# M1 vectorsWatch

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#1
Im unsure on how to go about doing q3. I thought it would be similar to q2 but its parallel to -i - 3j instead of just i or j where i made the other equal to zero, any help would be appreciated
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6 years ago
#2
(Original post by raiden95)
Im unsure on how to go about doing q3. I thought it would be similar to q2 but its parallel to -i - 3j instead of just i or j where i made the other equal to zero, any help would be appreciated
If something is parallel to -i-3j, then it can be written in the form r(-i-3j) for some, as yet, unknown number "r".

So you need to equate with r(-i-3j), and by equating coefficients get two simultaneous equations in lambda and r, that you can solve.
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#3
(Original post by ghostwalker)
If something is parallel to -i-3j, then it can be written in the form r(-i-3j) for some, as yet, unknown number "r".

So you need to equate with r(-i-3j), and by equating coefficients get two simultaneous equations in lambda and r, that you can solve.
Thanks I got the answer, is it more common to right r(-i-3j) than n(-i-3j) also what does the value of r represent?
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6 years ago
#4
(Original post by raiden95)
Thanks I got the answer, is it more common to right r(-i-3j) than n(-i-3j) also what does the value of r represent?
r is a real number and it represents the scalar multiple of -i-3j that's required to satisfy the given property.

It's usual to use greek letters, but I was being lazy. I initially used an "a", but then realised it clashed with the use of "a" as a vector, so I opted for "r".
Any letter will do, as long as it doesn't clash, and you state clearly what it is - see my previous post.
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