Calculus Homework: Plane Equations, Line Intersection, Error Analysis

This calculus assignment presents solutions to problems involving plane equations, line intersections, and error analysis. The first question explores the intersection of two planes, determining their normal vectors, deriving Cartesian equations, and finding the parametric vector form of their line of intersection. The second question investigates the existence of a unique solution to an equation using the Intermediate Value Theorem and Rolle's Theorem. The third question approximates the error in a calculation using the Mean Value Theorem. The solutions demonstrate a strong understanding of calculus concepts and problem-solving techniques.

QUESTION 1
π1 :
( x1
x2
x3
) =
( 1
−3
2 ) + λ1
( 2
−2
1 )+ λ2
( 3
0
3 )
π2 :
(x1
x2
x3
)=
( 1
0
−1 )+μ1
( 2
1
−2 )+ μ2
(3
3
0 )
a.
n1=¿ 3 ,0,3> x <2 ,−2,1> ¿
¿
|i j k
3 0 3
2 −2 1|=| 0 3
−2 1|i−|3 3
2 1| j+|3 0
2 −2|k=6 i+3 j−6 k
Factorizing to get unit normal vector,
n1=¿ 2,1,−2>¿
n2 =¿ 3,3,0> x<2,1,−2>¿
¿
|i j k
3 3 0
2 1 −2|=|3 0
1 −2|i−|3 0
2 −2| j+|3 3
2 1|k =−6 i+6 j−3 k
Factorizing to get unit normal vector,
n2 =¿ 2,−2 ,1>¿
The elements of the normal vectors of the planes are neither the same nor multiples of each other
hence their cross product is not a zero. Therefore, the two planes are not parallel and must
intersect at a given point or line.
b. Cartesian forms of the equations
For plane π1 , the parametric equations are:
x1=1+2 λ1 +3 λ2 …(i)
x2=−3−2 λ1 …(ii)
x3=2+ λ1+3 λ2 … (iii)
¿ ( i ) ,

2 λ1=x1−1−3 λ2
Replacing this∈ ( ii ) ,
x2=−3− ( x1−1−3 λ2 ) =−2−x1 +3 λ2
So that ,
3 λ2 =x2 +x1 +2∧λ1=−3
2 − x2
2
Replacing theabove two equations∈ ( iii ) ,
x3=2+ λ1+3 λ2=2− 3
2 − x2
2 +x2+x1 +2
x3= 5
2 + x2
2 + x1
ℜ−arranging∧simplyfying the above equation ,
x1+ x2
2 −x3 + 5
2 =0
The cartesian equation of π1 is;
2 x1 + x2−2 x3+5=0
For plane π2 , the parametric equations are:
x1=1+2 μ1+3 μ2 …(i)
x2=μ1 +3 μ2 …(ii)
x3=−1−2 μ1 …(iii )
¿ ( i ) ,
3 μ2=x1−1−2 μ1
Replacing this∈ ( ii ) ,
x2=μ1 + x1−1−2 μ1=x1−1−μ1
So that ,
μ1=x1−x2−1
Replacing theabove equation∈ ( iii ) ,
x3=−1−2(x1−x2 −1)
x3=−1−2 x1 +2 x2 +2 ¿
ℜ−arranging∧simplyfying the above equation , cartesianequation of π2 is ;
2 x1 −2 x2 + x3−1=0
c. Let x1=ω,
The cartesian equation s thus reduces ¿ ;
2 ω−2 x2 +x3−1=0
2 ω+ x2−2 x3+ 5=0
Expressing x2 ∈terms of ω ,
x3=1+2 x2 −2 ω

