Plane Equations in 3D Geometry

This assignment focuses on solving problems related to finding the equations of planes in three-dimensional space. It covers various scenarios such as planes passing through given points, planes parallel to specified lines, and planes defined by their normal vectors. The questions guide students through the process of deriving plane equations using cross products, direction ratios, and point-normal form.

CALCULUS
Student id
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Question 1
Let
Vector 1=a
Vector 2=v
Point located on the line ( a+ tv ) which is near to the origin in such as way that a ┴ v .
a ┴ vonly when the dot product would be equal to zero.
a . v =0
Line equation ¿ a+ t v
The distance (t , a+t v) from origin (0, 0) would be determined as shown below:
¿ √ ( a−c ) 2+ ¿ ¿
a=a , b=a+tv ,c =0 , d=0
¿ √ ( t−0 ) 2 +¿ ¿
¿ √ t2 +¿ ¿
Now,
let
1

f ( x )= √t2 +¿ ¿
It is essential to minimize the square of the distance and hence,
g ( x )=t2+¿
The next step is to determine the first derivative of f ( x ) .
d
dx f ( x )= d
dx ¿
¿ t +v (a+tv)
√t2+¿ ¿ ¿
Further, find the first derivative of g ( x )
d
dx g ( x ) = d
dx t2 +¿
¿ 2 t+¿
Let the value of denominator close to zero.
t2+ ¿0 (derivative is defined)
t=0 , a+tv=0∨t=−a
v
This is termed as critical point of the and therefore, point (−a
v ,0 ¿
is not close to the origin.
2

Question 2
All unit vectors which are perpendicular to the vector ( 5
12 )=?
Let the vector is a = 5i +12 j
Two vectors would be termed as perpendicular to each other when their dot product would be
equal to zero.
cos θ= a . b
|a||b|
θ=90 ° ( perpendicular )
a . b
|a||b|
=cos 90=0
a . b=0
Assume that vector (ai+bj) is perpendicular to (5i +12 j).
Now,
¿ ( ai+bj ) .(5 i+ 12 j)
3

5 a+12 b=0 … … …… . ( 1 )
In order to determine the unit vector which would be perpendicular to the (5i +12 j) hit and trial
method needs to be applied.
at a=12 , b=−5
( 5∗12 )+ ( 12∗−5 ) =0 satisfied
u1= 12i−5 j
√ ( 12 )2+ (−5 )2 =( 12i
13 − 5 j
13 )
At a=−12, b=5
¿
u2= −12 i+ 5 j
√ (−12 )2 + ( 5 )2 =(−12 i
13 + 5 j
13 )
Hence, the two unit vectors i.e. ( 12i
13 − 5 j
13 ) and ( −12i
13 + 5 j
13 ) are perpendicular to ( 5
12 ) .
Question 3
Angle between any of the two line segment from carbon atom to hydrogen atom =?
4

Let a be the vector along AC∧b be the vector along CD .
AC= ( 1
2 −0 )i+( 1
2 −0 ) j+ (1
2 −1 )k
AC=a= 1
2 i+ 1
2 j− 1
2 k
Similarly
CD=b= 1
2 i− 1
2 j+ 1
2 k
5

Now
cos θ= a . b
|a||b|
cos θ= ( 1
2 i+ 1
2 j− 1
2 k ) . ( 1
2 i− 1
2 j+ 1
2 k )
√ ( 1
2 )
2
+( 1
2 )
2
+(−1
2 )
2
√ (1
2 )
2
+( −1
2 )
2
+( 1
2 )
2
¿
1
4 − 1
4 − 1
4
√ 3
4 √ 3
4
¿
−1
4
3
4
¿− 4
12
cos θ=−1
3
θ=cos−1
(−1
3 )
θ=121.6 °
Hence, the angle would be 121.60.
6

Question 4
Given figure
Let the center of the systemO(0,0,0)
And the corner would be A ( ± 1, ± 1, ± 1 )
Hence, the by co-ordinate system as per the figure is highlighted below:
OA= (−1i+1 j+1 k )
OB= (−1 i+ 1 j−1 k )
OA .OB = ( −1i+1 j+1 k ) . ( −1 i+1 j−1 k )
¿ 1+1−1=1
Hence,
|OA|= √ (−1)2 +(1)2+(1)2= √ 3
7

|OB|= √ (−1)2+(1)2 +(−1)2 = √ 3
Now,
OA .OB =|OA||OB|cos β
cos β= OA .OB
|OA||OB|
cos β= 1
√3 × √3 = 1
3
β=cos−1
( 1
3 )
β=78.36 °
Question 5
It is given that P and Q points are located in R3.
The aim is to prove that R is the midpoint of PQ only when X is equidistant from P and Q and
XR is orthogonal to PQ.
Let the X = ( x , y , z )
Point X is equidistance from P and Q and hence, vector XR is orthogonal to PQ.
8

