9Eresolutes


 * Resolving Vectors**

Recall that **components** of a vector __add__ to give the vector.

Recall that if two components of a vector are __perpendicular__, they are called **resolutes**.

Finding the resolutes of a vector is called **resolving the vector**.

Resolving a vector

Given two vectors, u and v with an angle of theta between them. The goal is to resolve (split) **__v__** into two vector resolutes: -- one parallel to **__u__** (**OA**) -- and one perpendicular to **__u__** (**AB**)

ie we want to find **OA** and **AB** so that math \underline{v}=\overrightarrow{OA}+\overrightarrow{AB} \qquad. math

(this will take a few steps)

Scalar Resolute

The **scalar resolute** of **__v__** in the direction of **__u__** is math OA=| \overrightarrow{OA} | \qquad. math

Using trigonometry: math OA=|\underline{v}| \cos( \theta ) \qquad. math

math \\ \text{Use dot product: } \; \underline{u} \centerdot \underline{v}=|\underline{u}||\underline{v}|\cos( \theta ) \qquad .\\. \\ \underline{u} \centerdot \underline{v}=|\underline{u}|OA \qquad .\\. \\ OA=\dfrac{\underline{u} \centerdot \underline{v}}{|\underline{u}|} \qquad. math

math \\ OA=\dfrac{\underline{u}}{|\underline{u}|} \centerdot \underline{v} \qquad. \\ . \\ \text{Use unit vector: } \; \hat{\underline{u}}=\dfrac{\underline{u}}{|\underline{u}|} \qquad. \\ . \\ OA=\underline{\hat{u}} \centerdot \underline{v} \qquad \textit{ Scalar Resolute of v onto u} \qquad. math

Example 1

Given **__a__** = 7**__i__** + 2**__j__** and **__b__** = 5**__i__** – 12**__j__** Find the scalar resolute of **__a__** in the direction of **__b__**.

1. Find the magnitude of **__b__** math math
 * \underline{b}|=\sqrt{5^2+(-12)^2}=13 \qquad.

2. Hence the unit vector in the direction of **__b__** is math \underline{\hat{b}}=\frac{1}{13} \big( 5\underline{i}-12\underline{j} \big) \qquad. math

3. Hence the scalar resolute of **__a__** in the direction of **__b__** is: math \\ \underline{\hat{b}} \centerdot \underline{a}=\frac{1}{13} \big( 5\underline{i}-12\underline{j} \big) \centerdot \big( 7\underline{i}+2\underline{j} \big) \qquad. \\ . \\ \underline{\hat{b}} \centerdot \underline{a}=\frac{1}{13} \big( 35-24)=\frac{11}{13} \qquad. math

Vector Resolute Parallel to u The **vector resolute** of **__v__** in the direction of **__u__** is the vector **OA**

We know the __magnitude__ of **OA** is: math \underline{\hat{u}} \centerdot \underline{v} \qquad. math

And we know the __direction__ of **OA** is math \underline{\hat{u}} \qquad. math

Hence the vector resolute of **__v__** onto **__u__** is: math \overrightarrow{OA}=\big( \underline{\hat{u}} \centerdot \underline{v} \big) \, \underline{\hat{u}} \qquad. math

Example 1 (continued)

Given **__a__** = 7**__i__** + 2**__j__** and **__b__** = 5**__i__** – 12**__j__** Find the vector resolute of **__a__** in the direction of **__b__**.

From above, we know that the scalar resolute is: math \underline{\hat{b}} \centerdot \underline{a}=\frac{11}{13} \qquad. math

And we know the unit vector is: math \underline{\hat{b}}=\frac{1}{13} \big( 5\underline{i}-12\underline{j} \big) \qquad. math

Hence math \\ \overrightarrow{OA}=\big( \underline{\hat{b}} \centerdot \underline{a} \big) \, \underline{\hat{b}} \qquad. \\ .\\ \overrightarrow{OA}=\big( \frac{11}{13} \big) \times \frac{1}{13} \big( 5\underline{i}-12\underline{j} \big) \qquad. \\ . \\ \overrightarrow{OA}=\frac{11}{169} \big( 5\underline{i}-12\underline{j} \big) \qquad. math

Vector Resolute Perpendicular to u The vector resolute of **__v__** perpendicular to **__u__** is the vector **AB**

Using vector addition: math \\ \overrightarrow{OA}+\overrightarrow{AB}=\overrightarrow{OB} \qquad .\\. \\ \overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA} \qquad. \\ . \\ \overrightarrow{AB}=\underline{v}-\big( \underline{\hat{u}} \centerdot \underline{v} \big) \, \underline{\hat{u}} \qquad. math

Example 1 (continued)

Given **__a__** = 7**__i__** + 2**__j__** and **__b__** = 5**__i__** – 12**__j__** Find the vector resolute of **__a__** perpendicular to **__b__**.

From above, we know that: math \\ \overrightarrow{OA}=\big( \underline{\hat{b}} \centerdot \underline{a} \big) \, \underline{\hat{b}} = \overrightarrow{OA}=\frac{11}{169} \big( 5\underline{i}-12\underline{j} \big) \qquad. math

Hence math \\ \overrightarrow{AB}=\underline{a}-\big( \underline{\hat{b}} \centerdot \underline{a} \big) \, \underline{\hat{b}} \qquad .\\. \\ \overrightarrow{AB}=\big( 7\underline{i}+2\underline{j} \big) - \big( \frac{55}{169}\underline{i}-\frac{132}{169}\underline{j} \big) \qquad. \\ . \\ \overrightarrow{AB}=6\frac{114}{169}\underline{i}+2\frac{132}{169}\underline{j} \qquad. math

Thus we have resolved the vector **__a__** into resolutes, so math \\ \underline{a}=\overrightarrow{OA}+\overrightarrow{AB} \qquad. \\ . \\ \underline{a}=\big( \frac{55}{169}\underline{i}-\frac{132}{169}\underline{j} \big)+\big( 6\frac{114}{169}\underline{i}+2\frac{132}{169}\underline{j} \big) \qquad. math

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