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1. |
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a
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3a =
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b
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a + b =
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c
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2a + 3b =
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d
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b − a =
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e
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c − a − b =
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f
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a + (b − c) =
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g
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2(a + b) =
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h
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-6b =
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2.
(i) a = 3i + 2j + k
b = 5i − 2j + 4k
c = -2i + 3j
(ii) Use unit vectors to calculate
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a
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2a = 6i + 4j + 2k
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b
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b + c = 3i + j + 4k
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c
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4c + 2b = 2i + 8j + 8k
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d
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a − b = -2i + 4j − 3k
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e
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b − a + c = -j + 3k
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f
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b − (a + c) = 4i − 7j + 3k
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g
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3(b + c) = 9i + 3j + 12k
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h
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-3c = 6i − 9j
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3. 3r + 2s - t = 13i + j + 8k = 
4. AB = 
= 2i + j − 3k
5. PQ = 
and QR = 
thus PQ = 3QR therefore P, Q and R lie on the same line.
6. PQ = i + j − 6k
SR = i + j − 6k
As PQ and SR are the same displacement vector they must be parallel.
So PQ || SR and as they have the same lengths PQRS must be a parallelogram.
