# Thread: [SOLVED] vectors, i,j notation- dont understand!

1. ## [SOLVED] vectors, i,j notation- dont understand!

In this question, i and j are unit vectors in the direction of the positive x- and y-axes respectively.

Given that a = i – 2j and b = –3i + j find

(a) λ, if a + λb is parallel to – i – 3j

(b) μ, if μa + b is parallel to 2i + j.

in the solution bank it shows the answer worked out like this but I just dont get the working.

$xi + yj$ is parallel to $Xi + Yj$ if $\frac{x}{y} = \frac{X}{Y}$

$a + \lambda b = (i - 2j) + \lambda ( -3i + j )$

$= (1 - 3\lambda)i + ( -2 + \lambda)j$ I dont understand how they got this expression ?

This is parallel to – i – 3j if

$\frac{ -2 + \lambda}{1-3\lambda} = \frac{-3}{-1}$

2. Originally Posted by Tweety

In this question, i and j are unit vectors in the direction of the positive x- and y-axes respectively.

Given that a = i – 2j and b = –3i + j find

(a) λ, if a + λb is parallel to – i – 3j

(b) μ, if μa + b is parallel to 2i + j.

in the solution bank it shows the answer worked out like this but I just dont get the working.

$xi + yj$ is parallel to $Xi + Yj$ if $\frac{x}{y} = \frac{X}{Y}$

$a + \lambda b = (i - 2j) + \lambda ( -3i + j )$

$= (1 - 3\lambda)i + ( -2 + \lambda)j$ I dont understand how they got this expression ?

$a + \lambda b = (i - 2j) + \lambda ( -3i + j )$
$i - 2j + -3i\lambda + j\lambda
$

$i-3i\lambda -2j +j\lambda
$

Sum i and j vector separately
$= (1 - 3\lambda)i + ( -2 + \lambda)j$

3. Originally Posted by u2_wa
$a + \lambda b = (i - 2j) + \lambda ( -3i + j )$
$i - 2j + -3i\lambda + j\lambda
$

$i-3i\lambda -2j +j\lambda
$

Sum i and j vector separately
$= (1 - 3\lambda)i + ( -2 + \lambda)j$
thanks a lot!

4. Originally Posted by Tweety
$a + \lambda b = (i - 2j) + \lambda ( -3i + j )$
$= (1 - 3\lambda)i + ( -2 + \lambda)j$ I dont understand how they got this expression ? This is parallel to – i – 3j if
$\frac{ -2 + \lambda}{1-3\lambda} = \frac{-3}{-1}$
$({\color{red}i} - {\color{blue} 2j}) + \lambda ({\color{red}-3i} + {\color{blue}j} )$
Just collect like terms.