Question

(e) contains the lines (vec{r}_1(t) = langle 1, 0, -1
angle + t langle 1, 1, 1
angle) and (vec{r}_2(t) = langle 4, 3, 2
angle + t langle 1, 0, 0
angle).

Explanation:

To find the equation of the plane containing the two given lines, we follow these steps:

Step 1: Find two direction vectors and a vector between points on the lines

• The direction vector of \(\vec{r}_1(t)\) is \(\vec{v}_1=\langle1,1,1

angle\).

• The direction vector of \(\vec{r}_2(t)\) is \(\vec{v}_2=\langle1,0,0

angle\).

• We take a point on \(\vec{r}_1(t)\) when \(t = 0\): \(P=(1,0, - 1)\), and a point on \(\vec{r}_2(t)\) when \(t=0\): \(Q=(4,3,2)\). Then the vector \(\vec{PQ}=\langle4 - 1,3-0,2-(-1)

angle=\langle3,3,3
angle\)

Step 2: Find the normal vector of the plane

The normal vector \(\vec{n}\) of the plane is the cross - product of two non - parallel vectors in the plane. We can use \(\vec{v}_1\) and \(\vec{v}_2\) (or \(\vec{v}_1\) and \(\vec{PQ}\), since \(\vec{PQ} = 3\vec{v}_1\), so we use \(\vec{v}_1\) and \(\vec{v}_2\))

The cross - product formula for two vectors \(\vec{a}=\langle a_1,a_2,a_3
angle\) and \(\vec{b}=\langle b_1,b_2,b_3
angle\) is \(\vec{a}\times\vec{b}=

$$\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\a_1&a_2&a_3\\b_1&b_2&b_3\end{vmatrix}$$

=\vec{i}(a_2b_3 - a_3b_2)-\vec{j}(a_1b_3 - a_3b_1)+\vec{k}(a_1b_2 - a_2b_1)\)

For \(\vec{v}_1=\langle1,1,1
angle\) and \(\vec{v}_2=\langle1,0,0
angle\)

\[

$$\begin{align*} \vec{n}&=\vec{v}_1\times\vec{v}_2\\ &=\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\1&1&1\\1&0&0\end{vmatrix}\\ &=\vec{i}(1\times0 - 1\times0)-\vec{j}(1\times0 - 1\times1)+\vec{k}(1\times0 - 1\times1)\\ &=\vec{i}(0)-\vec{j}(- 1)+\vec{k}(-1)\\ &=\langle0,1, - 1 angle \end{align*}$$

\]

Step 3: Use the point - normal form of the plane equation

The point - normal form of a plane equation is \(A(x - x_0)+B(y - y_0)+C(z - z_0)=0\), where \((x_0,y_0,z_0)\) is a point on the plane and \(\langle A,B,C
angle\) is the normal vector of the plane.

We can use the point \(P=(1,0,-1)\) and the normal vector \(\vec{n}=\langle0,1,-1
angle\)

Substitute into the formula: \(0(x - 1)+1(y - 0)-1(z + 1)=0\)

Simplify the equation: \(y-z - 1=0\) or \(y - z=1\)

Answer:

The equation of the plane is \(y - z=1\) (or \(y-z - 1 = 0\))