# finding angles between vectorsWatch

#1
Hi guys
I'm struggling to understand part c of the question on the attachment. i know that the scalar multiple of two perpendicular angle is zero bur i don't know how to apple it on part c. any help would be much appreciated
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2 years ago
#2
a.b = magnitudes of two vectors * cos angle.

The magnitudes aren't zero but cos can be zero and cos is zero at 90 degrees.
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2 years ago
#3
(Original post by Alen.m)
Hi guys
I'm struggling to understand part c of the question on the attachment. i know that the scalar multiple of two perpendicular angle is zero bur i don't know how to apple it on part c. any help would be much appreciated
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2 years ago
#4
(Original post by notnek)
Just as an FYI - we tend to use (\mathbf{p}) for bold vectors in - that should save you some typing time.
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2 years ago
#5
this appears to be GCSE.

Modulus may confuse.
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2 years ago
#6
(Original post by Zacken)
Just as an FYI - we tend to use (\mathbf{p}) for bold vectors in - that should save you some typing time.
Yes I've always used that in mathematical writing but I couldn't remember it so I chose the lazy option!
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2 years ago
#7
(Original post by Mutleybm1996)
this appears to be GCSE.

Modulus may confuse.
Definitely not GCSE - look at part (b).
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2 years ago
#8
(Original post by notnek)
Definitely not GCSE - look at part (b).
oops.

It's quite a nice question actually, I remember those days fondly ^_^
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2 years ago
#9
this is not GCSE

this is a standard geometric proof involving vectors

see Q154
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#10
yeah the question is in c4 , thanks everyone managed to find a great explanation for it
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