area A A A A A A A. Also A 4 E is the length of the projection of A A 4 onto the line A 4 E. (See figure above.) A 4 E A A 4 X, where X is a unit direction vector for the line A 4 E. We can take A A A A X A A A. A vol A A A A 4 6 A A A A 6 A A 4 ( A A A A 4 ( A A A A ) A A A A A A ) 96

11 6 ( A A 9. We have CB [, 4, ] t, A A ) CD [,, 0] t, A A 4. CB CD i + j + k, AD [, 0, ] t. so the vector i + j + k is perpendicular to the plane BCD. Now the plane BCD has equation x + y + z 9, as B (,, ) is on the plane. Also the line through A normal to plane BCD has equation x y z + t ( + t) x + t, y + t, z ( + t). [We remark that this line meets plane BCD in a point E which is given by a value of t found by solving ( + t) + ( + t) + ( + t) 9. So t / and E (/, /, /).] The distance from A to plane BCD is To find the distance between lines AD and BC, we first note that (a) The equation of AD is P (b) The equation of BC is Q + t + s t + t + s + 4s s. ;

12 C E D B Then P Q [ + s t, + 4s, 4 s t] t and we find s and t by solving the equations P Q AD 0 and P Q BC 0, or ( + s t) + ( + 4s)0 + ( 4 s t) 0 ( + s t) + 4( + 4s) ( 4 s t) 0. t / s. Correspondingly, P ( /,, 7/) and Q (/, 0, /). Thus we have found the closest points P and Q on the respective lines AD and BC. Finally the shortest distance between the lines is P Q P Q. 98