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Consider the line through points \(A = (1, 1, 2)\)and \(B = (2, 3, 4)\).

This line can be parametrized as \((at +b, ct +d, et +f)\)for some choices of constants \(a,b, c, d, e\) and \(f\). What is the ordered pair \(\left(\dfrac{c}{a}, \dfrac{e}{a}\right)\)?

 May 4, 2020
 #1
avatar+483 
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in general, a line given two vectors can be parametrized as :

 

a + t*d 

where:

a = starting vector

t = any real number

d = the direction vector

In this case, we have (1,1,2) and (2,3,4)

the direction vector can be found through subtracting the two. We get:

 

(2,3,4) - (1,1,2) = (1,2,2)

with our starting point being (1,1,2)

We then get the line equation as:

(1,1,2) + t(1,2,2)

this then turns into:

(1,1,2) + (t,2t,2t)

Adding the two, we get:
(t+1,2t+1,2t+2)

a = 1

b = 1

c = 2

d = 1

e = 2

f = 2

c/a = 2/1 = 2

e/a = 2/1 = 2

the ordered pair (c/a, e/a) is then (2,2). Please correct me if I'm wrong, precalc has never been one of my strong suits. 

 May 5, 2020
 #2
avatar+39 
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Yes that makes sense! Thank you!!

yeliah  May 5, 2020

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