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such that the distance between Q and the line {P + tA | t ∈ R} is equal to 1.

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L. A. G. 2015. Second intermediate test, april 1, 2015

1. Let A = (3, −4) and Q = (1, 1). Find a point p ∈ V

2

such that the distance between Q and the line {P + tA | t ∈ R} is equal to 1.

Solution. First of all: in there is such a point ¯ P , then there are infinitely many (ant point P belonging to the line { ¯ P + tA | t ∈ R} will do the job).

A cartesian equation of a line parallel to the vector A = (3, −4) is 4x + 3y = c. We find c such that the distance between the line and Q is equal to 1: the distance is

|4+3−c|5

hence we want c ∈ R such that

|4+3−c|5

= 1 that is c = 2 and c = 12 For example, for c = 2 we get the line of cartesian equation 4x + 3y = 2. For example P = (1/2, 0) is a point of that line.

2. Let A = (1, 2, −3) and B = (3. − 2, 4). (a) is there a vector C ∈ V

3

, non-parallel to A or B, such that k C k= 1 and such that det

 A B C

 = 0?. (b) If the answer is yes, show an example of such a vector C.

Solution. (a) We have that

(1) det

 A B C

 = 0

if and only if {A, B, C} is a dependent set. Therefore, any linear combination of A and B, will do the job. hence the answer to the first question is yes.

(b) for example C = A + B = (4, 0, 1), is such that (1) holds. This C is not a unit vector, but dividing by its norm we get the unit vector

1

k C k C = 1

k C k (A + B) = 1

k C k A + 1 k C k B,

which is still a linear combination of A and B. This vector will do the job (as well as any other unit vector of L(A, B)). Therefore a vector as required is:

√ 1

17 (4, 0, 1)

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