the planes - supplementary problems

7/27/2019 The Planes - Supplementary Problems

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THE PLANESUPPLEMENTARY PROBLEMS

PROBLEM 59.a.Write the equation of the plane parallel to the -plane and units belowit.SOLUTION:

The -plane is just the plane covered all points where . Therefore, theequation of the plane parallel to the -plane and 3 units below it is or .VON JERIC F. ADONA

PROBLEM 59.b. Write the equation of the plane parallel to the yz-plane and having x-intercept 4.

SOLUTION:


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We write down the equation for this coordinate plane using the idea: (yz-plane).Since we have x-intercept 4, we write the equation parallel to yz-plane or .

VON JERIC F. ADONAPROBLEM 59.c.Write the equation of the plane perpendicular to the z-axis at pointSOLUTION:

VON JERIC F. ADONAPROBLEM 59.d.Write the equation of the plane parallel to xz-plane and 6 units behind it.SOLUTION:

Parallel to -plane . Therefore the equation is or .VON JERIC F. ADONA


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PROBLEM 61. Write the equation of the plane parallel to the -axis, -intercept is , -intercept is .SOLUTION:

Since the plane is parallel to -axis, equation only contains and intercepts only.

Multiply both sides by 6, we get MARVIN A. FRONDA

PROBLEM 62.Write the equation of the plane parallel to the z-axis, with xy-trace .SOLUTION:


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PROBLEM 63. Write the equations of the following planes:a. Through and perpendicular to a line whose direction numbers

are .b. Through and perpendicular to the line segment from to .c. Through the point and perpendicular to the line segment

joining this point to the point.d. Perpendicular to the line segment from to at its

midpoint.

SOLUTION:a. Through and perpendicular to a line whose direction numbers are .

Let be an arbitrary point on the plane.To get the equation of the plane, we need to get the dot product of the direction

vector from to another point on the plane, and the direction numbersto which it is perpendicular. It will be equal to 0 because the dot product of two vectorsis equal to 0.

VON JERIC F. ADONAThe equation is .


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b. Through and perpendicular to the line segment from to .

Let be an arbitrary point on the plane. be the direction vector of the line segment from to .Again, use dot product of two direction vectors and the dot product of two

perpendicular vectors is 0. First, we get the vector of the segment from to.

Dot product: c. Through the point and perpendicular to the line segment joining this point

to the point

.


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Let be an arbitrary point on the plane. be the direction vector of the line segment from to .Again, use dot product of two direction vectors and the dot product of two

perpendicular vectors is 0. First, we get the vector of the segment from

to

.

Dot product:

d. Perpendicular to the line segment from to at its midpoint.

Let be an arbitrary point on the plane. be the direction vector of the line segment from to .First, we must get the midpoint of the segment because the plane passes through

it. It is the point of intersection of the line segment and the desired plane and get the

direction vector of the line segment.

Dot product:

LIZZA MAE N. NAPUTO


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PROBLEM 64.Find the equation of the plane:a. Through the point: and parallel to the plane .

SOLUTIONS:

Through the point: Normal direction:

b. Through the point: and parallel to the plane

SOLUTION:

Through the point:Normal direction:


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c. Through the origin and parallel to the plane

SOLUTION:

Through the origin: Normal direction: d. Parallel to the plane and half as far from the origin.

SOLUTION:


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At

e. Parallel to the plane at a distance 3 from the origin.SOLUTION:

at (0,0,0)

D=

RENA CIELO A. ARIASPROBLEM 65.Find the equation of the plane:

a. Parallel to the plane and 2 units farther from the origin.


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SOLUTION:

Let be the equation of the desired plane andv be the normal vector of then, Distance of a point to a plane | | But the given plane is parallel to the desired plane. Therefore, they have the same normal

vector. We will have:

| | || || and Therefore the equations are and .b. Parallel to the plane and distant 4 units from point.

SOLUTION:


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Let be the equation of the desired plane andv be the normal vector of then, Distance of a point to a plane

|

|

But the given plane is parallel to the desired plane. Therefore, they have the same normal

vector. We will have: | | | | | |Solving for d:

Therefore, the equations are and .c. Parallel to the plane and distant 3 units from the point.

