Determine the general equation of plane I passing through the point (-4, 3,-1) and parallel to the plane 2x − y − 3z +4= 0. Also, sketch the graph of this plane.

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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[QUESTIONPART4] Kindly provide the correct and complete solution. (I will like your solution if it is complete + Please follow the sample format below)

 

Examples:
1. Let 2 be a plane containing the point Po(4, -2, -3) and having a normal vector
N = (2,-1,2).
The standard equation of the plane 2 is given by 2 (x-4)-1(y + 2) + 2(+3) = 0.
So, the general equation of the plane 2 is given by 2x-y+2z-4 = 0.
To sketch the graph of the plane, we use the intercepts:
2-intercept: 2
y-intercept:
2-intercept: 2
Po
2. In geometry, it is known that a plane is uniquely determined by any three noncollinear
points of the plane.
Let I be a plane containing the points P(2,3,0), Q(0,5, -1) and R(1,0,3).
To determine the equation of the plane, we need to obtain a normal vector-a vector that
is orthogonal to the plane. This can be obtained by solving for a vector that is orthogonal
to vectors determined by points P, Q and R.
We consider vectors PQ = (-2,2,-1) and PR = (-1,-3,3). A normal vector to the
plane should be orthogonal to both vectors PQ and PR. This can then be obtained using
cross product.
X
Let N = PQ PR. Vector N is orthogonal to the plane. So, N is a normal vector to I.
i j
Computing, N=PQ x PR = -2 2
k
-1=(3,7,8).
-1-3
3
Using N as a normal vector and Po any of the three given points, a standard equation
can be obtained. The resulting general equation of plane I is 3x + 7y +82-27 = 0.
Verify that points P, Q and R each satisfies the equation.
Alternatively, obtain a normal vector to the plane using the cross product of vectors QR
and QP. Use this normal vector to obtain a standard equation of the plane. Verify that
this leads to the same eral equation of plane I.
Transcribed Image Text:Examples: 1. Let 2 be a plane containing the point Po(4, -2, -3) and having a normal vector N = (2,-1,2). The standard equation of the plane 2 is given by 2 (x-4)-1(y + 2) + 2(+3) = 0. So, the general equation of the plane 2 is given by 2x-y+2z-4 = 0. To sketch the graph of the plane, we use the intercepts: 2-intercept: 2 y-intercept: 2-intercept: 2 Po 2. In geometry, it is known that a plane is uniquely determined by any three noncollinear points of the plane. Let I be a plane containing the points P(2,3,0), Q(0,5, -1) and R(1,0,3). To determine the equation of the plane, we need to obtain a normal vector-a vector that is orthogonal to the plane. This can be obtained by solving for a vector that is orthogonal to vectors determined by points P, Q and R. We consider vectors PQ = (-2,2,-1) and PR = (-1,-3,3). A normal vector to the plane should be orthogonal to both vectors PQ and PR. This can then be obtained using cross product. X Let N = PQ PR. Vector N is orthogonal to the plane. So, N is a normal vector to I. i j Computing, N=PQ x PR = -2 2 k -1=(3,7,8). -1-3 3 Using N as a normal vector and Po any of the three given points, a standard equation can be obtained. The resulting general equation of plane I is 3x + 7y +82-27 = 0. Verify that points P, Q and R each satisfies the equation. Alternatively, obtain a normal vector to the plane using the cross product of vectors QR and QP. Use this normal vector to obtain a standard equation of the plane. Verify that this leads to the same eral equation of plane I.
Determine the general equation of plane I passing through the point (-4,3,-1)
and parallel to the plane 2r-y-3z +4= 0. Also, sketch the graph of this plane.
Transcribed Image Text:Determine the general equation of plane I passing through the point (-4,3,-1) and parallel to the plane 2r-y-3z +4= 0. Also, sketch the graph of this plane.
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