Question

1. find the intersection of the line\\(vec{r}(t) = langle -1, 0, 1

angle + t langle 1, 1, 1
angle\\) and the plane:\\(3x + 2y + z = 10\\).

Explanation:

Step1: Parametrize line coordinates

Express $x, y, z$ from $\vec{r}(t)$:
$x = -1 + t$, $y = 0 + t$, $z = 1 + t$

Step2: Substitute into plane equation

Plug $x,y,z$ into $3x+2y+z=10$:
$$3(-1 + t) + 2(t) + (1 + t) = 10$$

Step3: Simplify and solve for $t$

Expand and combine like terms:
$-3 + 3t + 2t + 1 + t = 10$
$6t - 2 = 10$
$6t = 12$
$t = 2$

Step4: Find intersection coordinates

Substitute $t=2$ back to line equations:
$x = -1 + 2 = 1$, $y = 0 + 2 = 2$, $z = 1 + 2 = 3$

Answer:

The intersection point is $\langle 1, 2, 3
angle$