Question

in exercises 11 and 12, give the vector equation of the line passing through p and q.

1. p=(1, - 2), q=(3,0)
2. p=(0,1, - 1), q=(-2,1,3)

in exercises 13 and 14, give the vector equation of the plane passing through p, q, and r.

1. p=(1,1,1), q=(4,0,2), r=(0,1, - 1)
2. p=(1,1,0), q=(1,0,1), r=(0,1,1)

1. find parametric equations and an equation in vector form for the lines in r² with the following equations: (a) y = 3x - 1 (b) 3x + 2y = 5

Explanation:

Step1: Recall vector - equation of a line formula

The vector equation of a line passing through two points $P(x_1,y_1)$ and $Q(x_2,y_2)$ in $\mathbb{R}^2$ is $\vec{r}=\vec{r}_0 + t\vec{v}$, where $\vec{r}_0$ is the position - vector of a point on the line (we can take $\vec{r}_0=\overrightarrow{OP}$) and $\vec{v}=\overrightarrow{PQ}$. For points $P(x_1,y_1,z_1)$ and $Q(x_2,y_2,z_2)$ in $\mathbb{R}^3$, the formula is the same. First, find $\overrightarrow{PQ}=\langle x_2 - x_1,y_2 - y_1,z_2 - z_1
angle$ (in $\mathbb{R}^3$) or $\langle x_2 - x_1,y_2 - y_1
angle$ (in $\mathbb{R}^2$).

Step2: Solve Exercise 11

Given $P=(1, - 2)$ and $Q=(3,0)$ in $\mathbb{R}^2$. $\overrightarrow{PQ}=\langle3 - 1,0-( - 2)
angle=\langle2,2
angle$. Let $\vec{r}_0=\langle1,-2
angle$. The vector equation of the line is $\vec{r}=\langle1,-2
angle+t\langle2,2
angle=\langle1 + 2t,-2 + 2t
angle$, where $t\in\mathbb{R}$.

Step3: Solve Exercise 12

Given $P=(0,1,-1)$ and $Q=(-2,1,3)$ in $\mathbb{R}^3$. $\overrightarrow{PQ}=\langle-2-0,1 - 1,3-( - 1)
angle=\langle-2,0,4
angle$. Let $\vec{r}_0=\langle0,1,-1
angle$. The vector equation of the line is $\vec{r}=\langle0,1,-1
angle+t\langle-2,0,4
angle=\langle-2t,1, - 1+4t
angle$, where $t\in\mathbb{R}$.

Step4: Recall vector - equation of a plane formula

The vector equation of a plane passing through three non - collinear points $P$, $Q$, and $R$ is $\vec{r}=\vec{r}_0 + s\vec{u}+t\vec{v}$, where $\vec{r}_0$ is the position vector of a point on the plane (e.g., $\vec{r}_0=\overrightarrow{OP}$), and $\vec{u}=\overrightarrow{PQ}$, $\vec{v}=\overrightarrow{PR}$. First, find $\overrightarrow{PQ}$ and $\overrightarrow{PR}$.

Step5: Solve Exercise 13

Given $P=(1,1,1)$, $Q=(4,0,2)$, $R=(0,1,-1)$. $\overrightarrow{PQ}=\langle4 - 1,0 - 1,2 - 1
angle=\langle3,-1,1
angle$ and $\overrightarrow{PR}=\langle0 - 1,1 - 1,-1 - 1
angle=\langle-1,0,-2
angle$. Let $\vec{r}_0=\langle1,1,1
angle$. The vector equation of the plane is $\vec{r}=\langle1,1,1
angle+s\langle3,-1,1
angle+t\langle-1,0,-2
angle=\langle1 + 3s-t,1 - s,1 + s-2t
angle$, where $s,t\in\mathbb{R}$.

Step6: Solve Exercise 14

Given $P=(1,1,0)$, $Q=(1,0,1)$, $R=(0,1,1)$. $\overrightarrow{PQ}=\langle1 - 1,0 - 1,1 - 0
angle=\langle0,-1,1
angle$ and $\overrightarrow{PR}=\langle0 - 1,1 - 1,1 - 0
angle=\langle-1,0,1
angle$. Let $\vec{r}_0=\langle1,1,0
angle$. The vector equation of the plane is $\vec{r}=\langle1,1,0
angle+s\langle0,-1,1
angle+t\langle-1,0,1
angle=\langle1 - t,1 - s,s + t
angle$, where $s,t\in\mathbb{R}$.

Step7: Recall parametric equations of a line in $\mathbb{R}^2$

For a line $y = mx + c$ in $\mathbb{R}^2$, we can set $x=t$. Then $y=mt + c$. For a general linear equation $Ax+By = C$, we can solve for $y=\frac{C - Ax}{B}$ (if $B
eq0$) and set $x = t$.

Step8: Solve Exercise 15(a)

Given $y = 3x-1$. Let $x=t$. Then the parametric equations are $x=t$, $y = 3t-1$, and the vector equation is $\vec{r}=\langle t,3t - 1
angle=\langle0,-1
angle+t\langle1,3
angle$, where $t\in\mathbb{R}$.

Step9: Solve Exercise 15(b)

Given $3x + 2y=5$, so $y=\frac{5 - 3x}{2}$. Let $x=t$. Then the parametric equations are $x=t$, $y=\frac{5 - 3t}{2}$. The vector equation is $\vec{r}=\langle t,\frac{5 - 3t}{2}
angle=\langle0,\frac{5}{2}
angle+t\langle1,-\frac{3}{2}
angle$, where $t\in\mathbb{R}$.

Answer:

Exercise 11: $\vec{r}=\langle1 + 2t,-2 + 2t
angle,t\in\mathbb{R}$
Exercise 12: $\vec{r}=\langle-2t,1, - 1+4t
angle,t\in\mathbb{R}$
Exercise 13: $\vec{r}=\langle1 + 3s-t,1 - s,1 + s-2t
angle,s,t\in\mathbb{R}$
Exercise 14: $\vec{r}=\langle1 - t,1 - s,s + t
angle,s,t\in\mathbb{R}$
Exercise 15(a): Parametric equations: $x=t,y = 3t-1$; Vector equation: $\vec{r}=\langle0,-1
angle+t\langle1,3
angle,t\in\mathbb{R}$
Exercise 15(b): Parametric equations: $x=t,y=\frac{5 - 3t}{2}$; Vector equation: $\vec{r}=\langle0,\frac{5}{2}
angle+t\langle1,-\frac{3}{2}
angle,t\in\mathbb{R}$