# Thread: Find the values of k so that lines are perpendicular using symetric equations

1. ## Find the values of k so that lines are perpendicular using symetric equations

Find the value of k so that the lines and are perpendicular
I got the the vector equation for both lines
L1=(x,y,z) =(4,-4,-3) + k(3,0,2)
L2=(x,y,z) (-4,-8,-12) + k(0,-2,0)

I got the dot product of (3,0,2) and (0,-2,0) and it equals zero, so it is perpendicular. I just don't know what to after

2. ## Vector equations of lines

Hello JohnBlaze
Originally Posted by JohnBlaze
Find the value of k so that the lines and are perpendicular
I got the the vector equation for both lines
L1=(x,y,z) =(4,-4,-3) + k(3,0,2)
L2=(x,y,z) (-4,-8,-12) + k(0,-2,0)

I got the dot product of (3,0,2) and (0,-2,0) and it equals zero, so it is perpendicular. I just don't know what to after
Sorry, but you must have misunderstood the question. The one you have asked doesn't make sense. The equation

$\vec{r} = 4\vec{i} - 4\vec{j} -3\vec{k} + \lambda(3\vec{i} +0\vec{j} + 2\vec{k})$

represents a straight line passing through the point with position vector $4\vec{i} - 4\vec{j} -3\vec{k}$, parallel to the vector $3\vec{i} +0\vec{j} + 2\vec{k}$. By varying the values of $\lambda$, you can create the position vector of any point on the line. E.g.

• If $\lambda = 0$, you get the original point $\vec{r}=4\vec{i} - 4\vec{j} -3\vec{k}$ itself

• If $\lambda = 1$, you get the point $\vec{r}=7\vec{i} - 4\vec{j} -\vec{k}$

• If $\lambda = 2$, you get the point $\vec{r}=10\vec{i} - 4\vec{j} +\vec{k}$

... and so on.

Similarly, $\vec{r}=-4\vec{i} - 8\vec{j} -12\vec{k} + \lambda(0\vec{i} -2\vec{j} + 0\vec{k})$ represents, for different values of $\lambda$, all the points on a line passing through $-4\vec{i} - 8\vec{j} -12\vec{k}$ parallel to $0\vec{i} - 2\vec{j} +0\vec{k}$.

Can you see why what you've put here is meaningless? Would you like to look at the question again, and if you can't see how to do it, to post the original question here?