Find parametric equations of the line in R* passing through (10,0,~4) and (7,3,-1). 5 . Find parametric equations of the plane in R? containing the points (0,5,0), ...

Find parametric equations for the line. The line crosses the $z$ -axis where $z=4$ and crosses the $x y$ -plane where $x=2$ and $y=5$

Really given problem, we want to find the set of parametric equations for the line. So we know that the point we're dealing with is -452. And were given um that is going to be perpendicular to the plane, but we already know that the normal vector to the plane where the standard normal vector is going to be -1: one. Yeah, that's one option for the vector, but we know it's perpendicular. So now that we have our point and we have our vector, we can combine these to create our parametric equations for the given line. So we'll get negative four x equals negative four minus T. We'll get that Y is equal to five last two T. And we'll get that um Z is equal to two plus T. So that is going to be our final result.

So now what we want to do is you want to find the line that passes to the .12 or six. And the spot particular to the line X. That's too white. Plus G equals 2 5. So it's a bit different from what we have done before but it's the same thing. So this is again my this guy and my vector B is just the coefficients of X. Y. And G. So this is one three and one. So the victory question is A. Plus tv. This is 106 nasty Times 1, 3 and one. There's a little equation if you write it in terms of high. So this is I. Plus judo Jessica less six K. Cup. Last two times I cap. That's T. Jacob last cake up. And then for the parametric equation. Yeah With a parametric equation. This is X zero over one. Is why my sorry X -1 Over one equal to 1 -0/3 is equal to G -6/1. And then you get it to A. T. I should do that. Then we just solve it. I was just write it and blow here on the side. So excess D. Plus like one. Why is three D. And G. S. T. Plus six. This is a victory question. That's it.

For this exercise were given the point p 37 negative fire and to 73 negative five and want to find the equations for a line that passes through both points. We can start by calculating the vector P two because the line is parallel to this factor. So this is given by seven minus three three minus seven negative five. Mine is notified, which gives us for negative for zero, and this is a vector that is parallel to the line and Pisa point for which the light passes. Therefore, we confined the Parametric equation based on to these two things, and we have X is equal to three plus for tea. Why is equal to seven minus 40 and Z is equal to negative five plus zero t or just negative five. And these are the higher metric equation that we were looking for.

The problem, we want to find the set of parametric equations of the line Knowing that our line passes through the 0.-608. And um it can be parallel to the given line. Well, the line itself is important to us because we can extrapolate the vector from it. So we see 5 -2 t. That's going to give us an X. Component of the vector of negative two. The y component of the factor is going to be positive too. It's the coefficient of T and the coefficient of tea for Z is just gonna be zero. So we have this and we end up getting for our line um X for the parametric equations of the line, X equals negative six minus two. T. Y equals zero plus two. T N. z equals eight um plus zero key. So this is our final parametric equations.