Question

Find the point where the line $x=2+t, y=3-2t, z=4+t$ intersects the plane $x-y+z=6$.

Answer 1

$\left(\frac{11}{4}, \frac{3}{2}, \frac{19}{4}\right)$

Answer 2

The parametric equations of the line are:

$x = 2+t$ $y = 3-2t$ $z = 4+t$

The equation of the plane is: $x-y+z = 6$

To find the point of intersection, substitute the expressions for $x$, $y$, and $z$ from the line's parametric equations into the plane's equation: $(2+t) - (3-2t) + (4+t) = 6$

Now, simplify and solve for the parameter $t$: $2+t - 3+2t + 4+t = 6$ Combine the constant terms: $(2-3+4) = 3$ Combine the terms with $t$: $(t+2t+t) = 4t$ So the equation becomes: $3 + 4t = 6$

Subtract 3 from both sides: $4t = 6 - 3$ $4t = 3$

Divide by 4 to find the value of $t$: $t = \frac{3}{4}$

Now, substitute this value of $t$ back into the parametric equations of the line to find the coordinates of the intersection point $(x, y, z)$: $x = 2+t = 2 + \frac{3}{4} = \frac{8+3}{4} = \frac{11}{4}$ $y = 3-2t = 3 - 2\left(\frac{3}{4}\right) = 3 - \frac{6}{4} = 3 - \frac{3}{2} = \frac{6-3}{2} = \frac{3}{2}$ $z = 4+t = 4 + \frac{3}{4} = \frac{16+3}{4} = \frac{19}{4}$

Thus, the intersection point is $\left(\frac{11}{4}, \frac{3}{2}, \frac{19}{4}\right)$.