# Thread: vector (simplified b1 and b2 )

1. ## vector (simplified b1 and b2 )

we know that r=a +(/lambda b ). /lambda is a constant. the above is the ans given while the below is my working. my b1 and b2 is not the same with the ans. is the ans given acceptable ? (b1 and b2) here's only the partial working. i'm having problem with tht first part. accorfing to the ans given, the b1 and b2 are simplified . in this case, why the b1 and b2 can be simplified?
https://www.flickr.com/photos/123101...3/14283658984/

https://www.flickr.com/photos/123101...3/14097487299/

https://www.flickr.com/photos/123101...3/14097497469/

sorry i cant upload the photo directly, so i choose to upload it onto flickr

2. ## Re: vector (simplified b1 and b2 )

For line one, let's call the variable you use $t$ and the variable given in the answer $t'$. Then let's see when the two vectors are equal.

$\begin{pmatrix}2 \\ 2 \\ 4\end{pmatrix} + \begin{pmatrix}-\tfrac{1}{2} \\ \tfrac{1}{2} \\ 1\end{pmatrix}t = \begin{pmatrix}2 \\ 2 \\ 4\end{pmatrix} + \begin{pmatrix}-1 \\ 1 \\ 2\end{pmatrix}t'$ when $t=2t'$. Since the map $t \mapsto 2t$ is a bijection, you wind up with the same line either way.

Same thing for line 2.

3. ## Re: vector (simplified b1 and b2 )

Originally Posted by SlipEternal
For line one, let's call the variable you use $t$ and the variable given in the answer $t'$. Then let's see when the two vectors are equal.

$\begin{pmatrix}2 \\ 2 \\ 4\end{pmatrix} + \begin{pmatrix}-\tfrac{1}{2} \\ \tfrac{1}{2} \\ 1\end{pmatrix}t = \begin{pmatrix}2 \\ 2 \\ 4\end{pmatrix} + \begin{pmatrix}-1 \\ 1 \\ 2\end{pmatrix}t'$ when $t=2t'$. Since the map $t \mapsto 2t$ is a bijection, you wind up with the same line either way.

Same thing for line 2.
so if i i do my ans in this way. wouldnt my ans be wrong?

4. ## Re: vector (simplified b1 and b2 )

Originally Posted by delso
so if i i do my ans in this way. wouldnt my ans be wrong?
I just showed that you get the same line either way. Why would you get the wrong answer that way? If two line intersect and you have two different formulas for each line, it is still the same two lines. It doesn't matter that the formulas are not identical. They still represent the same lines, so they will still intersect in the same place.

5. ## Re: vector (simplified b1 and b2 )

do u mean t=2t" just different in terms of length ? but it's still the same line with same gradient?