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The equation of plane containing the line 2x-5y z=3 x y 4z = 5 and parallel to the plane x 3y 6z=1,is?
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The equation of plane containing the line 2x-5y z=3 x y 4z = 5 and p...
Equation of a plane containing a line and parallel to another plane
To find the equation of a plane containing a given line and parallel to a given plane, we can follow these steps:
Step 1: Given line equation
The line equation given is:
2x - 5y + z = 3
We can rewrite this equation in parametric form as:
x = t
y = (2t - 3)/5
z = (3 - 2t)/5
Step 2: Given plane equation
The equation of the plane parallel to the desired plane is given as:
x + 3y + 6z = 1
We can rewrite this equation in normal form as:
n · r = d
where n is the normal vector to the plane, r is the position vector of any point on the plane, and d is a constant.
So, the normal vector of the given plane is:
n = <1, 3,="" 6="">
Step 3: Finding the direction vector of the line
To find a direction vector of the line, we can subtract two points on the line. Let's choose two points:
Point 1: t = 0
x1 = 0, y1 = -3/5, z1 = 3/5
Point 2: t = 1
x2 = 1, y2 = -1/5, z2 = 1/5
The direction vector of the line is then:
v =
= <1 -="" 0,="" -1/5="" -="" (-3/5),="" 1/5="" -="" 3/5="">
= <1, 2/5,="" -2/5="">
Step 4: Finding the normal vector of the desired plane
Since the desired plane is parallel to the given plane, its normal vector will be the same as the normal vector of the given plane.
So, the normal vector of the desired plane is:
n = <1, 3,="" 6="">
Step 5: Finding a point on the desired plane
To find a point on the desired plane, we can choose any point on the given line. Let's choose t = 0:
x = 0
y = -3/5
z = 3/5
So, a point on the desired plane is:
P(0, -3/5, 3/5)
Step 6: Writing the equation of the desired plane
Using the point-normal form of a plane equation, we can write the equation of the desired plane as:
n · (r - r0) = 0
where n is the normal vector of the plane, r is the position vector of any point on the plane, and r0 is the position vector of the chosen point on the plane.
Substituting the values, we get:
<1, 3,="" 6=""> · ( - <0, -3/5,="" 3/5="">) = 0
1(x - 0) + 3(y + 3/5) + 6(z - 3/5) = 0
To find the equation of a plane containing a given line and parallel to a given plane, we can follow these steps:
Step 1: Given line equation
The line equation given is:
2x - 5y + z = 3
We can rewrite this equation in parametric form as:
x = t
y = (2t - 3)/5
z = (3 - 2t)/5
Step 2: Given plane equation
The equation of the plane parallel to the desired plane is given as:
x + 3y + 6z = 1
We can rewrite this equation in normal form as:
n · r = d
where n is the normal vector to the plane, r is the position vector of any point on the plane, and d is a constant.
So, the normal vector of the given plane is:
n = <1, 3,="" 6="">
Step 3: Finding the direction vector of the line
To find a direction vector of the line, we can subtract two points on the line. Let's choose two points:
Point 1: t = 0
x1 = 0, y1 = -3/5, z1 = 3/5
Point 2: t = 1
x2 = 1, y2 = -1/5, z2 = 1/5
The direction vector of the line is then:
v =
= <1 -="" 0,="" -1/5="" -="" (-3/5),="" 1/5="" -="" 3/5="">
= <1, 2/5,="" -2/5="">
Step 4: Finding the normal vector of the desired plane
Since the desired plane is parallel to the given plane, its normal vector will be the same as the normal vector of the given plane.
So, the normal vector of the desired plane is:
n = <1, 3,="" 6="">
Step 5: Finding a point on the desired plane
To find a point on the desired plane, we can choose any point on the given line. Let's choose t = 0:
x = 0
y = -3/5
z = 3/5
So, a point on the desired plane is:
P(0, -3/5, 3/5)
Step 6: Writing the equation of the desired plane
Using the point-normal form of a plane equation, we can write the equation of the desired plane as:
n · (r - r0) = 0
where n is the normal vector of the plane, r is the position vector of any point on the plane, and r0 is the position vector of the chosen point on the plane.
Substituting the values, we get:
<1, 3,="" 6=""> · (
1(x - 0) + 3(y + 3/5) + 6(z - 3/5) = 0
Community Answer
The equation of plane containing the line 2x-5y z=3 x y 4z = 5 and p...
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The equation of plane containing the line 2x-5y z=3 x y 4z = 5 and parallel to the plane x 3y 6z=1,is?
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The equation of plane containing the line 2x-5y z=3 x y 4z = 5 and parallel to the plane x 3y 6z=1,is? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The equation of plane containing the line 2x-5y z=3 x y 4z = 5 and parallel to the plane x 3y 6z=1,is? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The equation of plane containing the line 2x-5y z=3 x y 4z = 5 and parallel to the plane x 3y 6z=1,is?.
The equation of plane containing the line 2x-5y z=3 x y 4z = 5 and parallel to the plane x 3y 6z=1,is? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The equation of plane containing the line 2x-5y z=3 x y 4z = 5 and parallel to the plane x 3y 6z=1,is? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The equation of plane containing the line 2x-5y z=3 x y 4z = 5 and parallel to the plane x 3y 6z=1,is?.
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