XR . PQ=0…………….. (1)
Here,
XR=XO +¿
Hence,
XR . PQ=0
( XO +¿ ¿ . PQ=0
XO . PQ +¿ . PQ=0
( −OX ) . PQ +¿ . PQ=0
OX . PQ=¿ . PQ … … …… … …… .. ( 2 )
Further,
PQ=PO+ OQ
R is midpoint of PQ and hence, ¿= PQ
2 = OP+OQ
2
9

From equation (2)
OX . PQ=¿ . PQ
OX . PQ=( OP +OQ
2 ).(PO +OQ)
PO=−OP
OX . PQ=( OP +OQ
2 ) .( OQ−OP )
OX . PQ= 1
2 (OP2−OQ2)
This is the equation of the plan on which x lies and hence, the above assumptions that X is
equidistance from P and Q and hence, vector XR is orthogonal to PQ. Also, R is the midpoint of
PQ and therefore, P and Q points are located in R3 .
Question 6
Cross product
10

¿ (2
0
0 )× (−1
0
1 )
¿
¿ ( 2 0 0 ) × ( −1 0 1 )
Rule of cross product of the two vectors
¿ ( 0.1−0.0 0.(−1)−2.1 2.0−0.(−1) )
¿ ( 0−0 0−20−0 )
¿ ( 0−2 0 )
¿ (−
0
2
0 )
11

Question 7
Cross product
¿ (1
1
0 )× (0
1
1 )
¿
¿ ( 1 10 ) × ( 01 1 )
Rule of cross product of the two vectors
¿ ( 1.1−0.1 0.0−1.11.1−1.0 )
¿ ( 1−00−1 1−0 )
¿ ( 1−1 1 )
¿ (−
1
1
1)
12

Question 8
Given figure
Based on the above figure it can be concluded that option (b) and (c) would form a right-handed
system.
Answer: ( c ) { v , u , w } ∧ ( b ) { w , v , u }
13

Question 9
Given vectors are
(3
1
1 )∧ (−1
2
1 )
¿
¿ ( 3 11 ) ∧ ( −12 1 )
 The first step is to determine the cross product of these two vectors.
¿ ( 3 11 ) × (−1 2 1 )
Rule of cross product of the two vectors
¿ ( 1.1−1.2 1.(−1)−3.1 3.2−1.(−1) )
¿ ( 1−2−1−3 6+1 )
¿(−1−4 7)
14

¿ ( −
−1
4
7 )
 The first step is to determine the unit vector
u= (−1−4 7 )
√¿ ¿ ¿
u= (−1−4 7 )
√66
u=
( −1
√ 66 , −4
√66 , 7
√66 )
Therefore, it can be said that the vector orthogonal to the given vectors ( 3 11 ) ∧ ( −12 1 ) is
u=( −1
√ 66 , −4
√66 , 7
√66 )
Further, it is essential to note that the negative of the orthogonal vector is also termed as
orthogonal. Hence, −u=( 1
√ 66 , 4
√ 66 , −7
√ 66 ) is also termed as orthogonal vector.
Finally, the two vectors orthogonal to ( 3 11 )∧ (−12 1 ) are shown below:
u=
( −1
√ 66 , −4
√66 , 7
√66 )
−u=
( 1
√66 , 4
√ 66 , −7
√66 )
15

Question 10
Volume of parallelepiped spanned by vector =?
u=¿ 1,0,4>¿
v=¿ 1,3,1>¿
w=← 4,2,6>¿
16

It can be seen from the above figure that vector u and v are perpendicular to each other. Hence,
u × v=¿ 1,0,4>×<1,3,1>¿
Rule of cross product of the two vectors
¿ ( 0.1−4.3 4.1−1.11.3−0.1 )
¿( 0−12 4−1 3−0)
u × v= ( −12 3 3 )
Now,
Find dot product of u × v withvector w
( u × v ) . w= (−12 3 3 ) . (−4 2 6 )
Rule of dot product of the two vectors
¿ (−12 ) . (−4 )+ ( 3.2 )+ ( 3.6 )
¿ 48+ 6+18
¿ 72
17

Hence, the volume of parallelepiped spanned by vector is 72.
Question 11
Area of triangle =?
P=(1,1,5)
18

Q= ( 3 , 4,3 )
R=(1,5,7)
Needs to find PQ and PR from the above values
Let
Compute PQ∧PR
P Q= ( 3−1 , 4−1, 3−5 ) =(2 ,3 ,−2)
PR= ( 1−3 ,5−4 ,7−3 ) =(−2, 1 , 4)
Area of triangle would be determined as shown below:
∆= 1
2 ∨PQ × PR∨¿
PQ × PR=(2, 3 ,−2) ×(−2 ,1 , 4)
Rule of cross product of the two vectors
¿ ( 3.4− (−2 ) .1 (−2 ) .(−2)−2.4 2.1−3.(−2) )
¿ ( 12+2 4−8 2+6 )
¿ ( 14−4 8 )
19