SOLUTION:

Let be the equation of the desired plane andv be the normal vector of then, Distance of a point to a plane | | But the given plane is parallel to the desired plane. Therefore, they have the same normal

vector. We will have: | | | | | |


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Solving for d: Therefore the equations are

and

.GEMMA MONICA M. VILORIA


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PROBLEM 67. Find the equation of the plane through the and perpendicular tothe line of intersection of the planes SOLUTION:

[

]

= PROBLEM 68.Find the equation of the plane through and perpendicular to eachof the planes and .SOLUTION:

Letbe a normal vector of then we can take =


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Let be a normal vector of then we can take = Let n . Then is a vector in the direction of line.

(1, -4, 2)

BRYAN M. PAMITTAN

PROBLEM 69. Find the equation of the plane through and perpendicular to each ofthe planes and .SOLUTION:

Let We must first get the cross product of the given vectors. This will be the normal of the

desired plane.

Get the dot product of the normal vector and


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MARVIN A. FRONDA

PROBLEM 70.Find the equation of the plane through (4,1,0) and perpendicular to each ofthe planes and .SOLUTION:

To find the equation of the plane through the point and perpendicular to the planes, get

the normal vectors of the given plane.

Letbe the normal vector of be the normal vector of = N2 =

* +

* +


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VON JERIC F. ADONA

PROBLEM 71. Find the equation of the plane through

and perpendicular to each of

the planes and .SOLUTIONS:

Let

We must first get the cross product of the given vectors. This will be the normal of thedesired plane.

Get the dot product of the normal vector and MARVIN A. FRONDA


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PROBLEM 72. Find the equation of the plane perpendicular to each of the planes and at a distance from the origin.SOLUTION:

Let a,b,c be equation of the plane , . = (ai+bj+ck)

(3i-j+k)= 0

3a-b+c=0(ai+bj+ck) (i+5j+3k)=0

a+5b+3c=0 3a+15b+9c=0

MARYLYN A.ALEMAN


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PROBLEM 73. Find the equation of the plane perpendicular to each of the planes and at a distance 1 from the origin.SOLUTION:

We need to find the cross product of the normal vector of the two equations to find the plane

perpendicular to each of them; and

= < -8, 2, 16 >We substitute the cross product of the normal vectors in the constants of the general

equation of a plane and we get; To find d, the problem says that the equation of the plane perpendicular to each of the

planes has a distance of 1 from the origin. By the formula of the distance from a point to a

plane. We get, ; ; Going back to the equation of our perpendicular plane, we get

and BENITO JY M GARCIA


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PROBLEM 74. Find the equation of the plane through and andperpendicular the plane .SOLUTION:

Let

JULIE ANNE D. CENAR


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PROBLEM 75. Find the equation of the plane through and andperpendicular to the plane .SOLUTION:

We let

At point

, the required equation is

MA. CRIZELDA M. IMPERIAL


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PROBLEM 76. Find the equation of the plane through and andperpendicular to the plane .SOLUTION:

Get the direction vector of the two points, and. Let

be the normal vector of the plane

Let be the cross product ofand, and the normal vector of the desired plane. [ ]

Now, to get the equation of the plane, let

be an arbitrary point on the locus. The

equation will be obtained by getting the dot product of and the direction vector ofpoints and. This will be equal to zero because they are orthogonal (perpendicular).

NORBERTO O. YAMUGAN


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PROBLEM 77.Find the equation of the plane through and andperpendicular to the plane .SOLUTION:

Since were perpendicular to the plane , the normal vector of thisplane will be parallel to ours. Moreover, the points lie in ourplane and therefore is also parallel to our plane.

Hence, their cross product is normal to our plane. Since

is in our plane, we can immediately write down an

equation for our plane

RAFAEL M. LLARINAS

PROBLEM 78: Find the equation of the plane through and andperpendicular to the plane .SOLUTION:


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Find the two initial vectors of the graph. First, using the two points given. Second, by using

gradient on the given plane.

A (1,3,-2) and B (3,4,3) and: 7x3y + 5z4 = 0

= (3-1)i + (4-3)j + (3+2)k

= 2i + j + 5k = (7x3y + 5z4 = 0)= (7x3y + 5z = 4) + (7x3y + 5z = 4) + (7x3y + 5z = 4)= 7i3j + 5k

x= =

= (5 + 15)i(10 - 35)j + (-6 - 7)k

= 20i + 25j13k

Dot product the obtained vector from the cross product and one of the points to get the

equation of the plane:

(20i + 25j13k) = 0: 20(x-1) + 25(y-3) -13(z+2) = 0

: 20x - 20 + 25y75 - 13z - 26 = 0

: 20x + 25y13z121 = 0

LEO ANGELO LUNAPROBLEM 79. Find the equation of plane through the points

a. b. c. d. e.

