Question

Find the foot of perpendicular drawn from the point (-1, 3, -6) to the plane 2x+y-

2z+5=0. Also find the equation and length of the perpendicular.
please dont just answer for points its urgent ​

Answer 1

Step-by-step explanation:

Given Find the foot of perpendicular drawn from the point (-1, 3, -6) to the plane 2x+y-2z+5=0. Also find the equation and length of the perpendicular.  

• Let P be the point for (-1,3,-6)
• Let PQ be the perpendicular.
• So let the equation be 2x + y – 2z + 5 = 0
• Direction ratio of PQ is (2,1,-2)
• So equation of PQ will be
•        x + 1 / 2 = y – 3 / 1 = z + 6 / - 2 = λ
• Now coordinates of Q will be (2 λ – 1, λ + 3, -2 λ – 6)
•                          2(2 λ – 1) + (λ + 3) – 2(- 2 λ – 6) + 5 = 0
•                         4 λ – 2 + λ + 3 + 4 λ + 12 + 5 = 0
•                          9 λ + 18 = 0
•                             9 λ = - 18
•                          Or λ = - 2
• Now  coordinate of Q will be (-5, 1, -2)
• We need to find the length of perpendicular
• So PQ = √(x1 – x2)^2 + (y1 – y2)^2 + (z1 – z2)^2
• So the point of P was (-1 , 3, - 6)
• Now PQ = √(-5 + 1)^2 + (1 – 3)^2 + (- 2 – (-6))^2
•              = √16 + 4 + 16
•             = √36
•            = 6

Reference link will be

https://brainly.in/question/11747216