To find the intersection of the line $r(t) = \langle -1, 2, 3 \rangle + t \langle -3, 1, -2 \rangle$ and the plane $x + 3y - 2z = 8$, we substitute the parametric equation of the line into the equation of the plane and solve for $t$.

Step 1: Parametrize the line

The parametric equations for the line are: $x = -1 - 3t, \quad y = 2 + t, \quad z = 3 - 2t.$

Step 2: Substitute into the plane equation

The plane equation is: $x + 3y - 2z = 8.$

Substitute $x = -1 - 3t$, $y = 2 + t$, and $z = 3 - 2t$ into the equation: $(-1 - 3t) + 3(2 + t) - 2(3 - 2t) = 8.$

Step 3: Simplify the equation

$-1 - 3t + 6 + 3t - 6 + 4t = 8.$

Combine like terms: $-1 + 6 - 6 + 4t = 8.$

$-1 + 4t = 8.$

$4t = 9.$

$t = \frac{9}{4}.$

Step 4: Find the point of intersection

Substitute $t = \frac{9}{4}$ back into the parametric equations of the line to find the coordinates of the intersection point.

$x = -1 - 3\left(\frac{9}{4}\right) = -1 - \frac{27}{4} = \frac{-4}{4} - \frac{27}{4} = \frac{-31}{4}.$

$y = 2 + \frac{9}{4} = \frac{8}{4} + \frac{9}{4} = \frac{17}{4}.$

$z = 3 - 2\left(\frac{9}{4}\right) = 3 - \frac{18}{4} = \frac{12}{4} - \frac{18}{4} = \frac{-6}{4} = \frac{-3}{2}.$

Step 5: Write the intersection point

The intersection point is: $\left( \frac{-31}{4}, \frac{17}{4}, \frac{-3}{2} \right).$

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Related Questions:

1. How do you verify the intersection point lies on the plane and the line?
2. What is the geometric significance of this intersection?
3. Can you find the angle between the line and the plane?
4. What happens if the line is parallel to the plane?
5. How do you find the distance from a point to a plane?

Tip:

Always verify your calculations by substituting the intersection point back into both the plane and the line equations to confirm the solution is accurate.