SOLUTIONS:

x

y

z


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a. Substitute the points to the equation . We can generate threeequations and use elimination method to solve for a, c, d in terms of b.

We solve this system of equation and we get,

Substitute a, b, c, and d values to the general equation of the plane.

Divide both sides of the equation by b and multiply by 2, we get

b. Substitute the points to the equation . We can generate three

equations and use elimination method to solve for a, b, d in terms of c.

We solve this system of equation and we get, Substitute a, b, c, and d values to the general equation of the plane.


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Divide both sides of the equation by c, we get

c. Substitute the points to the equation . We can generate three



Substitute a, b, c, and d values to the general equation of the plane. Divide both sides of the equation by c and multiply by, we get


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d. Substitute the points to the equation . We can generate three



Substitute a, b, c, and d values to the general equation of the plane.

Divide both sides of the equation by c and multiply by, we get


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e.

EARL JANZEN F. FAMILARA

Problem 80. Discuss the locus of each of the following planes showing intercepts and traces.(a) 2x+4y+3z-12=0


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To get the traces of the plane:xy-trace yz-trace xz-trace

Let z=0 Let x=0 Let y=02x+4y+0-12=0 0+4y+3z-12=0 2x+0+3z-12=02x+4y=12 4y+3z=12 2x+3z=12

This is the graph of the traces that we have got from the plane 2x+4y+3z-12=0

We just need the two coordinates be equal to 0. For example, we need to find the x-

intercept, let the value of y and z into 0, and vice versa.x-intercept y-intercept z-intercept2x+0+0-12=0 0+4y+0-12=0 0+0+2z-12=02x=12 4y=12 3z=12X=6 y=3 z=4(6,0,0) (0,3,0) (0,0,4) RICHARD PAUL T. MARTIN

(b) 3x-5y+2z-30=0


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Traces of the plane:xy-trace yz-trace xz-traceLet z=0 Let x=0 Let y=03x-5y+0-30=0 0+5y+2z-30=0 3x-0+2z-30=03x-5y=30 5y+2z=30 3x+2z=30

This is the graph of the traces we have got from the plane 3x-5y+2z-30=0

Getting the intercepts of the plane 3x-5y+2z-30=0x-intercept y-intercept z-intercept

3x-0+0-30=0 0-5y+0-30=0 0-0+2z-30=03x=30 -5y=30 2z=30x=10 y=-6 z=15(10,0,0) (0,-6,0) (0,0,15) RICHARD PAUL T. MARTIN

(c)x+y=6

Traces of the plane: (since z=0, there is no need to take the xz-trace) z-trace=x+y=6.yz-trace xz-traceLet x=0 Let y=00+y=6 x+0=6y=6 x=6


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This is the graph of the traces we have got from x+y=6.

Getting the intercepts of the plane x+y=6:x-intercept y-intercept z-interceptx+0=6 0+y=6 0+0=6x=6 y=6 z=0(6,0,0) (0,6,0) (0,0,0) Richard Paul T. Martin

(d)

2y-3z=6

(since x=0, we dont need to get the yz-trace because yz-trace is equal to the plane 2y-3z=6). Traces of the plane 2y-3z=6:

xy-trace xz-trace

Let z=0 Let y=02y-3(0)=6 2(0)-3z=62y=6 -3z=6y=3 z=-2This is the traces that we have got from the plane 2y-3z=6


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Getting the intercepts of the plane 2y-3z=6:x-intercept y-intercept z-intercept0-0=6 2y-0=6 0-3z=6x=0 2y=6 -3z=6

y=3 z=-2(0,0,0) (0,3,0) (0,0,-2) RICHARD PAUL T. MARTIN

(e) 2x-z=0

(since y=0, we dont need to get the xz-trace because it is equal to the plane 2x-z=0).yz-trace xy-traceLet x=0 Let z=02(0)-z=0 2x-0=0z=0 x=0

Therefore, there is no traces for the graph 2x-z=0.Getting the intercepts of the plane 2x-z=0x-intercept y-intercept z-intercept2x-0=0 0-0=0 0-z=0x=0 y=0 z=0(0,0,0) (0,0,0) (0,0,0) RICHARD PAUL T. MARTIN


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(f)x-6=0

(Since y and z are equal to zero, we dont need to get the y and z traces because it is equal

to the plane x-6=0).Trace of the plane x-6=0:yz-traceLet x=00-6=0Therefore, no traces for the plane x-6=0.Getting the intercepts of the plane 2x-6=0:

x-intercept y-intercept z-interceptx-6=0 0-6=0 0-6=0x=6 y=0 z=0(6,0,0) (0,0,0) (0,0,0) RICHARD PAUL T. MARTINPROBLEM 81. Use the normal form and write the equations of each of the following planes,

given:

a. SOLUTION:


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Let , , .Find.