2 ω+ x2−2 x3+ 5=0
2 ω+ x2−2 ( 1+2 x2 −2 ω ) +5=0
3 x2=6 ω+3
∴ x2=1+2 ω
Expressing x3 ∈terms of ω ,
2 ω+ x2−2 x3+ 5=0
x2=−2 x3−2 ω−5
2 ω−2 x2 +x3−1=0
2 ω−2 ( −2 x3−2 ω−5 )+ x3−1=0
3 x3=9+6 ω
x3=3+2 ω
Hence the line of intersection takes the parametric vector form
L=¿ 0,1,3>+ ω<1,2,2>¿
d.
x1=1+2 μ1+3 μ2
x2=μ1 +3 μ2
x3=−1−2 μ1
Replacing the above equations∈the cartesian equation of plane1 ,
2 x1 + x2−2 x3+5=0
2 ( 1+2 μ1+3 μ2 ) + μ1 +3 μ2−2 ( −1−2 μ1 ) +5=0
9 μ1 +9 μ2 +9=0
μ1 + μ2 +1=0
L=¿ 1,1,0>+¿1 , 2 ,2> μ
e. The parametric equations, though slightly different, represents the same line due to the
same vector direction obtained in both cases.
f.
m=n1 x n2
¿<2,1 ,−2>x<2,−2,1>¿

¿
|i j k
2 1 −2
2 −2 1 |=| 1 −2
−2 1 |i−|2 −2
2 1 | j+|2 1
2 −2|k
¿−3 i−6 j−6 k
= <1, 2, 2>
¿ show that mis∥¿ the line , cross product of thetwo vector directions isevaluated
m x lc=¿−3 ,−6 ,−6> x <1,2,2>¿
¿
| i j k
−3 −6 −6
1 2 2 |=|−6 −6
2 2 |i−|−3 −6
1 2 | j+|−3 −6
1 2 |k
= <0, 0, 0>
Hence the vector mis∥¿ the line of ∩¿
g. The line intersects both the planes with a direction vector m=n1 x n2 because m is
perpendicular to both the normal vectors n1∧n2 and is therefore parallel to the two planes.

QUESTION 2
To show that the equation, e−7 x+6 cos ( 8 x ) =0 , has a unique solution over the interval [0, π
8 ],
Let e−7 x+6 cos ( 8 x ) be a given function say f ( x )
i .e . f ( x )=e−7 x +6 cos ( 8 x )
This function , f ( x ) ,is continous
the given interval
Step 1: ¿ the given interval ,the function f ( x )=0 is first shown ¿ haveat least
one solution .
Replacing thelower ∧theupper values of the interval given ,
f ( 0 ) =e−7 (0)+6 cos ( 0 ) =7
f ( π
8 )=e−7 ( π
8 )+ 6 cos ( π
8 )=5.607
Both f ( 0 ) ∧f ( π
8 ) are greater than zero∧thus there exist no f ( c ) =0 for some
c ∈theinterval [ 0 , π
8 ] … inter−mediate valuetheorem
Step 2: f ( x ) =0is then shown ¿ have at most one solution ∈theinterval [ 0 , π
8 ]
By first assuming that more than one solution exist for
f ( x )=0 ,two arbitrary values are picked ; say x=a , x=b∧a <b
Because of the continuity of f on [ a , b ] , its differentiability on ( a ,b ) ∧the that!
f ( a )=f ( b )=0; by Roll e' s theorem, for some d ∈theinterval ( a , b ) ,follows that
f ' ( d )=0
But ,

f ' ( x )=−7 e−7 x−48 sin ( 8 x ) <0 for all values of x ∈the interval [0 , π
8 ]
i .e .sin ( 8 x ) ∧e−7 x are both positive∈the intervalthus the∑ of their negationmust yield
negative values .
Thisis an implication that f ' ( d ) can never be 0;
f ' ( d ) ≠ 0
Therefore , a contradiction o ccur depicting that f ( x ) =0 has at most one solution∈¿
the interval [ 0 , π
8 ] . Coupling this reasoning withthe that!f ( x ) =0 has got at
at least one solution ,is reasonable ¿ conclude that theequation has aunique solution∈the
interval provided .
QUESTION 3
To approximate the error in ( 16.04 )
1
4 by ( 16 )
1
4 ,
Let a function f ( x ) =x4 such that f ( 16 ) =16
1
4 =2
∴ f ' ( x )= 1
4 x
−3
4
so that ,
f ' ( 16 ) = 1
4 ( 16 )
−3
4 = 1
32
Using mean value theorem ,
Error , F ( x )=f ( 16 ) + f ' ( 16 ) ( x−16 )
¿ 2+ 1
32 ( x−16 )
F ( 16.04 )=2+ 1
32 ( 16.04−16 )
¿ 2.00125