Question

Find the distance of the point (-2,3,-4) from the line (x+2) /3 = (2y+3) /4 = (3z+4) /5 measured parallel to the plane 4x+12y-3z+1=0.​

Answer 1

QUESTION :-

Find the distance of the point (-2,3,-4) from the line (x+2) /3 = (2y+3) /4 = (3z+4) /5 measured parallel to the plane 4x+12y-3z+1=0.

SOLUTION :-

=> Let us consider P = ( -2, 3, -4 )

=>Let π be the plane parallel to the given plane through P, and let Q be the point where the given line π. The distance required is then the length of the segment PQ.

=>Let us find the value of π:

=>We know that π is parallel to the given plane so the equation of π which differs only in the constant term.

=>4x + 12y – 3z + c = 0………………………(1)

=>Where c = some constant term

=>Where P is in π, its coordinates satisfy the equation

=> 4(-2) + 12(3) – 3(-4) + c = 0

=> -8 + 36 + 12 + c = 0

=> c = -40

=>So the equation for π is

=>4x + 12y – 3z – 40 = 0……………..(2)

=>Let us find Q:

=>For points on the line (x+2)/3=(2y+3)/4

=> 4x+8 = 6y+9

=> 6y = 4x-1 (x+2)/3=(3z+4)/5

=> 5x+10 = 9z+12

=> 9z = 5x-2

=>For points in plane π

=>4x + 12y – 3z – 40 = 0

=> 12x + 36y – 9z – 120 = 0

=>Point Q is on both line and π

=>12x + 36y – 9z – 120 = 0

=> 12x + 6(4x-1) – (5x-2) – 120 = 0

=> 12x + 24x – 6 -5x + 2 – 120 = 0

=> 31x = 124

=> x = 4

=>Then y = (4x-1)/6

=>y = 15/6

=>y = 5/2

=>z = (5x-2)/9

=>z = 18/9

=>z = 2

=>Thus Q = (4, 5/2, 2)

=>Let us calculate the length of PQ:

=>We know P = ( -2, 3, -4 ) and Q = ( 4, 5/2, 2 )

=>Length of PQ = √((-2 – 4)² + (3 – 5/2)² + (-4 – 2)²)

=>Length of PQ = √((-6)² + (1/2)² + (-6)²)

=>Length of PQ = √(36 + 1/4 + 36)

=>Length of PQ = √(72 + 1/4)

=>Length of PQ = √(288/4 + 1/4)

=>Length of PQ = √(289 / 4)

=>Length of PQ = 17 / 2.

The distance between coordinates (4,5/2,2) and (2,3,−4) is 17/2 unit.