Question 27 (Or 2nd) - CBSE Class 12 Sample Paper for 2019 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Last updated at April 16, 2024 by Teachoo

Question 27 (OR 2
nd
question)

Show that the line of intersection of the planes x + 2y + 3z = 8 and 2x + 3y + 4z = 11 is coplanar with the line (x + 1)/1 = (y + 1)/2 = (z + 1)/3 . Also find the equation of the plane containing them.

Transcript

Question 27 (OR 2nd question) Show that the line of intersection of the planes x + 2y + 3z = 8 and 2x + 3y + 4z = 11 is coplanar with the line (𝑥 + 1)/1 = (𝑦 + 1)/2 = (𝑧 + 1)/3 . Also find the equation of the plane containing them.
Given planes
x + 2y + 3z = 8
2x + 3y + 4z = 11
and line
(𝑥 + 1)/1 = (𝑦 + 1)/2 = (𝑧 + 1)/3
Let the plane passing through intersection of planes (1) and (2) be
(x + 2y + 3z – 8) + k(2x + 3y + 4z – 11) = 0
Now, we have to show
line of intersection of the planes (1) and (2) is coplanar with line (3)
Line of intersection of planes will lie in plane (4)
If line (3) is coplanar with line of intersection of planes,
Then,
Line (3)’s point (–1, –1, –1) passes through plane (4)
and Normal of Plane (4) is perpendicular to parallel vector of line (3)
Putting (–1, –1, –1) in equation of plane from (4)
(x + 2y + 3z – 8) + k(2x + 3y + 4z – 11) = 0
Putting x = –1, y = –1, z = –1
((–1) + 2(–1) + 3(–1) – 8) + k(2(–1) + 3(–1) + 4(–1) – 11) = 0
(–1 – 2 – 3 – 8) + k(–2 – 3 – 4 – 11) = 0
–14 – 20k = 0
–20k = 14
k =14/(−20)
k = (−7)/10
Putting k = (−7)/10 in (4)
(x + 2y + 3z – 8) + ((−7)/10)(2x + 3y + 4z – 11) = 0
10(x + 2y + 3z – 8) – 7(2x + 3y + 4z – 11) = 0
10x + 20y + 30z – 80 – 14x – 21y – 28z + 77 = 0
–4x – y + 2z – 3 = 0
–4x – y + 2z –3 = 0
4x + y – 2z + 3 = 0
Thus, equation of plane is 4x + y – 2z + 3 = 0
Now, we check if line (3) (𝑥 + 1)/1 = (𝑦 + 1)/2 = (𝑧 + 1)/3 is perpendicular to plane (5)
Direction ratios of normal of plane = 4, 1, –2
So, a1 = 4, b1 = 1, c1 = –2
Direction ratios of line = 1, 2, 3
So, a2 = 1, b2 = 2, c2 = 3
Finding
a1a2 + b1b2 + c1c2
= 4 × 1 + 1 × 2 + (–2) × 3
= 4 + 2 – 6
= 0
Since a1a2 + b1b2 + c1c2 = 0
Normal to the plane and line is perpendicular
Hence line (3) lies in the plane passing through intersection of planes
Therefore, the two lines are coplanar
and the equation of the plane containing them is 4x + y – 2z + 3 = 0

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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