Chapter 6, Problem 9RE

Solution

To find: The component form and magnitude of the vector $\overrightarrow{AC}+\overrightarrow{BD}$ .

The component form of vector $\overrightarrow{AC}+\overrightarrow{BD}$ is $\left(-8,-3\right)$ , and magnitude of the vector $\overrightarrow{AC}+\overrightarrow{BD}$ is $8.544$ .

Given information:

The given information is $A=\left(2,-1\right),B=\left(3,1\right),C=\left(-4,2\right)$ and $D=\left(1,-5\right)$ .

Formula used:

Head Minus Tail (HMT) Rule for Vectors:

If an arrow has initial point $\left(x_1,y_1\right)$ and terminal point $\left(x_2,y_2\right)$ , it represents the vector $\left(x_2-x_1,y_2-y_1\right)$ .

Vector addition:

Let $u=\left(u_1,u_2\right)$ and $v=\left(v_1,v_2\right)$ be vectors. The sum (or resultant) of the vectors $u$ and $v$ is the vector .

Magnitude (or Length) of a vector:

If represents a vector $(a,b)$ , then $\left|v\right|=\sqrt{a^2+b^2}$ .

Calculation:

Apply the HMT Rule, $\overrightarrow{AC}$ represents the vector:

  $\left(\left(-4\right)-2,2-\left(-1\right)\right)=\left(-6,3\right)$

And, $\overrightarrow{BD}$ represents the vector:

  $\left(\left(1-3\right),\left(-5\right)-\left(-1\right)\right)=\left(-2,-6\right)$ .

Use the Vector addition rule, to find $\overrightarrow{AC}+\overrightarrow{BD}$ .

  $\left(-6+\left(-2\right),3+\left(-6\right)\right)=\left(-8,-3\right)$ .

Use the Magnitude of a vector Rule to find $\left|\overrightarrow{AC}+\overrightarrow{BD}\right|$ .

  $\begin{array}{c}\left|\overrightarrow{AC}+\overrightarrow{BD}\right|=\sqrt{{\left(-8\right)}^2+{\left(-3\right)}^2}\\ =\sqrt{73}\\ \simeq 8.544\end{array}$

Hence, the component form of vector $\overrightarrow{AC}+\overrightarrow{BD}$ is $\left(-8,-3\right)$ , and magnitude of the vector $\overrightarrow{AC}+\overrightarrow{BD}$ is $8.544$ .