Chapter 6.1, Problem 61E

Solution

a.

To prove: $\overrightarrow{BA}=\overrightarrow{OA}-\overrightarrow{OB}$ .

Given information:

The $A$ and $B$ are two points in a plane.

Proof:

Consider the points,

  $\begin{array}{c}A=(a_1,a_2)\\ B=(b_1,b_2)\\ O=(0,0)\end{array}$

By Head Minus Tail Rule,

  $\begin{array}{c}\overrightarrow{BA}=\left(a_1-b_1,a_2-b_2\right)\\ \overrightarrow{OA}=\left(a_1-0,a_2-0\right)\\ =\left(a_1,a_2\right)\\ \overrightarrow{OB}=\left(b_1-0,b_2-0\right)\\ =\left(b_1,b_2\right)\end{array}$

Substituting the values,

  $\begin{array}{c}\overrightarrow{OA}-\overrightarrow{OB}=\overrightarrow{OA}+(-\overrightarrow{OB})\\ \overrightarrow{OA}-\overrightarrow{OB}=\left(a_1,a_2\right)+\left(-\left(b_1,b_2\right)\right)\\ \overrightarrow{OA}-\overrightarrow{OB}=\left(\left(a_1,a_2\right)+\left(-b_1,-b_2\right)\right)\\ \overrightarrow{OA}-\overrightarrow{OB}=\left(a_1+\left(-b_1\right),a_2+\left(-b_2\right)\right)\\ \overrightarrow{OA}-\overrightarrow{OB}=\left(a_1-b_1,a_2-b_2\right)\end{array}$

Hence, $\overrightarrow{BA}=\overrightarrow{OA}-\overrightarrow{OB}$ is proved.

b.

To prove: $\overrightarrow{OC}=x\overrightarrow{OA}+y\overrightarrow{OB}$ .

Given information:

  $x+y=1$

  $\frac{\left|\overrightarrow{BC}\right|}{\left|\overrightarrow{CA}\right|}=\frac{x}{y}$

Proof:

Let $C$ be the point on the line segment $BA$ .

  $\frac{\left|\overrightarrow{BC}\right|}{\left|\overrightarrow{CA}\right|}=\frac{x}{y}$

Here, $\left|\overrightarrow{BC}\right|$ and $\left|\overrightarrow{CA}\right|$ are collinear

  $\begin{array}{c}\left|\overrightarrow{BC}\right|=\frac{x}{y}\left|\overrightarrow{CA}\right|\\ =\frac{x}{y}.\overrightarrow{CA}\\ y\left|\overrightarrow{BC}\right|=x\overrightarrow{CA}\end{array}$

  

From the above graph,

  $\begin{array}{c}x\overrightarrow{OA}+y\overrightarrow{OB}=x\left(\overrightarrow{OC}+\overrightarrow{CA}\right)+y\left(\overrightarrow{OC}+\overrightarrow{CB}\right)\\ =x\overrightarrow{OC}+x\overrightarrow{CA}+y\overrightarrow{OC}+y\overrightarrow{CB}\\ =(\underset{1}{\underbrace{x+y}})\overrightarrow{OC}+\underset{y\overrightarrow{BC}}{\underbrace{x\overrightarrow{CA}}}+y\overrightarrow{CB}\\ =\overrightarrow{OC}+y\overrightarrow{BC}+y\overrightarrow{CB}\\ =\overrightarrow{OC}+y\underset{0}{\underbrace{\left(\overrightarrow{BC}+\overrightarrow{CB}\right)}}\\ =\overrightarrow{OC}\end{array}$

Hence, $x\overrightarrow{OA}+y\overrightarrow{OB}=\overrightarrow{OC}$ is proved.