Find the shortest distance between the lines r → = i ^ + 2 j ^ + 3 k ^ + λ ( i ^ - j ^ + k ^ ) and r → = 2 i ^ - j ^ - k ^ + μ ( - i ^ + j ^ - k ^ ) .

Given lines are r → = ( i + 2 j + 3 k ) + λ ( i - j + k ) r → = ( 2 i - j - k ) + μ ( - i + j - k ) ⇒ r → = ( 2 i - j - k ) + μ ' ( i - j + k ) Where μ ' = - μ Hence, both given lines are parallel to b → = i - j + k Here a 1 → = i + 2 j + 3 k ; a 2 → = 2 i - j - k Now a 2 → - a 1 → = ( 2 i - j - k ) - ( i + 2 j + 3 k ) = i - 3 j - 4 k ∴ ( a 2 → - a 1 → ) × b → = i j k 1 - 3 - 4 1 - 1 1 = i ( - 3 - 4 ) - j ( 1 + 4 ) + k ( - 1 + 3 ) = - 7 i - 5 j + 2 k | ( a 2 → - a 1 → ) | × b → | = ( - 7 ) 2 + ( - 5 ) 2 + 2 2 = 49 + 25 + 4 = 78 | b → | = 1 2 + 1 2 + 1 2 = 3 ∴ Shortest distance = | ( a 2 → - a 1 → ) × b → | | b → | = 78 3 = 26

Find the shortest distance between the lines r → = i ^ + 2 j ^ + 3 k ^ + λ ( i ^ - j ^ + k ^ ) and r → = 2 i ^ - j ^ - k ^ + μ ( - i ^ + j ^ - k ^ ) .

Given lines are r → = ( i + 2 j + 3 k ) + λ ( i - j + k ) r → = ( 2 i - j - k ) + μ ( - i + j - k ) ⇒ r → = ( 2 i - j - k ) + μ ' ( i - j + k ) Where μ ' = - μ Hence, both given lines are parallel to b → = i - j + k Here a 1 → = i + 2 j + 3 k ; a 2 → = 2 i - j - k Now a 2 → - a 1 → = ( 2 i - j - k ) - ( i + 2 j + 3 k ) = i - 3 j - 4 k ∴ ( a 2 → - a 1 → ) × b → = i j k 1 - 3 - 4 1 - 1 1 = i ( - 3 - 4 ) - j ( 1 + 4 ) + k ( - 1 + 3 ) = - 7 i - 5 j + 2 k | ( a 2 → - a 1 → ) | × b → | = ( - 7 ) 2 + ( - 5 ) 2 + 2 2 = 49 + 25 + 4 = 78 | b → | = 1 2 + 1 2 + 1 2 = 3 ∴ Shortest distance = | ( a 2 → - a 1 → ) × b → | | b → | = 78 3 = 26

Find the vector equation of the line passing through the point ( - 1 , 2 , 1 ) and parallel to the line r → = 2 i ^ + 3 j ^ - k ^ + λ ( i ^ - 2 j ^ + k ^ ) .Also, find the distance between these lines.

Equation of line be r → = 2 i + 3 j - k + λ ( i - 2 j + k ) ... ( 1 ) Comparing with r → = a 1 → + λ b → where a 1 → = 2 i + 3 j - k ; b → = i - 2 j + k Now vector equation of line through the point ( - 1 , 2 , 1 ) and parallel to line ( 1 ) be given by r → = a 2 → + λ b → where a 2 → = - i + 2 j + k and b → = i - 2 j + k ∴ a 2 → - a 1 → = ( - i + 2 j + k ) - ( 2 i + 3 j - k ) = - 3 i - j + 2 k b → × ( a 2 → - a 1 → ) = i j k 1 - 2 1 - 3 - 1 2 = i ( - 4 + 1 ) - j ( 2 + 3 ) + k ( - 1 - 6 ) = - 3 i - 5 j - 7 k ∴ | b → × ( a 2 → - a 1 → ) | = ( - 3 ) 2 + ( - 5 ) 2 + ( - 7 ) 2 = 9 + 25 + 49 = 83 and | b → | = 1 2 + ( - 2 ) 2 + 1 2 = 6 Thus required shortest distance between given lines = | b → × ( a 2 → - a 1 → ) | | b → | = 83 6 = 83 6 units

Find the vector equation of the line parallel to the line x - 1 5 = 3 - y 2 = z + 1 4 and passing through ( 3 , 0 , - 4 ) . Also find the distance between these two lines.

Given equation of line be, x - 1 5 = y - 3 - 2 = z + 1 4 ... ( 1 ) Since line ( 1 ) passes through line point ( 1 , 3 , - 1 ) and having direction ratios < 5 , - 2 , 4 > So the vector equation of line ( 1 ) be r → = a 1 → + λ b → where a 1 → = i + 3 j - k ; b → = 5 i - 2 j + 4 k Now vector equation of line through the point ( 3 , 0 , - 4 ) and parallel to given line ( 1 ) is given by r → = 3 i + 0 j - 4 k + λ ( 5 i - 2 j + 4 k ) and it is of the form r → = a 2 → + λ b → where a 2 → = 3 i + 0 j - 4 k ∴ a 2 → - a 1 → = ( 3 i + 0 j - 4 k ) - ( i + 3 j - k ) = 2 i - 3 j - 3 k b → × ( a 2 → - a 1 → ) = i j k 5 - 2 4 2 - 3 - 3 = i ( 6 + 12 ) - j ( - 15 - 8 ) + k ( - 15 + 4 ) = 18 i + 23 j - 11 k ∴ | b → × ( a 2 → - a 1 → ) | = 18 2 + 23 2 + ( - 11 ) 2 = 324 + 529 + 121 = 974 and | b → | = 5 2 + ( - 2 ) 2 + 4 2 = 25 + 4 + 16 = 45 Required shortest distance between given lines = | b → × ( a 2 → - a 1 → ) | | b → | = 974 45 units

Find the Cartesian equation of the plane: r → . ( i ^ + j ^ - k ^ ) = 2

Given vector equation of plane r → · i + j - k = 2 Since r → is the position vector of any point P(x,y,z) on the given plane r → = x i ^ + y j ^ + z k ^ Substituting this value of r → in the given plane, we get ( x i ^ + y j ^ + z k ^ ) ( i ^ + j ^ - k ^ ) = 2 x + y - z = 2 is the required Cartesian equation.

Find the cartesian equation of the plane r → · ( 5 i ^ + 7 j ^ - 11 k ^ ) = 14

Given vector equation of plane r → · ( 5 i ^ + 7 j ^ - 11 k ^ ) = 14 Since r → is the position vector of any point P(x, y, z) on the given plane r → = x i ^ + y j ^ + z k ^ Substituting this value of r → in the given plane, we get ( x i ^ + y j ^ + z k ^ ) ( 5 i ^ + 7 j ^ - 11 k ^ ) = 14 5 x + 7 y - 11 z = 14 is required Cartesian equation