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Chapter 11: Problem 65

Find the points (if they exist) at which the following planes and curves intersect. $$z=16 ; \mathbf{r}(t)=\langle t, 2 t, 4+3 t\rangle, \text { for }-\infty

Short Answer

Expert verified
Answer: (4, 8, 16)

Step by step solution

01

Write down the given plane equation and parametric curve equation.

The given plane has the equation: $$z = 16$$ The given parametric curve is defined by the vector function: $$\mathbf{r}(t) = \langle t, 2t, 4 + 3t \rangle, \text{ for } -\infty < t < \infty$$
02

Determine the coordinates of the curve in terms of t.

From the parametric equation of the curve, the coordinates of the curve are given by: $$x = t$$ $$y = 2t$$ $$z = 4 + 3t$$
03

Substitute the coordinates of the curve into the plane equation.

Now, we want to find the points where the curve intersects the plane. To do this, we need to substitute the coordinates of the curve into the plane equation. We get: $$4+3t = 16$$
04

Solve for t.

To determine if the points exist, we need to solve the equation obtained in Step 3 for t: $$3t = 12$$ $$t = 4$$
05

Find the coordinates of the intersection point.

Since we have found a valid value for t, we can now find the coordinates of the intersection point by substituting t back into the parametric equations of the curve: $$x = t = 4$$ $$y = 2t = 2(4) = 8$$ $$z = 4 + 3t = 4 + 3(4) = 16$$ So, the intersection point is given by: $$(4,8,16)$$ Thus, the plane and the curve intersect at the point \((4, 8, 16)\).

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Most popular questions from this chapter

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