If the lines x - 2 1 = y - 3 1 = z - 4 - k and x - 1 k = y - 4 2 = z - 5 1 are coplanar, find the value(s) of k .

The given lines are x - 2 1 = y - 3 1 = z - 4 - k ( 1 ) x - 1 k = y - 4 2 = z - 5 1 ( 2 ) On comparing equation ( 1 ) and ( 2 ) with lines x - x 1 a 1 = y - y 1 b 1 = z - z 1 c 1 and x - x 2 a 2 = y - y 2 b 2 = z - z 2 c 2 Now lines are coplanar if x 2 - x 1 y 2 - y 1 z 2 - z 1 a 1 b 1 c 1 a 2 b 2 c 2 = 0 Since lines ( 1 ) and ( 2 ) coplanar 1 - 2 4 - 3 5 - 4 1 1 - k k 2 1 = 0 ⇒ - 1 1 1 1 1 - k k 2 1 = 0 - 1 ( 1 + 2 k ) - 1 ( 1 + k 2 ) + 1 ( 2 - k ) = 0 ⇒ - 1 - 2 k - 1 - k 2 + 2 - k = 0 ⇒ - k ( 3 + k ) = 0 ⇒ k = 0 , - 3

If the lines x - 2 1 = y - 3 1 = z - 4 - k and x - 1 k = y - 4 2 = z - 5 1 are coplanar, find the value(s) of k .

The given lines are x - 2 1 = y - 3 1 = z - 4 - k ( 1 ) x - 1 k = y - 4 2 = z - 5 1 ( 2 ) On comparing equation ( 1 ) and ( 2 ) with lines x - x 1 a 1 = y - y 1 b 1 = z - z 1 c 1 and x - x 2 a 2 = y - y 2 b 2 = z - z 2 c 2 Now lines are coplanar if x 2 - x 1 y 2 - y 1 z 2 - z 1 a 1 b 1 c 1 a 2 b 2 c 2 = 0 Since lines ( 1 ) and ( 2 ) coplanar 1 - 2 4 - 3 5 - 4 1 1 - k k 2 1 = 0 ⇒ - 1 1 1 1 1 - k k 2 1 = 0 - 1 ( 1 + 2 k ) - 1 ( 1 + k 2 ) + 1 ( 2 - k ) = 0 ⇒ - 1 - 2 k - 1 - k 2 + 2 - k = 0 ⇒ - k ( 3 + k ) = 0 ⇒ k = 0 , - 3

If the lines x - 1 k = y - 3 - 4 = z - 2 5 and x - 2 3 = y - 3 k = z - 1 2 intersect at a point, find the values of k .

The given lines are x - 1 k = y - 3 - 4 = z - 2 5 ( 1 ) x - 2 3 = y - 3 k = z - 1 2 ( 2 ) On comparing equation ( 1 ) and equation ( 2 ) with lines x - x 1 a 1 = y - y 1 b 1 = z - z 1 c 1 and x - x 2 a 2 = y - y 2 b 2 = z - z 2 c 2 Now lines are coplanar if x 2 - x 1 y 2 - y 1 z 2 - z 1 a 1 b 1 c 1 a 2 b 2 c 2 = 0 Since lines ( 1 ) and ( 2 ) coplanar ∴ 2 - 1 3 - 2 1 - 3 k 2 3 3 k 2 = 0 ⇒ 1 1 - 2 k 2 3 3 k 2 = 0 Expanding along R 1 1 ( 4 - 3 k ) - 1 ( 2 k - 9 ) - 2 ( k 2 - 6 ) = 0 ⇒ 4 - 3 k - 2 k + 9 - 2 k 2 + 12 = 0 ⇒ 2 k 2 + 5 k - 25 = 0 ⇒ ( 2 k - 5 ) ( k + 5 ) = 0 ⇒ k = 5 2 , - 5

Show that the lines x + 3 - 3 = y - 1 1 = z - 5 5 and x + 1 - 1 = y - 2 2 = z - 5 5 are coplanar. Also find the equation of the plane containing the lines.

We know that x - x 1 l 1 = y - y 1 m 1 = z - z 1 n 1 and x - x 2 l 2 = y - y 2 m 2 = z - z 2 n 2 are co-planer if x 2 - x 1 y 2 - y 1 z 2 - z 1 l 1 m 1 n 1 l 2 m 2 n 2 = 0 Here, l 1 = - 3 ; m 1 = 1 ; n 1 = 5 ; l 2 = - 1 ; m 2 = 2 ; n 2 = 5 x 1 = - 3 ; , y 1 = 1 ; z 1 = 5 ; x 2 = - 1 ; y 2 = 2 ; z 2 = 5 Here, x 2 - x 1 y 2 - y 1 z 2 - z 1 l 1 m 1 n 1 l 2 m 2 n 2 = 2 1 0 - 3 1 5 - 1 2 5 Operate R 1 → R 1 + R 2 = - 1 2 5 - 3 1 5 - 1 2 5 R 1 and R 3 are identical. Thus, given lines are coplanar. Equation of plane containing given lines be given by x - x 1 y - y 1 z - z 1 l 1 m 1 n 1 l 2 m 2 n 2 Here, x 1 = - 3 ; , y 1 = 1 ; z 1 = 5 ; Equation of plane containing the given lines is x + 3 y - 1 z - 5 - 3 1 5 - 1 2 5 = 0 Expanding along R 1 ( x + 3 ) ( - 5 ) - ( y - 1 ) ( - 10 ) + ( z - 5 ) ( - 5 ) = 0 ⇒ - 5 ( x + 3 - 2 y + 2 + z - 5 ) = 0 ⇒ x - 2 y + z = 0 is the required equation of the plane.