Chapter 14.1, Problem 14OE

To determine

To explain: The vector joining two lettered points of aparallelogram $ABCD$ is equivalent to the given vector $\overrightarrow{AD}+\overrightarrow{DC}+\overrightarrow{CX}+\overrightarrow{XB}$ .

  

Answer to Problem 14OE

The statement “The vector joining two lettered points of a parallelogram $ABCD$ of The statement is $AB$ .

Explanation of Solution

Given information:

The vector joining two lettered points of a parallelogram $ABCD$ is equivalent to the given vector $\overrightarrow{AD}+\overrightarrow{DC}+\overrightarrow{CX}+\overrightarrow{XB}$ .

Consider the provided statement “A parallelogram $ABCD$ .

Parallelogram has a property that opposite sides of a parallelogram are parallel to each other.

Parallelogram has four sides and four angles, sometimes sides and angles are equal and sometimes they may different.

Each diagonal bisects each other and each diagonal divides the quadrilateral into two equal congruent triangles.

Thus the, $\overrightarrow{AD}+\overrightarrow{DC}+\overrightarrow{CX}+\overrightarrow{XB}$ ,

Here, The two vectors from a triangle to show the sum vector.

The sum angle is that angle made by when two vectors make a triangle in sequence, the third side that closing the triangle, in the opposite direction of the sequence represents the sum or resultant of two vectors .

  $AD+DC=AC$

The two vectors from a triangle to show the sum vector.

The sum angle is that angle made by when two vectors make a triangle in sequence, the third side that closing the triangle, in the opposite direction of the sequence represents the sum or resultant of two vectors .

  $CX+XB=CB$

Substitute the above both values in the given equation $\overrightarrow{AD}+\overrightarrow{DC}+\overrightarrow{CX}+\overrightarrow{XB}$ .

  $\begin{array}{c}\overrightarrow{AD}+\overrightarrow{DC}+\overrightarrow{CX}+\overrightarrow{XB}\\ AD+DC=AC\\ CX+XB=CB\\ AC+CB=AB\end{array}$

Thus, the statement, The statement “The vector joining two lettered points of a parallelogram $ABCD$ of $\overrightarrow{AD}+\overrightarrow{DC}+\overrightarrow{CX}+\overrightarrow{XB}$ is $AB$ .