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 May 29, 2023 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.

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

Question 36 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.
Given plane passes through A (2, 1, 2) and B (4, –2, 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 = (𝐴𝑃) ⃗ × (𝐴𝐵) ⃗
It is also given that
Required plane is perpendicular to 𝑟 ⃗ . (𝑖 ̂ − 2𝑘 ̂) = 5
So, its normal = 𝑛 ⃗ = 𝑖 ̂ − 2𝑘 ̂
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
((𝐴𝑃) ⃗ × (𝐴𝐵) ⃗). 𝑛 ⃗ = 0
𝑛 ⃗ . ((𝐴𝑃) ⃗ × (𝐴𝐵) ⃗) = 0
This is scalar triple product 𝑎 ⃗.(𝑏 ⃗×𝑐 ⃗ )= [■8(𝑎 ⃗&𝑏 ⃗&𝑐 ⃗ )]
[■8(𝑛 ⃗&(𝐴𝑃) ⃗&(𝐴𝐵) ⃗ )] = 0
Now,
Points are A (2, 1, 2), B (4, −2, 1) and P(x, y, z)
Therefore,
(𝐴𝑃) ⃗ = (𝑥−2) 𝑖 ̂+(𝑦−1) 𝑗 ̂+(𝑧−2) 𝑘 ̂
(𝐴𝐵) ⃗ = (4−2) 𝑖 ̂+(−2−1) 𝑗 ̂+(1−2) 𝑘 ̂ = = 2𝑖 ̂−3𝑗 ̂−𝑘 ̂
And the other plane is
𝑟 ⃗ . (𝑖 ̂ − 2𝑘 ̂) = 5
So, its normal = 𝑛 ⃗ = 𝑖 ̂ − 2𝑘 ̂ = 𝑖 ̂ + 0𝑗 ̂ − 2𝑘 ̂
Putting these values in (3)
[■8(𝑛 ⃗&(𝐴𝑃) ⃗&(𝐴𝐵) ⃗ )] = 0
|■8(1&0&−2@(𝑥−2)&(𝑦−1)&(𝑧−2)@2&−3&−1)| = 0
1(–(y – 1) + 3(z – 2)) + 0 – 2 (–3(x – 2) – 2 (y – 1)) = 0
–(y – 1) + 3(z – 2) + 6(x – 2) + 4 (y – 1) = 0
–y + 1 + 3z – 6 + 6x – 12 + 4y – 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
(𝑥 − 𝑥_1)/(𝑥_2 − 𝑥_1 ) = (𝑦 − 𝑦_1)/(𝑦_2 − 𝑦_1 ) = (𝑧 − 𝑧_1)/(𝑧_2 − 𝑧_1 )
(𝑥 − 3)/(5 − 3) = (𝑦 − 4)/(1 − 4) = (𝑧 − 1)/(6 − 1)
(𝑥 − 3)/2 = (𝑦 − 4)/(−3) = (𝑧 − 1)/5
Now, we need to find intersection point of line and plane
Equation of line is
(𝑥 − 3)/2 = (𝑦 − 4)/(−3) = (𝑧 − 1)/5 = k
So, point on the line is
x = 2k + 3, y = –3k + 4, z = 5k + 1
Putting these values in equation of plane
2x + y + z = 7
2(2k + 3) + (–3k + 4) + (5k + 1) = 7
4k + 6 – 3k + 4 + 5k + 1 = 7
6k + 11 = 7
6k = 7 – 11
6k = –4
k = (−2)/3
Thus, coordinates of points are
x = 2k + 3 = 2 × ((−2)/3) + 3 = (−4)/3+3 = (−4 + 9)/3 = 5/3
y = –3k + 4 = –3 × ((−2)/3) + 4 = 6/3+4 = 2 + 4 = 6
z = 5k + 1 = 5 × ((−2)/3) + 1 = (−10)/3+1 = (−10 + 3)/3 = (−7)/3
Thus, the required point is (5/3,6,(−7)/3)

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

Hi, it looks like you're using AdBlock :(

Displaying ads are our only source of revenue. To help Teachoo create more content, and view the ad-free version of Teachooo... please purchase Teachoo Black subscription.

Please login to view more pages. It's free :)

Teachoo gives you a better experience when you're logged in. Please login :)

Solve all your doubts with Teachoo Black!

Teachoo answers all your questions if you are a Black user!