Last name: name: 1 Quizz 1 (Notes, books, and calculators are not authorized)

```Last name:
name:
1
Quizz 1 (Notes, books, and calculators are not authorized)
Show all your work in the blank space you are given on the exam sheet. Always justify your
Question 1: Find a parametric representation of the line in R4 that passes through P =
(1, 1, 3, 3) and is orthogonal to the hyperplane 2x1 + 4x2 + 6x3 − 8x4 = 5.
A normal vector to the hyperplane 2x1 + 4x2 + 6x3 − 8x4 = 5 is u := (2, 4, 6, −8). The line in
question is the collection of points X = (x1 , x2 , x3 , x4 ) so that X − P is parallel to u. This means
that there is a real number t so that
X − P = tu = (2t, 4t, 6t, −8t).
The parametric representation of the line is
x1 = 1 + 2t,
x2 = 1 + 4t,
x3 = 3 + 6t,
x4 = 3 − 8t.
Question 2: Find a unit vector orthogonal to u = (1, 2, 3) and v = (1, −1, 2).
It is a known property of the cross product that the following vector w := u &times; v is orthogonal to u
and v. By definition
w = (2 &times; 2 − 3 &times; (−1), 3 &times; 1 − 2 &times; 1, 1 &times; (−1) − 2 &times; 1) = (7, 1, −3).
By normalizing w we obtain the desired vector:
1
w
= √ (7, 1, −3).
kwk
59
(1)
2
Quizz 1, September 7, 2010
Question 3: Find an equation of the hyperplane in R4 that passes through P = (3, −4, 1, −2)
and is normal to u = (1, 1, −1, −1).
By definition H is the collection of points X = (x1 , x2 , x3 , x4 ) so that (X − P )&middot;u. This means
0 =(X − P )&middot;(1, 1, −1, −1) = (x1 − 3, x2 + 4, x3 − 1, x4 + 2)&middot;(1, 1, −1, −1)
= x1 − 3 + x2 + 4 − (x3 − 1) − (x4 + 2) = x1 + x2 − x3 − x4 .
The equation of the hyperplane in question
0 = x1 + x2 − x3 − x4 .
Question 4: Prove the triangle inequality: ku + vk ≤ kuk + kvk.
By definition of the dot product, we have the folowing equality:
ku + vk2 := (u + v)2 = (u + v)&middot;(u + v) = u&middot;u + u&middot;v + v&middot;u + v&middot;u.
Using the Cauchy-Schwarz inequality, we obtain
ku + vk2 ≤ kuk2 + kukkvk + kvkkuk + kvk2 = (kuk + kvk)2 .
Taking the square root gives the triangle inequality:
ku + vk ≤ kuk + kvk.
```