# Thread: Finding scalars from vectors

1. ## Finding scalars from vectors

(Note: the bold letters are vectors.)

"Find the scalars, α and β and such that
(i) 3i + j is parallel to αi - 4j
(ii) 3i + β(j - k) is parallel to i - 4j+ 4k
(iii) 3i + γ(j + 3k) has the same length as 12i - 5k"

I first assumed that since the vectors are parallel, they would be equal to each other. So, as for (i), and assuming that v = 3i + j, I found |v| as assumed that would be the length and thought that finding the length of the other vector through solving it that way would be the right approach. However, I realised that it isn't.

I did have a peep at the answers although, that's all that was there so that was hardly any help. (If it is needed - (i)-12 (ii)-12 (iii) ±4). I tried looking over my course notes and even the appropriate section from my textbook but, I can't find anything that is similar to this question.

2. Originally Posted by cottontails
(Note: the bold letters are vectors.)

"Find the scalars, α and β and such that
(i) 3i + j is parallel to αi - 4j
$\displaystyle 3i + j = k(\alpha i - 4 j)$ where k is a scalar.

Therefore:

$\displaystyle 3 = k \alpha$ .... (1)

$\displaystyle 1 = -4k$ .... (2)

Solve equations (1) and (2) simultaneously to get $\displaystyle \alpha$.

3. I now no longer have any difficulties with (i) and (ii) however, with (iii), I cannot get the correct answer.
My working (using t instead of k, since there is a k vector in the equation):
-> t(3i + γj + 3γk) = 12i - 5k
--> 3t=12, γt=0, 3γt=-5
Whenever I solve the equations simultaneously so, t=4 and substituting that into 3γt=-5, I get γ=-5/12.
Although, am I doing something wrong there since shouldn't that also equal the same (-5/12) with substituting t=4 in γt=0?

4. Originally Posted by cottontails
[snip]

(iii) 3i + γ(j + 3k) has the same length as 12i - 5k

[snip]
Originally Posted by cottontails
I now no longer have any difficulties with (i) and (ii) however, with (iii), I cannot get the correct answer.

My working (using t instead of k, since there is a k vector in the equation):

-> t(3i + γj + 3γk) = 12i - 5k

--> 3t=12, γt=0, 3γt=-5

Whenever I solve the equations simultaneously so, t=4 and substituting that into 3γt=-5, I get γ=-5/12.

Although, am I doing something wrong there since shouldn't that also equal the same (-5/12) with substituting t=4 in γt=0?
Get the length, that is, magnitude, of 3i + γ(j + 3k) and equate it to the length, that is, magnitude, of 12i - 5k. Then solve for γ. I don't see what the trouble can be here, apart from not reading the question properly.