Question 1. Find the equation of the plane which is parallel to 2x – 3y + z = 0 and passes through the point (1, –1, 2).
Solution:
We know that the equation of a plane parallel to 2x – 3y + z = 0 is given by:
2x – 3y + z + λ = 0
Since the plane passes through the point (1, –1, 2), we have:
2(1) – 3(–1) + 2 + λ = 0
⇒ λ = –7
On substituting the value of λ in the equation, we have:
2x – 3y + z + (-7) = 0
2x – 3y + z – 7= 0 is the required equation.
Question 2. Find the equation of the plane through (3, 4, –1) which is parallel to the plane
Solution:
The given plane passes through the vector
. Thus,
(3)(2) + (4)(-3) + (-1)(5) + λ = 0
⇒ λ = 11
On substituting the value of λ in the equation, we have:
is the required equation.
Question 3. Find the equation of the plane passing through the line of intersection of the planes 2x – 7y + 4z – 3 = 0 and 3x – 5y + 4z + 11 = 0 and the point (–2, 1, 3).
Solution:
The equation of the plane passing through the line of intersection of the given planes is:
(2x – 7y + 4z – 3) + λ(3x – 5y + 4z + 11) = 0⇒ x(2 + 3λ) + y(–7 – 5λ) + z(4 + 4λ) – 3 + 11λ = 0
Also, since the plane passes through the point (–2, 1, 3), we have:
(–2)(2 + 3λ) + (1)(–7 – 5λ) + (3)(4 + 4λ) – 3 + 11λ = 0
⇒ λ = 1/6
On substituting the value of λ in the equation, we have:
x(2 + 3(1/6)) + y(–7 – 5(1/6)) + z(4 + 4(1/6)) – 3 + 11(1/6) = 0
15x – 47y + 28z = 7 is the required equation.
Question 4. Find the equation of the plane passing through the point
Solution:
The equation of the plane passing through the line of intersection of the given planes is:
Also, since the plane passes through point
, we have:
⇒ λ = 6
On substituting the value of λ in the equation, we have:
is the required equation.
Question 5. Find the equation of the plane passing through the intersection of 2x – y = 0 and 3z – y = 0 and perpendicular to 4x + 5y – 3z = 8.
Solution:
The equation of the plane passing through the line of intersection of the given planes is:
2x – y + λ(3z – y) = 0
⇒ 2x + y(–1 – λ) + z(3λ) = 0
Since the planes are perpendicular, we have:
2(4) + (–5)(–1 – λ) + (–3)(3λ) = 0
⇒ λ = 3/14
On substituting the value of λ in the equation, we have:
2x + y(–1 – 3/14) + z(3(3/14)) = 0
28x – 17y + 9z = 0 is the required equation.
Question 6. Find the equation of the plane which contains the line of intersection of planes x + 2y + 3z – 4 = 0 and 2x + y – z + 5 = 0 and is perpendicular to the plane 5x + 3y – 6z + 8 = 0.
Solution:
The equation of the plane passing through the line of intersection of the given planes is:
x + 2y + 3z – 4 + λ(2x + y – z + 5) = 0
⇒ x(1 + 2λ) + y(2 + λ) + z(3 – λ) – 4 + 5λ = 0
Since the planes are perpendicular, we have:
5(1 + 2λ) + 3(2 + λ) + (–6)(3 – λ) = 0
⇒ λ = 7/19
On substituting the value of λ in the equation, we have:
x(1 + 2(7/19)) + y(2 + 7/19) + z(3 – 7/19) – 4 + 5(7/19) = 0
33x + 45y + 50z – 41 = 0 is the required equation.
Question 7. Find the equation of the plane passing through the line of intersection of the planes x + 2y + 3z + 4 = 0 and x – y + z + 3 = 0 and passing through the origin.
Solution:
The equation of the plane passing through the line of intersection of the given planes is:
x + 2y + 3z + 4 + λ(x – y + z + 3) = 0
⇒ x(1 + λ) + y(2 – λ) + z(3 + λ) + 4 + 3λ = 0
Also, since the plane passes through the origin, we have:
0(1 + λ) + 0(2 – λ) + 0(3 + λ) + 4 + 3λ = 0
⇒ λ = -4/3
On substituting the value of λ in the equation, we have:
x(1 + (-4/3)) + y(2 – (-4/3)) + z(3 + (-4/3)) + 4 + 3(-4/3) = 0
x – 10y – 5z = 0 is the required equation.
Question 8. Find the vector equation in scalar product form of the plane containing the line of intersection of the planes x – 3y + 2z – 5 = 0 and 2x – y + 3z – 1 = 0 and passing through (1, –2, 3).
Solution:
The equation of the plane passing through the line of intersection of the given planes is:
x – 3y + 2z – 5 + λ(2x – y + 3z – 1) = 0
⇒ x(1 + 2λ) + y(–3 – λ) + z(2 + 3λ) – 5 – λ = 0
Also, since the plane passes through the origin, we have:
1(1 + 2λ) + (–2)(–3 – λ) + 3(2 + 3λ) – 5 – λ = 0
⇒ λ = -2/3
On substituting the value of λ in the equation, we have:
x(1 + 2(-2/3)) + y(–3 – (-2/3)) + z(2 + 3(-2/3)) – 5 – (-2/3) = 0
is the required equation.
