Question 36 - CBSE Class 12 Sample Paper for 2020 Boards

Last updated at Oct. 9, 2019 by Teachoo

Find the equation of a plane passing through the points A (2, 1, 2) and B (4, −2, 1) and perpendicular to plane r . (i − 2k) = 5. Also, find the coordinates of the point, where the line passing through the points (3, 4, 1) and (5, 1, 6) crosses the plane thus obtained.

Question 36 Find the equation of a plane passing through the points A (2, 1, 2) and B (4, โ2, 1) and perpendicular to plane ๐ โ . (๐ ฬ โ 2๐ ฬ) = 5. Also, find the coordinates of the point, where the line passing through the points (3, 4, 1) and (5, 1, 6) crosses the plane thus obtained.
Given plane passes through A (2, 1, 2) and B (4, โ2, 1)
Let point P (x, y, z) be any point in the required plane
Now,
AP lies in the required plane
Also,
AB lies in the required plane
Therefore,
Normal of required plane = (๐ด๐) โ ร (๐ด๐ต) โ
It is also given that
Required plane is perpendicular to ๐ โ . (๐ ฬ โ 2๐ ฬ) = 5
So, its normal = ๐ โ = ๐ ฬ โ 2๐ ฬ
Since required plane and other plane are perpendicular
Both their normals will be perpendicular
So, their dot product is 0
i.e. Normal of Required Plane . Normal of other plane = 0
((๐ด๐) โ ร (๐ด๐ต) โ). ๐ โ = 0
๐ โ . ((๐ด๐) โ ร (๐ด๐ต) โ) = 0
This is scalar triple product ๐ โ.(๐ โร๐ โ )= [โ 8(๐ โ&๐ โ&๐ โ )]
[โ 8(๐ โ&(๐ด๐) โ&(๐ด๐ต) โ )] = 0
Now,
Points are A (2, 1, 2), B (4, โ2, 1) and P(x, y, z)
Therefore,
(๐ด๐) โ = (๐ฅโ2) ๐ ฬ+(๐ฆโ1) ๐ ฬ+(๐งโ2) ๐ ฬ
(๐ด๐ต) โ = (4โ2) ๐ ฬ+(โ2โ1) ๐ ฬ+(1โ2) ๐ ฬ = = 2๐ ฬโ3๐ ฬโ๐ ฬ
And the other plane is
๐ โ . (๐ ฬ โ 2๐ ฬ) = 5
So, its normal = ๐ โ = ๐ ฬ โ 2๐ ฬ = ๐ ฬ + 0๐ ฬ โ 2๐ ฬ
Putting these values in (3)
[โ 8(๐ โ&(๐ด๐) โ&(๐ด๐ต) โ )] = 0
|โ 8(1&0&โ2@(๐ฅโ2)&(๐ฆโ1)&(๐งโ2)@2&โ3&โ1)| = 0
1(โ(y โ 1) + 3(z โ 2)) + 0 โ 2 (โ3(x โ 2) โ 2 (y โ 1)) = 0
โ(y โ 1) + 3(z โ 2) + 6(x โ 2) + 4 (y โ 1) = 0
โy + 1 + 3z โ 6 + 6x โ 12 + 4y โ 4 = 0
6x + 3y + 3z = 21
2x + y + z = 7
So, the equation of plane is 2x + y + z = 7
Now, the next part of the question is
Also, find the coordinates of the point, where the line passing through the points (3, 4, 1) and (5, 1, 6) crosses the plane thus obtained.
Equation of line passing through (3, 4, 1) and (5, 1, 6) is
(๐ฅ โ ๐ฅ_1)/(๐ฅ_2 โ ๐ฅ_1 ) = (๐ฆ โ ๐ฆ_1)/(๐ฆ_2 โ ๐ฆ_1 ) = (๐ง โ ๐ง_1)/(๐ง_2 โ ๐ง_1 )
(๐ฅ โ 3)/(5 โ 3) = (๐ฆ โ 4)/(1 โ 4) = (๐ง โ 1)/(6 โ 1)
(๐ฅ โ 3)/2 = (๐ฆ โ 4)/(โ3) = (๐ง โ 1)/5
Now, we need to find intersection point of line and plane
Equation of line is
(๐ฅ โ 3)/2 = (๐ฆ โ 4)/(โ3) = (๐ง โ 1)/5 = k
So, point on the line is
x = 2k + 3, y = โ3k + 4, z = 5k + 1
Putting these values in equation of plane
2x + y + z = 7
2(2k + 3) + (โ3k + 4) + (5k + 1) = 7
4k + 6 โ 3k + 4 + 5k + 1 = 7
6k + 11 = 7
6k = 7 โ 11
6k = โ4
k = (โ2)/3
Thus, coordinates of points are
x = 2k + 3 = 2 ร ((โ2)/3) + 3 = (โ4)/3+3 = (โ4 + 9)/3 = 5/3
y = โ3k + 4 = โ3 ร ((โ2)/3) + 4 = 6/3+4 = 2 + 4 = 6
z = 5k + 1 = 5 ร ((โ2)/3) + 1 = (โ10)/3+1 = (โ10 + 3)/3 = (โ7)/3
Thus, the required point is (5/3,6,(โ7)/3)

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.