Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
12.2k views
in 3D Coordinate Geometry by (55.4k points)
closed by

Find the length and the foot of perpendicular from the point (1, 3/2, 2) to the plane 2x – 2y + 4z + 5 = 0.

1 Answer

+1 vote
by (50.2k points)
selected by
 
Best answer

Given plane is 2x – 2y + 4z + 5 = 0 and point (1, 3/2, 2)

The direction ratios of the normal to the plane are 2, -2, 4

So, the equation of the line passing through (1, 3/2, 2) and direction ratios are equal to the direction ratios of the normal to the plane i.e. 2, -2, 4 is

Now, any point in the plane is 2λ + 1, -2λ + 3/2, 4λ + 2

Since, the point lies in the plane, then

2(2λ + 1) – 2(-2λ + 3/2) + 4(4λ + 2) + 5 = 0

4λ + 2 + 4λ – 3 + 16λ + 8 + 5 = 0

24λ + 12 = 0 λ = ½

So, the coordinates of the point in the plane are

2(-1/2) + 1, -2(-1/2) + 3/2, 4(-1/2) + 2 = 0, 5/2, 0

Thus, the foot of the perpendicular is (0, 5/2, 0) and the required length

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...