Now,
|PQ × PR|= √ ( 14 ) 2+ ( −4 ) 2+ ( 8 ) 2= √ 276=16.61
∆= 1
2 |PQ × PR|= 1
2 ×16.61=8.306
Hence, area of triangle is 8.306.
Question 12
Equation of the plane in scalar form =?
With normal vector n=
(−
2
4
1 )
Passing through the point ( 1
3 , 2
3 , 1 )
Let the P is the point passes through, R which is an arbitrary point (x, y, z) on the plane.
n . ( R−P )
20

(−
2
4
1 ) . ( x− 1
3 , y− 2
3 , z−1 )=0
(−
2
4
1 ) ( 3 x−1
3 , 3 y −2
3 , z−1 )=0
2. ( 3 x−1
3 )−4 (3 y −2
3 )+1 ( z −1 )=0
6 x−2
3 +−12 y+8
3 + z−1=0
6 x−2−12 y +8+ 3 z−3=0
6 x−12 y +3 z+3=0
Hence the equation of plane in the form of ax +by +cz =d is shown below:
6 x−12 y +3 z=−3
Question 13
Equation of plane passing through three points =?
P= (5,1 , 1 ), Q= (1,1,2 ), R= ( 2,1,1 )
PQ=Q−P= ( 1−5 1−1 2−1 ) = ( −4 , 0 ,1 )
PR=R−P= ( 2−5 1−11−1 ) =(−3 , 0 , 0)
21

Both PQ and PR lie in the plane and therefore, their cross product i.e. PQ× PRis orthogonal to
the plane and hence, it can be said that it is a normal vector.
PQ × PR= (−4 , 0 ,1 ) × (−3 ,0 , 0 )
Rule of cross product of the two vectors
¿ ¿
¿ ( 0−0−3−0−0−0 )
n=PQ × PR=(0−3 0)
Point P= ( 5,1 , 1 ) and normal vector n, would be used to find the equation of plane.
0 ( x−5 ) −3 ( y−1 ) +0 ( z−1 ) =0
−3 y +3=0
Hence, the equation of plane would be 0x-3y +0z +3 = 0
Question 14
Equation of plane =?
Plane is passing through the point ¿(4 , 1 , 9)
Plane is parallel to x + y + z=3
22

The normal vector to the plane (which is || to the plane) would be determined as highlighted
below:
n= ( 1,1,1 )
The equation of plane which is passing through the point (x0 , y0 , z0) and is having direction ratio
of normal vector (A, B, C) is shown below:
A ( x−x0 ) + B ( y− y0 ) +C ( z−z0 )=0…….(1)
In present case ( A , B , C )=Normal vector= (1,1,1 ) (x0 , y0 , z0) = Plane is passing through the
point ¿(4,1,9)
Equation (1) A ( x−x0 ) + B ( y− y0 ) +C ( z−z0 )=0
1 ( x−4 ) +1 ( y −1 ) +1 ( z−9 )=0
x−4+ y −1+ z−9=0
x + y + z−14=0
x + y + z=−14
Equation of plane x + y + z=14
Question 15
Equation of the plane =?
Given line equations
23

r1 ( t ) =
( t
2t
3 t )
r2 ( t ) =
( 3t
t
8t )
 Cross product of the line equations (Normal vector)
r1 ( t ) ×r 2 ( t ) =
( t
2t
3 t ) × ( 3t
t
8t )
o r
r1 ( t ) ×r 2 ( t ) = ( 1 23 ) × ( 3 1 8 )
Rule of cross product of the two vectors
r1 ( t ) ×r 2 ( t ) = ( 2.8−3.13.3−1.8 1.1−2.3 )
r1 ( t ) ×r 2 ( t ) = ( 13 1−5 )
 Point of plane would be computed as
¿ ( 2 ,1 , 0 )
24

Equation of plane
13 ( x−2 ) +1 ( y−1 ) −5 ( z−0 ) =0
13 x+ y−5 z=27
25

Question 16
Let
 Plane 1 :2 x−4 y−z=3Or 2 x−4 y−z−3=0
Compare with a1 x+ b1 y+ c1 z +d1=0
a1=2 ,b1=−4 , c1=−1 , d1 =−3
 Plane 2 :−6 x+ 12 y +3 z=1Or −6 x +12 y +3 z−1=0
Compare with a2 x +b2 y+ c2 z +d2=0
a2=−6 , b2=12 , c2=3 , d2=−1
These two planes would be termed as parallel to each other only when the below highlighted
condition is satisfied.
a1
a2
= b1
b2
= c1
c2
2
−6 =−4
12 =−1
3
Simplification
−1
3 =−1
3 =−1
3
The condition is satisfied and thus, the given planes are parallel.
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