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Math 1141 Assignment 2019 Question 1. You are given two planes in parametric form 1.1. Find vectors N1 and N2 that are normal to Π1 and Π2 respectively and explain without any further calculations how you can tell that the two planes intersect in the form of a line. Answer. To find the normal to a plane, take the cross product between two vectors that are parallel to the plane using the formula for the cross product. One simple way to remember the cross product is, To find the first coordinate we ignore the first row ( a 1 ¿ b1 ) then we are left with a 2 , a 3 ,b 2∧b 3 . Then we simply subtract a 3 b 2 from a 2 b 3 . This makes it easier to see which coordinates are involved in a certain row. To find the second coordinate, ignore the second row and then subtract a 1 b 3 from a 3 b 1 . To find the third coordinate, ignore the third row and then subtract a 2 b 1 from a 1 b 2 . This is an example of the first one:

( )( )(

b1 ¿ a 2 b 3 − a3 b 2 a1 a 2 × b 2 = a 3 b 1−a1 b3 a3 b3 a1 b 2−a2 b1

∴

)

The normal to each plane is given by

( )()(

)( )

−1 ( 3) −2(1) −5 1 2 × N 1= −1 = 1 1= 4−3 −2 1+2 3 3

( )( )(

)( )

1 (−1 ) −3(−2 ) 1 2 5 N 2= 1 × 3 = 2 (−2 ) −1(−1 ) = −3 −2 −1 1 1(3 )−1 (2)

By definition two non-zero vectors a and b are parallel if and only if there is a nonzero real number λ such that a = λb.

By considering the two normals:

( )( )

5 −5 λ 1 ≠ −3 1 3

for any λ ∈ R .

Now we can conclude that these two planes are not parallel to each other, hence they must intersect in the form of a line. 1.2. Find Cartesian equations for Π1 and Π2. Answer. The Cartesian form of a plane is given by : a x 1+ b x2 +c x 3=d

where the coefficients of the variables is the normal to the plane. The Cartesian form of Π1 is: −¿ 5�1 + �2 + 3�3 = � 1

By substituting in (0, −¿ 3, 1) to the above equation we can find �1 . The value of �1 is: −¿ 5(0) + (-3) + 3(1) = �1 = 0

The Cartesian equation for Π1 is: −5 �1 + �2 + 3�3 = 0

Now, the Cartesian equation of Π2 is: 5�1 − 3�2 + �3 = �2 Similarly, by substituting in (0, −¿ 1, −¿ 3) into the above equation we can find �2 . The value of �2 is: 5(0) – 3(–1) + (–3) = �2 = 0 The Cartesian equation for Π2 is:

5�1 – 3�2 + �3 = 0 1.3. Assign � to either �1,2,�3 and then use the Cartesian forms to write the other two variables in terms of � and hence write down a parametric vector form of the line of intersection L. Answer. Let line of intersection of Π1 and Π2 be L. By assigning � to �2, let �2 = �. ∴ By substituting �2 = � into the two Cartesian planes derived earlier we get, −5 �1 + � + 3�3 = 0

5�1 – 3 � + �3 = 0 Subtract these two to solve for omega. −¿ 2 � + 4 �3 = 0 1

�

�1 = 2

1

�

1

�

�3 = 2 Sub into the Cartesian form of the plane.

Collecting all our values, �1 = 2 �2 = � 1

�3 = 2

�

is a relationship between �1, �2, �3 with a common parameter �. Since there is no number without a multiple of �, A would be 0 in the equation x = A+ωv. ∴ x = ωv ∴ The form of the line of the two intersections is given by:

0.5 L=ω 1 0.5

( )

1.4. For your second method, substitute expressions for �1, �2, �3 from the parametric form of Π2 into your Cartesian equation for Π1 and hence find a parametric vector form of the line of intersection L. Answer. The Cartesian equation for Π1 is: −5 �1 + �2 + 3�3 = 0

By comparing the components of the plane Π2 in parametric vector form, we get the following parametric equations �1 = � 1 + 2�2 �2 = �3 =

−¿ 1 + �1 + 3�2

−3 − 2�1 −¿

�2

By substituting in the values of �1, �2, �3 from the above equations into the plane and expanding and collecting like terms, −5 �1 −¿

10�2 −¿ 1 + �1 + 3�2 −9 − 6�1 −¿ 3�2 = 0 −¿ 10 �1 −¿ 10�2 −¿ 10=0

Divide through by – 10 �1 + �2 −¿ 1=0 �2 = 1 −¿

�1

Use this relationship in the parametric form of the plane the parameters were taken form to get the line

( ) ()

()

0 1 2 L= −1 + μ1 1 +( 1−μ 1 ) 3 −3 −2 −1

2 1 L= 2 + μ 1 2 1 −4

( ) ()

1.5. Show that the line in part 1.3. and 1.4. are the same. Answer. As said earlier, by definition two non-zero vectors a and b are parallel if and only if there is a non-zero real number λ such that a = λb.

( ) () ( )()

In this case

0.5 1 1 =λ 2 0.5 1

, where λ is 2

0.5 0.5 1 =2 1 0.5 0.5

Therefore, lines in 1.3. and 1.4. are parallel to each other and go through a common point. m=N 1× N 2 and show that m is parallel to the line you found in parts 1.3 1.6.Find and 1.4 The cross product of the two normals is given by

( )( )( ) −5 5 10 1 × −3 = 20 1 3 10

This is parallel to the line since,

( )()

0.5 10 20 1 = 20 0.5 10

1.7 Geometric interpretation of the result of 1.6 Since the vector perpendicular to the normals is parallel to the line, the line is thus parallel to both planes. This can be imagined visually using two sheets of paper intersecting one another. The line of intersection must be parallel to both planes. Question 2. Show that the equation −3 x

e +4 cos(9 x) π has a unique solution for x∈[0, ]. 9

Answer. Let

=0

�(�) = e−3 x +4 cos(9 x) Using Intermediate Value Theorem for x = 0 (0 ) (0) = e−3 + 4 cos( 9 ( 0 ) ) =1+4 =5 >0

Now for x =

π , 9

�(

π −3( ) 9

π )= 9

e

+4 cos(9(

π )) 9

−π

= e 3 +4 cos(9 π ) −π 3

¿ , e + 4 cos(9 π )

∴ (0) > 0, where � (...