**14** In the following cases, find the distance of each of the given points from the corresponding given plane

POINT | PLANE |

a. (0, 0, 0) |
3x – 4y + 12 z = 3 |

b. (3, – 2, 1) |
2x – y + 2z + 3 = 0 |

c. (2, 3, – 5) |
x + 2y – 2z = 9 |

d. (– 6, 0, 0) |
2x – 3y + 6z – 2 = 0 |

We know that the distance between a point and a plane is given by,
.......................(1)
So, calculating for each case;
(a) Point and Plane
Therefore,
(b) Point and Plane
Therefore,
(c) Point and Plane
Therefore,
(d) Point and Plane
Therefore,

Two planes
whose direction ratios are and whose direction ratios are ,
are said to Parallel:
If,
and Perpendicular:
If,
And the angle between is given by the relation,
So, given two planes
Here,
and
So, applying each condition to check:
Parallel check:
Clearly, the given planes are NOT parallel as .
Perpendicular check:
.
Clearly, the given planes are...

Two planes
whose direction ratios are and whose direction ratios are ,
are said to Parallel:
If,
and Perpendicular:
If,
And the angle between is given by the relation,
So, given two planes
Here,
and
So, applying each condition to check:
Parallel check:
Therefore
Thus, the given planes are parallel to each other.

Two planes
whose direction ratios are and whose direction ratios are ,
are said to Parallel:
If,
and Perpendicular:
If,
And the angle between is given by the relation,
So, given two planes
Here,
and
So, applying each condition to check:
Parallel check:
Thus, the given planes are parallel as .

Two planes
whose direction ratios are and whose direction ratios are ,
are said to Parallel:
If,
and Perpendicular:
If,
And the angle between is given by the relation,
So, given two planes
Here,
and
So, applying each condition to check:
Perpendicular check:
.
Thus, the given planes are perpendicular to each other.

Two planes
whose direction ratios are and whose direction ratios are ,
are said to Parallel:
If,
and Perpendicular:
If,
And the angle between is given by the relation,
So, given two planes
Here,
and
So, applying each condition to check:
Parallel check:
Clearly, the given planes are NOT parallel.
Perpendicular check:
.
Clearly, the given planes are NOT perpendicular.
Then...

Given two vector equations of plane
and .
Here, and
The formula for finding the angle between two planes,
.............................(1)
and
Now, we can substitute the values in the angle formula (1) to get,
or
or

The equation of the plane through the intersection of the given two planes, and is given in Cartesian form as;
or ..................(1)
So, the direction ratios of (1) plane are which are .
Then, the plane in equation (1) is perpendicular to whose direction ratios are .
As planes are perpendicular then,
we get,
or
or
Then we will substitute the values of in the...

Here and
and and
Hence, using the relation , we get
or ..............(1)
where, is some real number.
Taking , we get
or
or .............(2)
Given that the plane passes through the point , it must satisfy (2), i.e.,
or
Putting the values of in (1), we get
or
or
which is the required vector...

The equation of any plane through the intersection of the planes,
Can be written in the form of; , where
So, the plane passes through the point , will satisfy the above equation.
That implies
Now, substituting the value of in the equation above we get the final equation of the plane;
is the required equation of the plane.

Given that the plane is parallel to the ZOX plane.
So, we have the equation of plane ZOX as .
And an intercept of 3 on the y-axis
Intercept form of a plane given by;
So, here the plane would be parallel to the x and z-axes both.
we have any plane parallel to it is of the form, .
Equation of the plane required is .

Given plane
We have to find the intercepts that this plane would make so,
Making it look like intercept form first:
By dividing both sides of the equation by 5 (as we have to make the R.H.S =1) , we get then,
So, as we know that from the equation of a plane in intercept form, where a,b,c are the intercepts cut off by the plane at x,y, and z-axes respectively.
Therefore after comparison, we...

The equation of the plane which passes through the three points is given by;
Determinant method,
As determinant value is not equal to zero hence there must be a plane that passes through the points A, B, and C.
Finding the equation of the plane through the points,
After substituting the values in the determinant we get,
So, this is the required Cartesian equation of the plane.

The equation of the plane which passes through the three points is given by;
Determinant method,
Or,
Here, these three points A, B, C are collinear points.
Hence there will be an infinite number of planes possible which passing through the given points.

(a) (1, 1, – 1), (6, 4, – 5), (– 4, – 2, 3)
A,B,C are collinear points so there are infinite number of the planes that passes through three points

Given the point and the normal vector which is perpendicular to the plane is
The position vector of point A is
So, the vector equation of the plane would be given by,
Or
where is the position vector of any arbitrary point in the plane.
Therefore, the equation we get,
So, this is the required Cartesian equation of the plane.

Given the point and the normal vector which is perpendicular to the plane is
The position vector of point A is
So, the vector equation of the plane would be given by,
Or
where is the position vector of any arbitrary point in the plane.
Therefore, the equation we get,
or
So, this is the required Cartesian equation of the plane.

Let the coordinates of the foot of perpendicular P from the origin to the plane be
Given plane equation .
or written as
The direction ratios of the normal of the plane are .
Therefore
So, now dividing both sides of the equation by we will obtain,
This equation is similar to where, are the directions cosines of normal to the plane and d is the distance of normal from the origin.
Then...

Let the coordinates of the foot of perpendicular P from the origin to the plane be
Given plane equation .
The direction ratios of the normal of the plane are .
Therefore
So, now dividing both sides of the equation by we will obtain,
This equation is similar to where, are the directions cosines of normal to the plane and d is the distance of normal from the origin.
Then finding the...

Let the coordinates of the foot of perpendicular P from the origin to the plane be
Given a plane equation ,
Or,
The direction ratios of the normal of the plane are .
Therefore
So, now dividing both sides of the equation by we will obtain,
This equation is similar to where, are the directions cosines of normal to the plane and d is the distance of normal from the origin.
Then finding the...

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