 |
VOOZH | about |
|
VOOZH | about |
Chapter 29 of RD Sharma's Class 12 Mathematics textbook focuses on "The Plane," a fundamental concept in three-dimensional geometry. Exercise 29.3 specifically deals with finding the equations of planes under various conditions. This set of solutions provides step-by-step explanations for problems involving planes passing through points, parallel to given lines, and perpendicular to specific vectors.
General equation of a plane: Ax + By + Cz + D = 0
Distance formula between a point and a plane: |Ax₀ + By₀ + Cz₀ + D| / √(A² + B² + C²)
Angle between two planes: cos θ = |A₁A₂ + B₁B₂ + C₁C₂| / √((A₁² + B₁² + C₁²)(A₂² + B₂² + C₂²))
Condition for parallel planes: A₁/A₂ = B₁/B₂ = C₁/C₂
Condition for perpendicular planes: A₁A₂ + B₁B₂ + C₁C₂ = 0
Solution:
As we know that the vector equation of a plane passing through a point and normal to is
⇒
Here, and
So, the equation of the required equation is
⇒ = (4)(2) + (2)(-1) + (-3)(1)
⇒ = 8 - 2 - 3
⇒
Hence, the required equation is
Solution:
Given vector equation of a plane,
Since denotes the position vector of an arbitrary point (x, y, z) on the plane.
Therefore, putting
⇒
⇒ (x)(12) + (y)(-3) + (z)(4) = -5
⇒ 12x - 3y + 4z + 5 = 0
So, this is the required cartesian equation of the plane.
Solution:
Given vector equation of a plane,
Since denotes the position vector of an arbitrary point (x, y, z) on the plane.
Therefore, putting
⇒
⇒ (x)(-1) + (y)(1) + (z)(2) = 9
⇒ -x + y + 2z = 9
So, this is the required cartesian equation of the plane.
Solution:
Vector equation of XY-Plane:
The XY- Plane passes through origin whose position vector is and
perpendicular to Z-axis whose position vector is
So the equation of the XY plane is
⇒
Vector equation of XZ-Plane:
The XZ- Plane passes through origin whose position vector is
and perpendicular to Y-axis whose position vector is
So the equation of the XZ plane is
⇒
Vector equation of YZ-Plane:
The YZ- Plane passes through origin whose position vector is
and perpendicular to X-axis whose position vector is
So the equation of the YZ plane is
⇒
Hence, the vector equation of the coordinates planes.
XY-Plane =
XZ-Plane =
YZ-Plane =
Solution:
Given equation of plane is,
2x - y + 2z = 8
So,
Therefore, the vector equation of the plane is
Solution:
Given equation of plane is,
x + y - z = 5
So,
Therefore, Vector equation of the plane is
Solution:
Given equation of plane is,
x + y = 3
So,
Therefore, the vector equation of the plane is
Solution:
As we know that the vector equation of a plane passing through a point and normal to is
.....(1)
So, according to the question it is given that the plane passes through the point (1, -1, 1) and
normal to the line joining the points A(1, 2, 5) and B(-1, 3, 1).
So,
= Position vector of - Position vector of
=
=
Now, from eq (1), we get
⇒
⇒
⇒
⇒
⇒ ...(2)
Now, Putting in eq(2), we get
2x - y + 4z = 7
So, the vector equation is the plane is
and the cartesian equation of the plane is 2x - y + 4z = 7
Solution:
Given that = √3 and makes equal angle with coordinate axes.
Let us considered has direction cosine as u, v and w, and
it makes angle of α, β and γ with the coordinate axes.
So, α = β = γ
cos α = cos β = cos γ
Assuming u = v = w = p
We know that
u2 + v2 + w2 = 1
p2 + p2 + p2 = 1
P2 = 1/3
P= ±1/√3
So,
u = ±1/√3
cos α = ±1/√3
Now, α = cos-1(-1/√3)
It gives, α is an obtuse angle so, neglect it.
