CBSE Class 12 Sample Paper for 2022 Boards (For Term 2)

Class 12
Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

## Find the foot of the perpendicular from the point (1, 2, 0) upon the plane x – 3y + 2z = 9. Hence, find the distance of the point (1, 2, 0) from the given plane.

This question is similar to Question 37 (Choice 2) CBSE Class 12 - Sample Paper 2021 Boards

### Transcript

Question 13 Find the foot of the perpendicular from the point (1, 2, 0) upon the plane x β 3y + 2z = 9. Hence, find the distance of the point (1, 2, 0) from the given plane. Let point P(x1, y1, z1) be foot of perpendicular from point X (1, 2, 0) Since perpendicular to plane is parallel to normal vector Vector (πΏπ·) β is parallel to normal vector π β Given equation of the plane is x β 3y + 2z = 9 So, Normal vector = π β = π Μ β 3π Μ + 2π Μ Since, (πΏπ·) β and π β are parallel their direction ratios are proportional. Finding direction ratios (πΏπ·) β = (x1 β 1)π Μ + (y1 β 2)π Μ + (z1 β 0)π Μ Direction ratios = x1 β 1, y1 β 2, z1 β΄ a1 = x1 β 1 , b1 = y1 β 2, c1 = z1 π β = 1π Μ β 3π Μ + 2π Μ Direction ratios = 1, β3, 2 β΄ a2 = 1 , b2 = β3, c2 = 2 Since, (πΏπ·) β and π β are parallel their direction ratios are proportional. Finding direction ratios Direction ratios are proportional π_1/π_2 = π_1/π_2 = π_1/π_2 = k (π₯_1 β 1)/1 = (π¦_1 β 2)/( β3) = π§_1/2 = k Thus, x1 = k + 1, y1 = β3k + 2, z1 = 2k Also, point P(x1, y1, z1) lies in the plane. Putting P (k + 1, β3k + 2, 2k) in equation of plane x β 3y + 2z = 9 (k + 1) β 3(β3k + 2) + 2(2k) = 9 k + 1 + 9k β 6 + 4k = 9 k + 9k + 4k + 1 β 6 = 9 14k β 5 = 9 14k = 9 + 5 14k = 14 β΄ k = 1 Thus, x1 = k + 1 = 1 + 1 = 2 y1 = β3k + 2 = β3(1) + 2 = β1 z1 = 2k = β2(1) = 2 Therefore, coordinate of foot of perpendicular are P (2, β1, 2) Length of perpendicular X (1, 2, 0) and P (2, β1, 2) Let of Perpendicular is length of PX PX = β((2β1)^2+(β1β2)^2+(2β0)^2 ) PX = β(1^2+(β3)^2+2^2 ) PX = β(1+9+4) PX = βππ units