Find the vector equation of the plane passing through the point having position vector i ^ + j ^ + k ^ and perpendicular to the vector 4 i ^ + 5 j ^ + 6 k ^ .

Given, The plane passing through the point having position vector i ^ + j ^ + k ^ and perpendicular to the vector 4 i ^ + 5 j ^ + 6 k ^ , Now let a → = i ^ + j ^ + k ^ and given perpendicular vector n → = 4 i ^ + 5 j ^ + 6 k ^ , Now we know that equation of plane passing through point a → and perpendicular to vector n → is given by r → · n → = a → · n → , Now putting the value in equation and taking dot product we get, r → · 4 i ^ + 5 j ^ + 6 k ^ = i ^ + j ^ + k ^ · 4 i ^ + 5 j ^ + 6 k ^ ⇒ r → · 4 i ^ + 5 j ^ + 6 k ^ = 4 + 5 + 6 ⇒ r → · 4 i ^ + 5 j ^ + 6 k ^ = 15 Which is a required vector equation.

Find the Cartesian equation of a line passing through the points 3 , - 2 , - 5 and 3 , - 2 , 6 .

Let points A 3 , - 2 , - 5 and B 3 , - 2 , 6 . Equation of line passing through two points x 1 , y 1 , z 1 and x 2 , y 2 , z 2 is x - x 1 x 2 - x 1 = y - y 1 y 2 - y 1 = z - z 1 z 2 - z 1 . Equation of line passing through points A and B is x - 3 3 - 3 = y - - 2 - 2 - - 2 = z - - 5 6 - - 5 ⇒ x - 3 0 = y + 2 0 = z + 5 11

Find the shortest distance between the lines x + 1 7 = y + 1 - 6 = z + 1 1 and x - 3 1 = y - 5 - 2 = z - 7 1 .

We know that the shortest distance between the two 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 is given by, d = x 2 - x 1 y 2 - y 1 z 2 - z 1 a 1 b 1 c 1 a 2 b 2 c 2 b 1 c 2 - b 2 c 1 2 + c 1 a 2 - c 2 a 1 2 + a 1 b 2 - a 2 b 1 2 . . . . . . . i Given equations of line are x + 1 7 = y + 1 - 6 = z + 1 1 and x - 3 1 = y - 5 - 2 = z - 7 1 . Comparing with the given equations, we get x 1 = - 1 , y 1 = - 1 , z 1 = - 1 a 1 = 7 , b 1 = - 6 , c 1 = 1 x 2 = 3 , y 2 = 5 , z 2 = 7 a 2 = 1 , b 2 = - 2 , c 2 = 1 Therefore, x 2 - x 1 y 2 - y 1 z 2 - z 1 a 1 b 1 c 1 a 2 b 2 c 2 = 4 6 8 7 - 6 1 1 - 2 1 = 4 - 6 + 2 - 6 7 - 1 + 8 - 14 + 6 = - 16 - 36 - 64 = - 116 ⇒ ( b 1 c 2 - b 2 c 1 ) 2 + ( c 1 a 2 - c 2 a 1 ) 2 + ( a 1 b 2 - a 2 b 1 ) 2 = ( - 6 + 2 ) 2 + ( 1 - 7 ) 2 + ( - 14 + 6 ) 2 = 16 + 36 + 64 = 116 = 2 29 Substituting all the values in equation i , we get d = - 116 2 29 = - 58 29 = - 2 × 29 29 = 2 29 units The distance between the lines x + 1 7 = y + 1 - 6 = z + 1 1 and x - 3 1 = y - 5 - 2 = z - 7 1 is 2 29 units.