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Maths is this correct?

(edited 6 years ago)

The vector 9i+qj is parallel to the vector 2i-j. Find the value of the constant q.
(9+2)i+(q+1)j
Note: (11)i + (q+1)j
9=r•2
r=2/9
q=r•2
r=2/q
rq=2
q=2/r

It might have been quicker to use the fact that if a vector 'a' is parallel to vector b then vector a = k vector b wher k is a constant.

9i + qj = k (2i - j) can you see how to complete this?

Reply 2

Shaanv

I would do 9/q=2/(-1)

Then solve

Original post by Musicanor

The vector 9i qj is parallel to the vector 2i-j. Find the value of the constant q.
(9 2)i (q 1)j
Note: (11)i (q 1)j
9=r•2
r=2/9
q=r•2
r=2/q
rq=2
q=2/r

(edited 6 years ago)

Reply 3

Musicanor

Original post by Shaanv

I would do 9/q=2/(-1)

Then solve

9=2/-1 x q
9= 2q/-1
9 x-1 =2q
-9= 2q
-9/2=q

Reply 4

Musicanor

Original post by Muttley79

It might have been quicker to use the fact that if a vector 'a' is parallel to vector b then vector a = k vector b wher k is a constant.

9i + qj = k (2i - j) can you see how to complete this?

9i+qj=k(2i-j)
9i+qj= 2ki-kj
i(9-2k)= j(q-k)

Reply 5

Muttley79

Original post by Musicanor

9i+qj=k(2i-j)
9i+qj= 2ki-kj
i(9-2k)= j(q-k)

I'd compare the i components to get the value of k then you can work out q.

Reply 6

Musicanor

Original post by Muttley79

I'd compare the i components to get the value of k thegn you can work out q.

7i+qj=-kj
7i=-kj-qj
7i=j(-k-q)
Am I on the right track?

Reply 7

thotproduct

It's probably quicker to look at the fact that these vectors are parallel if the vector is a scalar multiple of the other, from that information alone, it is quicker to work out the scalar multiple and thus your j component.