Now, α = cos-1(1/√3)
It gives, α is an acute angle, so
cos α = ±1/√3
u = v = w = 1/√3
So,
= √3
Now, and
As we know that the vector equation of a plane passing through a point and normal to is
⇒
⇒
⇒ ....(1)
Now, Putting in eq(1), we get
⇒
(x)(1) + (y)(1) + (z)(1) = 2
⇒ x + y + z = 2
Hence, the vector equation of the plane is
and the cartesian equation of the plane is x + y + z = 2
Solution:
According to the question it is given that the coordinates of the foot of the
perpendicular drawn from the origin O to a plane is P(12, -4, 3)
Thus, we can say that the required plane is passing through P(12, -4, 3) and perpendicular to OP.
As we know that the vector equation of a plane passing through a point and normal to is
.......(1)
Here,
On putting the values of and in equation (1)
⇒
⇒
⇒
⇒ ........(2)
Now, on putting in eq(2), we get
⇒
(x)(12) + (y)(-4) + (z)(3) = 169
⇒ 12x - 4y + 3z = 169
Hence, the vector equation of the plane is
and the cartesian equation of the plane is 12x- 4y+ 3z = 169
Solution:
According to the question it is given that the plane is passing through P(2, 3, 1)
having (5, 3, 2) as the direction ratios of the normal to the plane.
As we know that the vector equation of a plane passing through a point and normal to is
.....(i)
So,
Now put all these values in equation (i),
.........(2)
Now, putting, in eq(2), we get
(x)(5) + (y)(3) + (z)(2) = 21
5x + 3y + 2z = 21
Hence, this is the required equation of plane
Solution:
According to the question it is given that P is the point (2,3,-1) and
the plane is passing through P and OP is the vector normal to the plane.
So,
As we know that the vector equation of a plane passing through a point and normal to is
.....(i)
Here,
So, = Position vector of P - Position vector of O
=
Now put, the value of and in equation (i),
........(2)
Now putting, , in eq(2), we get
(x)(2) + (y)(3) + (z)(−1) = 14
2x + 3y − z = 14
So, this is the required equation of plane.
Solution:
According to the question
The equation os plane = 2x + y -2z = 3
Now divide both sides of the equation by 3, we get
......(i)
Here, if a, b, c are the intercepts by a plane on the coordinates axes,
then the equation of the plane is:
......(ii)
On comparing the equation (i) and (ii), we get the value of a, b, and c
So, from the given equation of plane,
Hence, the vector normal to the plane is,
= √{(2)2 + (1)2 + (-2)2}
= √(4 + 1 + 4)
= √9
Hence, the unit vector perpendicular to
Hence, the direction cosine of normal to the plane = 2/3, 1/3, -2/3
Question 1. Find the equation of the plane passing through the points (1, -1, 2), (2, 3, 1), and (-1, 2, 4).
Question 2. Determine the equation of the plane passing through the point (3, -2, 1) and perpendicular to the vector <2, 1, -3>.
Question 3. Find the equation of the plane passing through the point (1, 2, -1) and parallel to the plane 2x + 3y - z = 5.
Question 4. Write the equation of the plane passing through the point (0, 2, -1) and containing the line x = t, y = 2t + 1, z = 3t - 2.
Question 5. Find the equation of the plane passing through the point (2, -1, 3) and perpendicular to the planes x - 2y + 3z = 4 and 3x + y - z = 2.
Question 6. Determine the equation of the plane passing through the points (a, 0, 0), (0, b, 0), and (0, 0, c), where a, b, and c are non-zero real numbers.
Question 7. Find the equation of the plane passing through the point (1, -1, 2) and parallel to the vectors <1, 2, -1> and <2, -1, 3>.
Question 8. Write the equation of the plane that is perpendicular to the line x = 2 + t, y = 1 - 2t, z = 3 + t and passes through the point (1, 1, 1).
Question 9. Find the equation of the plane passing through the point (2, 3, -1) and containing the line of intersection of the planes 2x + y - z = 3 and x - y + 2z = 1.
Question 10. Determine the equation of the plane passing through the origin and making equal angles with the coordinate axes.
Also Read,
In this exercise, we explored various methods to find equations of planes under different conditions. We utilized the point-normal form of plane equations, vector operations, and geometric relationships between points, lines, and planes. These problems demonstrate the practical application of vector algebra and analytical geometry in three-dimensional space. Understanding these concepts is crucial for solving more complex problems in spatial geometry and developing a strong foundation for advanced mathematics and physics.
