If A ( 4 , 8 , 12 ) , B ( 2 , 4 , 6 ) , C ( 3 , 5 , 4 ) and D ( 5 , 8 , 5 ) are four points, show that the lines A B ↔ and C D ↔ intersect.

Given points are A ( 4 , 8 , 12 ) , B ( 2 , 4 , 6 ) , C ( 3 , 5 , 4 ) , D ( 5 , 8 , 5 ) Equation of A B ↔ is x - 2 2 = y - 4 4 = z - 6 6 = t ⇒ x - 2 1 = y - 4 2 = z - 6 3 = t ...(i) Any point on (i) is P ( 2 + t , 4 + 2 t , 6 + 3 t ) Equation of C D ↔ is x - 3 2 = y - 5 3 = z - 4 1 = s ...(ii) Any point on (ii) is Q ( 3 + 2 s , 5 + 3 s , 4 + s ) Lines are intersecting, P = Q ⇒ ( 2 + t , 4 + 2 t , 6 + 3 t ) = ( 3 + 2 s , 5 + 3 s , 4 + s ) ⇒ t - 2 s = 1 ...(iii), 2 t - 3 s = 1 ...(iv), 3 t - s = - 2 ...(v) Solving (iii) and (iv), t = - 1 , s = - 1 Substituting t , s values in (v), 3 ( - 1 ) - ( - 1 ) = - 2 ⇒ - 2 = - 2 Therefore, (v) is satisfied by the values of t and s . Hence, the lines are intersecting lines.

Find the point of intersection of the lines A B → and C D → where A = ( 7 , - 6 , 1 ) , B = ( 17 , - 18 , - 3 ) , C = ( 1 , 4 , - 5 ) and D = ( 3 , - 4 , 11 ) .

Given points are A = ( 7 , - 6 , 1 ) , B = ( 17 , - 18 , - 3 ) , C = ( 1 , 4 , - 5 ) , D = ( 3 , - 4 , 11 ) Equation of A B ↔ is x - 7 17 - 10 = y + 6 - 18 + 6 = z - 1 - 3 - 1 = t ⇒ x - 7 10 = y + 6 - 12 = z - 1 - 4 = t ...(i) Any point on (i) is P ( 10 t + 7 , - 12 t - 6 , - 4 t + 1 ) Equation of C D ↔ is x - 1 3 - 1 = y - 4 - 4 - 4 = z + 5 11 + 5 = s ⇒ x - 1 2 = y - 4 - 8 = z + 5 16 = s ...(ii) Any point on (ii) is Q ( 2 s + 1 , - 8 s + 4 , 16 s - 5 ) Lines are intersecting, P = Q ⇒ ( 10 t + 7 , - 12 t - 6 , - 4 t + 1 ) = 2 s + 1 , - 8 s + 4 , 16 s - 5 ⇒ 10 t - 2 s = - 6 ...(iii), - 12 t + 8 s = 10 ...(iv), - 4 t - 16 s = - 6 ...(v) Solving (iii) and (iv), t = - 1 2 , s = 1 2 Substituting t , s values in (v), - 4 - 1 2 - 16 1 2 = - 6 ⇒ - 6 = - 6 Therefore, (v) is satisfied by the values of t and s . Hence, the lines are intersecting lines. Substitute t = - 1 2 in point P , then P = 10 - 1 2 + 7 , - 12 - 1 2 - 6 , - 4 - 1 2 + 1 ∴ P = ( 2 , 0 , 3 )

A ( 3 , 2 , 0 ) , B ( 5 , 3 , 2 ) , C ( - 9 , 6 , - 3 ) are vertices of a triangle. A D ¯ , the bisector of ∠ B A C meets B C ¯ at D . Find the coordinates of D .

A ( 3 , 2 , 0 ) , B ( 5 , 3 , 2 ) , C ( - 9 , 6 , - 3 ) are the vertices of Δ A B C . The distance between A x 1 , y 1 and B x 2 , y 2 is A B = x 1 - x 2 2 + y 1 - y 2 2 + z 1 - z 2 2 . A B = ( 3 - 5 ) 2 + ( 2 - 3 ) 2 + ( 0 - 2 ) 2 = 4 + 1 + 4 = 3 A C = ( 3 + 9 ) 2 + ( 2 - 6 ) 2 + ( 0 + 3 ) 2 = 144 + 16 + 9 = 169 = 13 Let, A D be the bisector of ∠ B A C . Then D divides B C in the ratio A B : A C i.e., 3 : 13 ∴ Coordinates of D are 3 ( - 9 ) + 13 · 5 3 + 13 , 3 · 6 + 13 · 3 3 + 13 , 3 ( - 3 ) + 13 · 2 3 + 13 = 38 16 , 57 16 , 17 16