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: Extract parametric equations

From the line \(\vec{r}(t)=\langle -1,0,1
angle + t\langle 1,1,1
angle\), we get the parametric equations:
\(x = - 1 + t\), \(y = 0 + t=t\), \(z = 1 + t\)

Step2: Substitute into plane equation

Substitute \(x=-1 + t\), \(y = t\), \(z = 1 + t\) into the plane equation \(3x + 2y+z = 10\):
\(3(-1 + t)+2t+(1 + t)=10\)

Step3: Solve for t

Expand and simplify the left - hand side:
\(-3 + 3t+2t + 1+t=10\)
Combine like terms: \((3t + 2t+t)+(-3 + 1)=10\)
\(6t-2 = 10\)
Add 2 to both sides: \(6t=10 + 2=12\)
Divide both sides by 6: \(t=\frac{12}{6}=2\)

Step4: Find intersection point

Substitute \(t = 2\) back into the parametric equations:
For \(x\): \(x=-1+2 = 1\)
For \(y\): \(y = 2\)
For \(z\): \(z=1 + 2=3\)

Answer:

The intersection point is \((1,2,3)\)