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Chapter 12: Problem 61

Find the point (if it exists) at which the following planes and lines intersect. $$z=4 ; \mathbf{r}(t)=\langle 2 t+1,-t+4, t-6\rangle$$

Short Answer

Expert verified
Answer: The intersection point is (21, -6, 4).

Step by step solution

01

Write out the components of the line equation

The line equation is given in vector form as \(\mathbf{r}(t) = \langle 2t+1, -t+4, t-6 \rangle\). We can write out the components of this equation as follows: $$x(t) = 2t + 1,$$ $$y(t) = -t + 4,$$ $$z(t) = t - 6.$$
02

Substitute the line equation components into the plane equation

We know that the plane equation is given by \(z=4\). The line will intersect the plane when the \(z\)-component of the line equation is equal to 4. So, we can substitute \(z(t)\) from the line equation into the plane equation to get: $$t-6 = 4$$
03

Solve for the parameter \(t\)

Now, we need to determine the value of \(t\) that satisfies the equation \(t-6=4\). Solving for \(t\), we get: $$t = 10$$
04

Find the intersection point using the value of \(t\)

Now that we have found the value of \(t\), which is 10, we can substitute it back into the line equation components to get the intersection point: $$x(10) = 2(10) + 1 = 21$$ $$y(10) = -(10) + 4 = -6$$ $$z(10) = (10) - 6 = 4$$ Thus, the intersection point is \((21, -6, 4)\).

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