Question

Find the distance of the point (−1, −5, −­10) from the point of intersection of the line and the plane .

Answer 1

The given point is ( −1,−5,−10 ).

The equation of given line is,

r → =( 2 i ^ − j ^ +2 k ^ )+λ( 3 i ^ +4 j ^ +2 k ^ )(1)

The equation of given plane is,

r → ⋅( i ^ − j ^ + k ^ )=5(2)

Substitute the value of equation (1) in equation (2).

[ ( 2 i ^ − j ^ +2 k ^ )+λ( 3 i ^ +4 j ^ +2 k ^ ) ]⋅( i ^ − j ^ + k ^ )=5 [ ( 3λ+2 ) i ^ −( 4λ−1 ) j ^ +( 2λ+2 ) k ^ ]⋅( i ^ − j ^ + k ^ )=5

Compare the value on both the sides,

( 3λ+2 )−( 4λ−1 )+( 2λ+2 )=5 3λ+2−4λ+1+2λ+2=5 λ+5=5 λ=0

Substitute the value of λ in equation (1).

r → =( 2 i ^ − j ^ +2 k ^ )+0⋅( 3 i ^ +4 j ^ +2 k ^ ) r → =( 2 i ^ − j ^ +2 k ^ )

The above equation shows the position vector of the point of intersection of the line and the plane.

The coordinates of the point of intersection of the line and the plane are ( 2,−1,2 ).

The formula for the distance between two points having coordinates ( x 1 , y 1 , z 1 ) and ( x 2 , y 2 , z 2 ) is given by,

d= ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 + ( z 2 − z 1 ) 2 (3)

Substitute the values of ( x 1 , y 1 , z 1 ) and ( x 2 , y 2 , z 2 ) as ( 2,−1,2 )and ( −1,−5,−10 ) in equation (3).

d= ( −1−2 ) 2 + ( −5+1 ) 2 + ( −10−2 ) 2 = ( −3 ) 2 + ( −4 ) 2 + ( −12 ) 2 = 9+16+144 = 169

Further simplify the above expression.

d=13

Thus, the distance of the point ( −1,−5,−10 ) from the intersection of line r → =( 2 i ^ − j ^ +2 k ^ )+λ( 3 i ^ +4 j ^ +2 k ^ ) and the plane r → ⋅( i ^ − j ^ + k ^ )=5 is 13.