# For these three points: A = (1, 0, 2) ; B = (0, 3, -1) ; C = (0, 0, 4). 1) Find Vector BA 2) Let...

## Question:

For these three points: {eq}\displaystyle A = (1,\ 0,\ 2)\ ;\ B = (0,\ 3, \ -1)\ ;\ C = (0,\ 0,\ 4) {/eq}.

1) Find Vector {eq}BA {/eq}

2) Let {eq}D {/eq} be the point closest to {eq}A {/eq}, for {eq}D {/eq} lying between {eq}B {/eq} and {eq}C {/eq}. (That is, {eq}D {/eq} is on the line segment connecting {eq}B {/eq} and {eq}C {/eq}.) Find the distance from {eq}D {/eq} to {eq}B {/eq}?

## Parametric Equation of a Line

A line segment connecting two points {eq}\textbf{B} {/eq} and {eq}\textbf{C} {/eq} in three dimensional space can be represented parametrically by a function {eq}\textbf{r}:\left[0,1\right]\rightarrow\mathbb{R}^3 {/eq} defined as {eq}\textbf{r}(t)=(1-t)\textbf{B}+t\textbf{C} {/eq}. This kind of parametric function will be used to help answer the question. The answer also makes use of the fact that the shortest distance from a point to a line is along a second line perpendicular to the first line. Two lines are perpendicular if their direction vectors are perpendicular. This means that the dot product of their direction vectors is equal to zero. Moreover, the shortest distance from a point to a line can be determined using calculus. All the concepts are discussed in the answer that follows.

The problem is restated using slightly different notation. Given the points {eq}\textbf{A}=(1,0,2) {/eq}, {eq}\textbf{B}=(0,3,-1) {/eq} and {eq}\textbf{C}=(0,0,4) {/eq} do the following.

1) Find the vector {eq}\overrightarrow{BA}=\textbf{A}-\textbf{B} {/eq}.

2) Let {eq}\textbf{D} {/eq} be the point closest to {eq}\textbf{A} {/eq} where {eq}\textbf{D} {/eq} is a point on the line segment connecting {eq}\textbf{B} {/eq} and {eq}\textbf{C} {/eq}. Find the distance from {eq}\textbf{D} {/eq} to {eq}\textbf{B} {/eq}.

Consider part 1). The vector {eq}\overrightarrow{BA} {/eq} is calculated as

{eq}\begin{eqnarray*} \overrightarrow{BA} &=& \textbf{A}-\textbf{B} \\ &=& (1,0,2)-(0,3,-1) \\ &=& (1,-3,3). \end{eqnarray*} {/eq}

Consider part 2). The parametric function {eq}\textbf{r}:\left[0,1\right]\rightarrow\mathbb{R}^3 {/eq} that represents the line segment connecting {eq}\textbf{B} {/eq} and {eq}\textbf{C} {/eq} is defined as

{eq}\begin{eqnarray*} \textbf{r}(t) &=& (1-t)\textbf{B}+t\textbf{C} \\ &=& (1-t)(0,3,-1)+t(0,0,4) \\ &=& (0,3-3t,-1+t)+(0,0,4t) \\ &=& (0,3-3t,-1+5t). \end{eqnarray*} {/eq}

The point {eq}\textbf{D} {/eq} on the line segment connecting {eq}\textbf{B} {/eq} and {eq}\textbf{C} {/eq} can be found in two ways.

Consider the first way. The point {eq}\textbf{D} {/eq} is on the line containing {eq}\textbf{A} {/eq} and the line segment connecting {eq}\textbf{B} {/eq} and {eq}\textbf{C} {/eq}. These lines are perpendicular to each other. Since the point {eq}\textbf{D} {/eq} is on the line segment connecting {eq}\textbf{B} {/eq} and {eq}\textbf{C} {/eq} then {eq}\textbf{D}=(0,3-3t,-1+5t) {/eq} for some {eq}t {/eq}. Once {eq}t {/eq} is obtained the point {eq}\textbf{D} {/eq} can be found. A direction vector for the line containing {eq}\textbf{A} {/eq} and {eq}\textbf{D} {/eq} is

{eq}\begin{eqnarray*} \textbf{u} &=& \textbf{A}-\textbf{D} \\ &=& (1,0,2)-(0,3-3t,-1+5t) \\ &=& (1,-3+3t,3-5t). \end{eqnarray*} {/eq}

A direction vector for line segment connecting {eq}\textbf{B} {/eq} and {eq}\textbf{C} {/eq} is {eq}\textbf{v}=(0,-3,5) {/eq}. Since the lines are perpendicular, their direction vectors are perpendicular. Consequently, the dot product of their direction vectors is zero. That is

{eq}\begin{eqnarray*} 0 &=& \textbf{u}\cdot\textbf{v} \\ &=& (1,-3+3t,3-5t)\cdot(0,-3,5) \\ &=& 1(0)+(-3+3t)(-3)+(3-5t)(5) \\ &=& 9-9t+15-25t \\ &=& 24-34t \\ \frac{24}{34} &=& t \\ \frac{12}{17} &=& t. \end{eqnarray*} {/eq}

A second way to find {eq}t {/eq}, and hence the point {eq}\textbf{D} {/eq}, is to find the point on line segment connecting {eq}\textbf{B} {/eq} and {eq}\textbf{C} {/eq} closest to point {eq}\textbf{A} {/eq} using calculus. The closest point has the minimum squared distance. Therefore, we want to minimize

{eq}\begin{eqnarray*} d(t) &=& (0-1)^2+(3-3t-0)^2+(-1+5t-2)^2 \\ &=& 1+(3-3t)^2+(5t-3)^2 \\ &=& 1+9-18t+9t^2+25t^2-30t+9 \\ &=& 19-48t+34t^2. \end{eqnarray*} {/eq}

Use the second derivative test

{eq}\begin{eqnarray*} 0 &=& d'(t) \\ &=& -48+68t \\ \frac{48}{68} &=& t \\ \frac{12}{17} &=& t. \end{eqnarray*} {/eq}

Since {eq}d''(t)=68>0 {/eq} and {eq}d'\left(\frac{12}{17}\right)=0 {/eq}, the function has a minimum at {eq}t=\frac{12}{17} {/eq}. This agrees with the first approach. Now that {eq}t {/eq} has been obtained the point {eq}\textbf{D} {/eq} can be found as follows

{eq}\begin{eqnarray*} \textbf{D} &=& \left(1,-3+3\left(\frac{12}{17}\right),3-5\left(\frac{12}{17}\right)\right) \\ &=& \left(1,-\frac{15}{17},-\frac{9}{17}\right). \end{eqnarray*} {/eq}

Finally, the distance {eq}d {/eq} from from {eq}\textbf{D} {/eq} to {eq}\textbf{B} {/eq} is

{eq}\begin{eqnarray*} d &=& \sqrt{(1-0)^2+\left(-\frac{15}{17}-3\right)^2+\left(-\frac{9}{17}+1\right)^2} \\ &=& \sqrt{1+\left(-\frac{66}{17}\right)^2+\left(\frac{8}{17}\right)^2} \\ &=& \sqrt{\left(\frac{17}{17}\right)^2+\left(-\frac{66}{17}\right)^2+\left(\frac{8}{17}\right)^2} \\ &=& \frac{\sqrt{17^2+(-66)^2+8^2}}{17} \\ &=& \frac{\sqrt{4709}}{17} \\ &=& \frac{\sqrt{17(277)}}{17} \\ &=& \sqrt{\frac{277}{17}}. \end{eqnarray*} {/eq}