Question 9. Find the equation of the plane which contains the line of intersection of planes x + 2y + 3z – 4 = 0 and 2x + y – z + 5 = 0 and is perpendicular to the plane 5x + 3y + 6z + 8 = 0.
Solution:
The equation of the plane passing through the line of intersection of the given planes is:
x + 2y + 3z – 4 + λ(2x + y – z + 5) = 0
⇒ x(1 + 2λ) + y(2 + λ) + z(3 – λ) – 4 + 5λ = 0
We know that two planes are perpendicular when
⇒ 5(1 + 2λ) + 3(2 + λ) + 6(3 – λ) = 0
⇒ λ = -29/7
On substituting the value of λ in the equation, we have:
x(1 + 2(-29/7)) + y(2 + (-29/7)) + z(3 – (-29/7)) – 4 + 5(-29/7) = 0
51x + 15y – 50z + 173 = 0 is the required equation.
Question 10. Find the equation of the plane passing through the line of intersection of the planes
Solution:
The equation of the plane passing through the line of intersection of the given planes is:
x(1 + 3λ) + y(3 + λ) – 4zλ + 6 = 0
Distance from plane to the origin = 1
⇒
⇒ λ = ±1
Hence, 4x + 2y – 4z + 6 = 0 and –2x + 2y + 4z + 6 = 0 are the required equations.
Question 11. Find the equation of the plane passing through the line of intersection of the planes 2x + 3y – z + 1 = 0 and x + y – 2z + 3 = 0 and perpendicular to the plane 3x – 2y – z – 4 = 0.
Solution:
The equation of the plane passing through the line of intersection of the given planes is:
2x + 3y – z + 1 + λ(x + y – 2z + 3) = 0
⇒ x(2 + λ) + y(3 + λ) + z(–1 – 2λ) + 1 + 3λ = 0
We know that two planes are perpendicular when
⇒ 3(2 + λ) + (–1)(3 + λ) + (–2)(–1 – 2λ) = 0
⇒ λ = -5/6
On substituting the value of λ in the equation, we have:
x(2 + (-5/6)) + y(3 + (-5/6)) + z(–1 – 2(-5/6)) + 1 + 3(-5/6) = 0
7x + 13y + 4z – 9 = 0 is the required equation.
Question 12. Find the equation of the plane that contains the line of intersection of the planes
Solution:
The equation of the plane passing through the line of intersection of the given planes is:
⇒
We know that two planes are perpendicular if
⇒
⇒ 5(1 + 2λ) + 3(2 + λ) + (–6)(3 – λ) = 0
⇒ λ = 7/19
On substituting the value of λ in the equation, we have:
33x + 45y + 50z – 41 = 0 is the required equation.
Question 13. Find the vector equation of the plane passing through the intersection of planes
Solution:
The equation of the plane passing through the line of intersection of the given planes is:
⇒ x(1 + 2λ) + y(1 + 3λ) +z(1 + 4λ) = 6 – 5λ
Also, since the plane passes through the point(1, 1, 1), we have:
1(1 + 2λ) + 1(1 + 3λ) +1(1 + 4λ) = 6 – 5λ
⇒ λ = 3/14
On substituting the value of λ in the equation, we have:
x(1 + 2(3/14)) + y(1 + 3(3/14)) +z(1 + 4(3/14)) = 6 – 5(3/14)
is the required equation.
Question 14. Find the equation of the plane passing through the intersection of the planes
Solution:
The equation of the plane passing through the line of intersection of the given planes is:
⇒
Also, since the plane passes through the point (2, 1, 3) we have:
9λ = –7
⇒ λ = -7/9
Substituting the value of λ in the equation, we have:
is the required equation.
Question 15. Find the equation of the plane passing through the intersection of the planes 3x – y + 2z = 4 and x + y + z = 2 and the point (2, 2, 1).
Solution:
The equation of the plane passing through the line of intersection of the given planes is:
3x – y + 2z – 4 + λ(x + y + z – 2) = 0
Also, since the plane passes through the point (2, 2, 1), we have:
λ = -2/3
On substituting the value of λ in the equation, we have:
3x – y + 2z – 4 + (-2/3)(x + y + z – 2) = 0
7x – 5y + 4z = 0 is the required equation.
Question 16. Find the vector equation of the plane through the line of intersection of the planes x + 2y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x – y + z = 0.
Solution:
The equation of the plane passing through the line of intersection of the given planes is:
x + 2y + z – 1 + λ(2x + 3y + 4z – 5) = 0
⇒ x(1 + 2λ) + y(1 + 3λ) +z(1 + 4λ) = 1 + 5λ
We know that two planes are perpendicular when
⇒ 1(1 + 2λ) + (–1)(1 + 3λ) + 1(1 + 4λ) = 1 + 5λ
⇒ λ = -1/3
On substituting the value of λ in the equation, we have:
x(1 + 2(-1/3)) + y(1 + 3(-1/3)) + z(1 + 4(-1/3)) = 1 + 5(-1/3)
x – z + 2 = 0 is the required equation.
Question 17. Find the equation of the plane passing through (a, b, c) and parallel to the plane
Solution:
Equation of the family of planes parallel to the given plane =
Since the plane passes through (a, b, c), we have:
a + b + c = d
Substituting the above equation in the equation of family of planes we have:
Hence, x + y + z = a + b + c is the required equation of the plane.