()

() ()

b. SOLUTION:

Let , , .


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Find.

() () c. The foot of the normal from the origin to the plane is point .

SOLUTION:

Let , then


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d.

SOLUTION:

Let , , .Find.

()


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e. SOLUTION:

RAYMOND M. CORPUZ


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PROBLEM 82. Reduce each of the following equations to normal form and thus determinethe direction cosines and length of the normal.

a. b.

c.

SOLUTIONS:To find the direction cosines and the length of the normal, we reduce the given equation to The sign of the radical has the same sign as of D.

a.

The direction cosines are and the length is .b.

The direction cosines are and the length is .

c.


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The direction cosines are and the length is .MA. CRIZELDA M. IMPERIAL

PROBLEM 83. Determine the perpendicular distance from the point to the plane in each ofthe following:

a. Point plane

To find the distance of a point from the plane, use | | | | b. Point

plane


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To find the distance of a point from the plane, use | |

| | c. Point plane

To find the distance of a point from the plane, use | | | |


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d. Point plane

x

To find the distance of a point from the plane, use | | | |

ZEAREEN A. VERGARAPROBLEM 84.Find the acute angle between each of the following planes:

(a) , (b) ,

SOLUTION:

(a)Given the equations ,


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Let be a normal vector to the first plane.Let be a normal vector to the second plane. Then,

=

and

=

= = =

=

= = 60

PROBLEM 84.Find the acute angle between each of the following planes:(a) , (b)

,

SOLUTION:

(b) Given the equations , Let be a normal vector to the first plane.Let be a normal vector to the second plane. Then, = and =


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= = = = = = 97.82123551Since is not an acute angle, we should get its supplement. Thus, we write,180 - = 180 - 97.82123551 82.17876449

Therefore, the acute angle between and planes is 82.17876449.

KENNEDY B. GAPUZPROBLEM 85. Find the point of intersection of the planes 2x-y-2z=5, 4x+y+3z=1, 8x-y+z=5.SOLUTION:

To find the point of intersection, we need to use the elimination method.Eq.(1) Eq.(2) Eq.(3) Eq.(4) Eq.(5)Then we use equation 4 and 5 to get the value of z. we need to multiply this to -2 to eliminate the value of x.


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Substitute the value of z to any of the equations 4 or 5. And substitute x and z to get the

value of y.

Therefore, The point of intersections are

POI(x,y,z)=(3/2,4,-3) RICHARD PAUL T. MARTINPROBLEM 86 Find the points of intersection of the planes:

a.

Use elimination to get the coordinates of the point of intersection.

(1) (2) (3)

Use (1) and (2) to derive (4)

(1) (2)+ (4)

Use (2) and (3) to derive the (5).

(1) (2) (5)

Use (4) and (5) to get the value of.

Substitute the value to (5)


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Lastly, substitute both

and

values to (1).

The point of intersection of the three planes is at

b. , ,

The point of intersection of the three

planes is at .


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c.

=7

The point of intersection of the three planes is at

.

LIZZA MAE N. NAPUTO


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PROBLEM 87. Find the equation of the plane passing through the line of intersection of theplanes , and the point.

SOLUTION:

Let

At point, the required equation is

MA. CRIZELDA M. IMPERIAL


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PROBLEM 88. Find the equation of the plane passing through the line of intersection of theplanes and the point (1,2,3).SOLUTION:

We need to find the line of intersection of the two planes.

Let n1be the normal of the plane n2be the normal of the plane n as the normal of the line of intersection, where

n1 =

n2 =

We need to find the cross product of n1 and n2 to get the line of intersection.

