p = 2i + 3j + 4k

q = i + 3j + 2k

r = 2i + j +ak

If p.(q+r) is 6 greater than q.(p+r) find the value ofa.

This is what I've done so far.

p.(q+r) = 4a + 26

q.(p+r) = 2a + 24

Can you tell me the next step of this? Thanks

Answer: 2

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- April 29th 2007, 09:25 AMr_mathsVectors
p = 2i + 3j + 4k

q = i + 3j + 2k

r = 2i + j +*a*k

If p.(q+r) is 6 greater than q.(p+r) find the value of*a*.

This is what I've done so far.

p.(q+r) = 4a + 26

q.(p+r) = 2a + 24

Can you tell me the next step of this? Thanks

Answer: 2 - April 29th 2007, 12:08 PMJhevon
i don't really like to work in the i-j-k notation, so let's write

p = <2,3,4>

q = <1,3,2>

r = <2,1,a>

i suppose your dot means the dot product

Now, p.(q + r) = <2,3,4>.(<1,3,2> + <2,1,a>)

....................= <2,3,4>.<3,4,2+a>

....................= 6 + 12 + 8 + 4a

....................= 26 + 4a

Now, q.(p + r) = <1,3,2>.(<2,3,4> + <2,1,a>)

....................= <1,3,2>.<4,4,4+a>

....................= 4 + 12 + 8 + 2a

....................= 24 + 2a

Now we are told that p.(q+r) is 6 greater than q.(p+r), that is,

q.(p + r) + 6 = p.(q + r)

=> 24 + 2a + 6 = 26 + 4a

=> 2a = 4

=> a = 2 - April 30th 2007, 08:03 AMr_maths
Thanks, i added 6 on to p.(q + r) instead of q.(p + r)

- April 30th 2007, 09:27 AMJhevon