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Solution To Test 1 - Linear Algebra - FALL Semester 2019

1. The document provides the solution to 5 problems in linear algebra. It gives the step-by-step working for computing vectors, planes, and solving systems of linear equations. 2. It finds the direction vector and parametric/symmetric equations of a line passing through two points. 3. It determines an equation of a plane parallel to a given line by finding a normal vector perpendicular to the direction vector. 4. It uses Gauss-Jordan elimination to solve a system of 3 linear equations, obtaining a solution in vector form.

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Juan Osorio
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113 views3 pages

Solution To Test 1 - Linear Algebra - FALL Semester 2019

1. The document provides the solution to 5 problems in linear algebra. It gives the step-by-step working for computing vectors, planes, and solving systems of linear equations. 2. It finds the direction vector and parametric/symmetric equations of a line passing through two points. 3. It determines an equation of a plane parallel to a given line by finding a normal vector perpendicular to the direction vector. 4. It uses Gauss-Jordan elimination to solve a system of 3 linear equations, obtaining a solution in vector form.

Uploaded by

Juan Osorio
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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Solution to Test 1 - Linear Algebra - FALL Semester 2019

Please find below an outline of the solution. If something is not clear please talk to
me.
   
1 3
1. Let ~u = −4 and ~v =  1 .
   

α α

(a) Compute
√ the length of the vector ~u.
k~uk = 17 + α2 .
(b) Compute 2~ u + 7~
 v .
23
2~u + 7~v = −1.
 

9α
(c) Compute proj~v ~u.  
3
2 −1
proj~v ~u = k~~uv·~kv2 ~v = αα2 +10  1 .
 

α
(d) Find all values α for which ~u and ~v are orthogonal.
~u · ~v = α2 − 1 = 0 ⇐⇒ α = ±1.
(e) For which values of α is the angle between ~u and ~v acute and for which values is it
obtuse?
◦
The angle between two vectors is acute iff it is less than 90 , so iff the dot product is
positive. So we have to solve ~u · ~v = α2 − 1 > 0. This yields α ∈ (−∞, −1) ∪ (1, ∞).
Analogously for the obtuse angle we have to solve ~u · ~v < 0, which gives α ∈ (−1, 1).

2. Let P = (1, 0, −4), Q = (3, 4, 0), and R = (0, −2, 1) be three points in R3 .

(a) Find an equation for the plane containing the points P, Q and R.
To find the equation of the plane, we have to find the normal vector. For that we use
the three given points to find two
 vectors
 in the plane and compute the cross product
2
between them. We can use ~n = −1. Further we compute ~x · ~n = p~ · ~n, which gives
 

0
the equation of the plane we seek 2x − y = 2.
(b) Find the distance of the above plane from the origin.
d = k~n2 k = √25 .
(c) Give an equation of the plane through the origin parallel to the above plane.
2x − y = 0.
3. Let a line go through the points (1, 2, 3) and (−1, 2, −2).

(a) Find the vector equation of the line.


Using the two given points we find the direction vector of the line. We obtain
   
1 −2
~x = 2 + t  0  .
   

3 −5

(b) Find the parametric equation of the line.



 x = 1 − 2t
y=2
z = 3 − 5t.

(c) Find the symmetric equation of the line.


x−1
−2
= z−3
−5
, y = 2.

4. Find an equation of a plane parallel to the line



 x = 3 + 2t
y=t
z = 1 − 3t.

A plane with a normal vector ~n is parallel to a line with


 adirection vector ~v if they are
2
perpendicular. The direction vector of this line is ~v =  1 . We need a vector ~n so that
 

  −3
1
~n · ~v = 0. For example we can take as ~n = 1 and as the required plane is x + y + z = 1.
 

5. (a) Use Gauss-Jordan elimination to solve the system of linear equations:



 x+y−z =7
4x − y + 5z = 4
6x + y + 3z = 18.

First we write the augmented matrix:


 
1 1 −1 7
 4 −1 5 4 
6 1 3 18

Then we execute the following row operations: R2 → R2 − 4R1 , R3 → R3 − 6R1 , R3 →


R3 − R2 , R2 → − 51 R2 , and R1 → R1 − R2 , to obtain:
 
1 0 4/5 11/5
 0 1 −9/5 24/5 
0 0 0 0
We read up the corresponding system:

x + 4/5z = 11/5
y − 9/5z = 24/5.

Letting z = t and rearranging we obtain the solution:


   
11/5 −4/5
~x = 24/5 + t  9/5  .
   

0 1

The solution set is a line.


(b) Determine whether the following augmented matrices are in reduced row echelon form
(rref),
 in row echelon form (ref) or neither:  
1 0 0 b1   1 3 2 5 b1
1 0 0 5 b1
(i)  0 1 0 b2  (ii) (iii)  0 1 3 6 b2 
0 0 1 2 b2
3 0 1 b3 0 0 0 0 b3
(i) neither (ii) rref (iii) ref

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