Equations of Planes: Solve xz-intersection Line

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SUMMARY

The discussion focuses on finding the intersection line of the parametric equation \(\mathbf{r} = (0,0,5) + s(4,1,0) + t(2,0,2)\) with the xz-coordinate plane. By setting the y-coordinate to 0, the equations simplify to x = 4s + 2t and z = 5 + 2t. This results in the line equation y = 2x + 5, which describes the intersection in the xz-plane. The solution effectively demonstrates the relationship between parametric and Cartesian coordinates in three-dimensional space.

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Cyto
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I need some help on this question...

[tex]\rightharpoonup{r}[/tex] = (0,0,5) + s(4,1,0) + t(2,0,2) which intersects the xz-coordinate plane in what line?
 
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First write the equation for XZ plane let it be u and the given plane be v

Then the point of intersection will be

u+kv = 0 where k is any constant which is to be found.

The above equations are in cartesian coordinate
 


The equation given is in parametric form, which means it represents a line in three-dimensional space. To find the intersection of this line with the xz-coordinate plane, we can set the y-coordinate to 0 and solve for the values of x and z.

Setting y = 0, we get:

x = 0 + 4s + 2t
z = 5 + 0s + 2t

Simplifying these equations, we get:

x = 4s + 2t
z = 5 + 2t

This represents a line in the xz-plane with a slope of 4/2 = 2 and a y-intercept of 5. Therefore, the line of intersection is y = 2x + 5.

I hope this helps to clarify the concept. If you need further assistance, please let me know.
 

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