Question 1. Find the vector equation of the following planes in scalar product form
Solution:
(i)
Here,
We know that,
represent a plane passing through a point having position vector and parallel to vectors and . Here,
The given plane is perpendicular to a vector
We know that vector equation of plane in scalar product form is,
—(Equation-1) Put
and in (Equation-1),
The equation is required form is,
(ii)
Here,
We know that,
represent a plane passing through a point having position vector and parallel to vectors and Here,
The given plane is perpendicular to a vector
We know that, vector equation of a plane is scalar product is,
—(Equation-1) Put value of
and in (Equation-1)
Multiplying both the sides by (-1),
The equation in the required form,
(iii)
Given, equation of plane,
We know that,
is the equation of a plane passing through point and parallel to and . Here,
The given plane is perpendicular to a vector
We know that, equation of plane in scalar product form is given by,
Dividing by 3, we get
Equation in required form is,
(iv)
Plane is passing through
and parallel to b and
Question 2. Find the cartesian form of the equation of the following planes:
Solution:
(i)
Here, given equation of plane is,
We know that,
represents the equation of a plane passing through a vector and parallel to vector and . Here,
Given plane is perpendicular to vector
We know that, equation of plane in the scalar product form,
—Equation-1 Put the value of
and in Equation-1,
Put
(x)(-3) + (y)(3) + (z)(-3) = -6
-3x + 3y – 3z = -6
Dividing by (-3), we get
x – y + z = 2
Equation in required form is,
x – y + z = 2
(ii)
Given, equation of plane,
We know that,
represents the equation of a plane passing through the vector and parallel to vector and Here,
The given plane is perpendicular to vector
We know that, equation of plane in scalar product form is given by,
—Equation-1 Put, the value of
and in equation-1
Put
(x)(0) + (y)(-4) + (z)(2) = -2
-4y + 2z = -2
The equation in required form is,
2y – z = 1
Question 3. Find the vector equation of the following planes in non-parametric form:
Solution:
(i)
Given, equation of plane is,
We know that,
represents the equation of a plane passing through a point and parallel to vector and . Given,
The given plane is perpendicular to
Vector equation of plane in non-parametric form is.
= (0)(2) + (3)(-5) + (0)(-1)
= 0 – 15 + 0
The required form of equation is,
(ii)
Given, equation of plane is,
We know that,
represents the equation of a plane passing through a vector and parallel to vector and . Here,
The given plane is perpendicular to vector
We know that, equation of a plane in non-parametric form is given by,
= (2)(20) + (2)(8) – (-1)(-12)
=40 + 16 + 12
Dividing by 4,
Equation of plane in required form is,