Question

The line $x=1+2t, y=2-3t, z=3-t$ intersects the plane $2x + y - z =7$. What is the intersection point?

Answer 1

(7, -7, 0)

Answer 2

The line is given by the parametric equations:

$x = 1+2t$ $y = 2-3t$ $z = 3-t$

The equation of the plane is: $2x + y - z = 7$

To find the intersection point, we substitute the expressions for $x$, $y$, and $z$ from the line's equations into the plane's equation: $2(1+2t) + (2-3t) - (3-t) = 7$

Now, expand and simplify the equation to solve for $t$: $2 + 4t + 2 - 3t - 3 + t = 7$ Combine the constant terms: $(2 + 2 - 3) + (4t - 3t + t) = 7$ $1 + 2t = 7$

Subtract 1 from both sides: $2t = 7 - 1$ $2t = 6$

Divide by 2 to find the value of $t$: $t = \frac{6}{2}$ $t = 3$

Now substitute the value of $t=3$ back into the parametric equations of the line to find the coordinates of the intersection point: $x = 1+2(3) = 1+6 = 7$ $y = 2-3(3) = 2-9 = -7$ $z = 3-3 = 0$

Thus, the intersection point is $(7, -7, 0)$.