With a point in the plane and the normal vector to the plane we can write an

equation of the plane. Using the point (1,2,3) we can now derive the equation by using

the standard form of the equation of the plane,

By subsitution, Therefore the equation of the plane passing through the line of intersection of the planes and the point (1,2,3) is

REISHSIER V. ABONITA

(1,2,3)


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PROBLEM 89. Find the equation of the plane passing through the line of intersection of theplanes

SOLUTION :

Substitute: Then: so, so, Let x = 0 , substitute: -6( ) + so,

Then, , , so,

Let A be the vector from (6, 0, 0) to

then,

Let B be the vector from ( 6, 0, 0) to then The normal to the plane is =

,General form of plane is Then,

= = then, ) 5

Lowest form : MARK FERDINAD SIOSON


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PROBLEM 90. Find the equations of the bisectors of the angles between the planes .SOLUTION:

Use ||

= ||

.|| = || = ` = Equation 1: =

= Equation 2: = =

MARK FERDINAD SIOSON


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PROBLEM 91. Find the equations of the bisectors of the angles between the planes and .SOLUTION:

To find the equations of the bisectors, we use the distance formula of a point to a plane.

Equations of the bisectors are loci of a point that is equidistant from the plane.

Let an arbitrary point on the locus. be the distance from the first plane. be the distance from the second plane. | | | | | | | |

| | | |

RAYMOND M. CORPUZ


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PROBLEM 92. Find the equations of the bisectors of the angles between the planes and .SOLUTION:

To get the equations of the bisector, get the locus of the point where the distance

from the two planes are equal.

Let be an arbitrary point on the locus be the distance from P to plane be the distance from P to plane We use:

| | | | Substituting values: | | | |

Therefore the equations of the bisectors of the angles are and .

GEMMA MONICA M. VILORIA


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PROBLEM 93.Write the equation of each of the following planes in the intercept form:a) b)

c)

SOLUTION:The intercept form of a plane a)

b)


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c)

4

REISHSIER V. ABONITAPROBLEM 94. Write the equations of the following planes whose intercepts are:

a. b. c.

SOLUTIONS:


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a. Use the intercept form of equation of a plane.

Substitute the intercepts to a, b, and c respectively.

Multiply both sides of the equation by -30 (Least Common Denominator). We get

b.

Use the intercept form of equation of a plane.

Substitute the intercepts to a and b respectively.

Multiply both sides of the equation by 6 (Least Common Denominator). We get


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c.

Use the intercept form of equation of a plane.

Substitute the intercept to a.

By cross multiplication, we get

ALDRIN V. PONCEPROBLEM 95. Show that the following planes are the faces of a parallelepiped: , , , .SOLUTION:


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Let A be the plane with normal vector = B be the plane with normal vector = C be the plane with normal vector = D be the plane

with normal vector =

By definition of parallel vectors: A and B are said to be parallel if and only if

=( ) Parallel

Parallel

Find the angle between the non parallel planes

| |

||||

|||||| ()() = , or |||||| ()() = , or |||||| ()() = , or |||||| ()() = , or Since planes A and C are parallel to each other, planes B and D are also parallel to eachother, and the angles between the planes AB, AD, BC, CD are all equal, therefore the

planes A, B, C and D are faces of a parallelepiped.

GEMMA MONICA M. VILORIA


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PROBLEM 96. Write the equation of the locus of a point whose distance from the plane is always twice its distance from SOLUTION:

Let be the equation of the desired plane.If the distance of a plane from is always twice its distance from , then by using the formula: Distance of a point to a plane

| | We will have the proportion: and Simplifying, and and

Therefore the equations will be: and GEMMA MONICA M. VILORIA

PROBLEM 97. Find the perpendicular distance between the parallel planes and SOLUTION:Because the two planes are parallel, they have the same normal vector.

| | | |


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||

units

EARL JANZEN F. FAMILARAPROBLEM 98. Find the perpendicular distance between the planes

and .SOLUTION:We take the normal of both equations. Since normal of two planes are equal, therefore the

two planes are parallel.

|| unitsEARL JANZEN F. FAMILARA

PROBLEM 99. Describe the locus of the equation SOLUTION.

Simplifying the equation we get,( Then, either = 0 or = 0 pass thru the origin


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Therefore we can state that the locus of the equation is twoplanes with a point of intersection in the origin.

BENITO JY M. GARCIAPROBLEM 100. Describe the locus of the equation SOLUTION:

Simplifying the equation we get, Therefore we can state that is a hyperboloid with the radius 2 and

lies in the xy plane.

REISHSIER V. ABONITA

z

y

x


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PROBLEM 101. Write the equation of the locus of point whose distance from the plane is equal to the its distance from the point.SOLUTION:

Let be an arbitrary point on the locus.

be the distance of

to the plane

be the distance of from the pointTo find the distance of a point to a given plane , we use | | | |

| |

| |

To find the distance between two points, from the point, If , then | |

the planes - supplementary problems

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