Question 36 - CBSE Class 12 Sample Paper for 2020 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Last updated at Oct. 24, 2019 by Teachoo

Find the equation of a plane passing through the points A (2, 1, 2) and B (4,Β β2, 1) and perpendicular to plane r . (i β 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, β2, 1) and perpendicular to plane π β . (π Μ β 2π Μ) =. 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, β2, 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 = (π΄π) β Γ (π΄π΅) β
It is also given that
Required plane is perpendicular to π β . (π Μ β 2π Μ) = 5
So, its normal = π β = π Μ β 2π Μ
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
((π΄π) β Γ (π΄π΅) β). π β = 0
π β . ((π΄π) β Γ (π΄π΅) β) = 0
This is scalar triple product π β.(π βΓπ β )= [β 8(π β&π β&π β )]
[β 8(π β&(π΄π) β&(π΄π΅) β )] = 0
Now,
Points are A (2, 1, 2), B (4, β2, 1) and P(x, y, z)
Therefore,
(π΄π) β = (π₯β2) π Μ+(π¦β1) π Μ+(π§β2) π Μ
(π΄π΅) β = (4β2) π Μ+(β2β1) π Μ+(1β2) π Μ = = 2π Μβ3π Μβπ Μ
And the other plane is
π β . (π Μ β 2π Μ) = 5
So, its normal = π β = π Μ β 2π Μ = π Μ + 0π Μ β 2π Μ
Putting these values in (3)
[β 8(π β&(π΄π) β&(π΄π΅) β )] = 0
|β 8(1&0&β2@(π₯β2)&(π¦β1)&(π§β2)@2&β3&β1)| = 0
1(β(y β 1) + 3(z β 2)) + 0 β 2 (β3(x β 2) β 2 (y β 1)) = 0
β(y β 1) + 3(z β 2) + 6(x β 2) + 4 (y β 1) = 0
βy + 1 + 3z β 6 + 6x β 12 + 4y β 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
(π₯ β π₯_1)/(π₯_2 β π₯_1 ) = (π¦ β π¦_1)/(π¦_2 β π¦_1 ) = (π§ β π§_1)/(π§_2 β π§_1 )
(π₯ β 3)/(5 β 3) = (π¦ β 4)/(1 β 4) = (π§ β 1)/(6 β 1)
(π₯ β 3)/2 = (π¦ β 4)/(β3) = (π§ β 1)/5
Now, we need to find intersection point of line and plane
Equation of line is
(π₯ β 3)/2 = (π¦ β 4)/(β3) = (π§ β 1)/5 = k
So, point on the line is
x = 2k + 3, y = β3k + 4, z = 5k + 1
Putting these values in equation of plane
2x + y + z = 7
2(2k + 3) + (β3k + 4) + (5k + 1) = 7
4k + 6 β 3k + 4 + 5k + 1 = 7
6k + 11 = 7
6k = 7 β 11
6k = β4
k = (β2)/3
Thus, coordinates of points are
x = 2k + 3 = 2 Γ ((β2)/3) + 3 = (β4)/3+3 = (β4 + 9)/3 = 5/3
y = β3k + 4 = β3 Γ ((β2)/3) + 4 = 6/3+4 = 2 + 4 = 6
z = 5k + 1 = 5 Γ ((β2)/3) + 1 = (β10)/3+1 = (β10 + 3)/3 = (β7)/3
Thus, the required point is (5/3,6,(β7)/3)

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.