1. Finding a vector

Along the line 3x+4y=-2 in the direction of decreasing y what is the vector of this line ? I don't understand how they say the vector is (4,-3)

2. Originally Posted by nyasha
Along the line 3x+4y=-2 in the direction of decreasing y what is the vector of this line ?
$3x + 4y = -2$

$4y = -3x - 2$

$y = -\frac{3}{4}x - \frac{1}{2}$.

So for every $4$ units you have gone to the right, you have gone $3$ units down.

This would suggest that this line would have direction the same as $4\mathbf{i} - 3\mathbf{j}$.

But we don't know the length of this vector. So multiply the vector by some parameter $t$.

Therefore the vector that goes in the direction of this line that also has the same length as this line can be written as

$t(4\mathbf{i} - 3\mathbf{j})$.

Now we just need a point that lies on the line so we know where to place this vector. The $y$ intercept will suffice.

Therefore, the vector form of this line is

$\left(0\mathbf{i} - \frac{1}{2}\mathbf{j}\right) + t(4\mathbf{i} - 3\mathbf{j})$

or if you like

$4t\mathbf{i} - \left(\frac{1}{2} + 3t\right)\mathbf{j}$.

3. Originally Posted by Prove It
$3x + 4y = -2$

$4y = -3x - 2$

$y = -\frac{3}{4}x - \frac{1}{2}$.

So for every $4$ units you have gone to the right, you have gone $3$ units down.

This would suggest that this line would have direction the same as $4\mathbf{i} - 3\mathbf{j}$.

But we don't know the length of this vector. So multiply the vector by some parameter $t$.

Therefore the vector that goes in the direction of this line that also has the same length as this line can be written as

$t(4\mathbf{i} - 3\mathbf{j})$.

Now we just need a point that lies on the line so we know where to place this vector. The $y$ intercept will suffice.

Therefore, the vector form of this line is

$\left(0\mathbf{i} - \frac{1}{2}\mathbf{j}\right) + t(4\mathbf{i} - 3\mathbf{j})$

or if you like

$4t\mathbf{i} - \left(\frac{1}{2} + 3t\right)\mathbf{j}$.

How did u get 4i-3j ?

4. Originally Posted by nyasha
How did u get 4i-3j ?
4 units to the right implies $4\mathbf{i}$.

3 units down implies $-3\mathbf{j}$.

5. Here's another way to do this:

The equation of the line is 3x+4y= 2. If x= 0, 4y= 2 so y= 1/2. One point on the line is (0, 1/2). If y= 0, 3x= 2 so x= 2/3. Another point on that line is (2/3, 0).

Since 0< 1/2, going in the direction of decreasing y means going from (0, 1/2) to (2/3, 0). A vector from (0,1/2) to (2/3, 0) is $(2/3- 0)\vec{i}+ (0- 1/2)\vec{j}= (2/3)\vec{i}- (1/2)\vec{j}$.

Since multiplying a vector by any number gives another vector in that direction, we can get rid of the fractions by multiplying the vector by 6: $6((2/3)\vec{i}- (1/2)\vec{j}= 4\vec{i}- 3\vec{j}$.