Question

A. In a unit vector notation, what is $$\overrightarrow{r}=\overrightarrow{a}-\overrightarrow{b}+\overrightarrow{c}$$when $$\overrightarrow{a}=5\hat{i}+4\hat{k}$$, $$\overrightarrow{b}=-2\hat{i}+2\hat{j}+3\hat{k}$$ and $$\overrightarrow{c}=4\hat{i}+3\hat{j}+2\hat{k}$$,
B. Calculate the angle between $$\overrightarrow{r}$$and the positive – axis.
C. What is the component of $$\overrightarrow{a}$$ along the direction of $$\overrightarrow{b}$$
D. What is the perpendicular to the direction of $$\overrightarrow{b}$$ but in the plane of $$\overrightarrow{b}$$ and $$\overrightarrow{a}$$?

Answer 1

Hint: To answer this question we will first define what is a unit vector. Next step is to answer each part of this question with a detailed explanation. We will first find answer A by substituting values of , $$\overrightarrow{a}$$, $$\overrightarrow{b}$$and $$\overrightarrow{c}$$in $$\overrightarrow{r}$$and then finding the unit vector using its formula. For part b and c and d there are direct formulas available, hence we will simply apply those formulas and solve the questions.

Formula Used:
Unit vector= $$\hat{u}={\displaystyle \frac{u}{|u|}}$$ where u is a vector.
Angle between vectors:
$$\cos \theta ={\displaystyle \frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}}$$ where a and b are given vectors.
For component of $$\overrightarrow{a}$$ perpendicular to $$\overrightarrow{b}$$ In the plane of $$\overrightarrow{a}$$ and $$\overrightarrow{b}$$ we use: $${\displaystyle \frac{\overrightarrow{a}\times \left(\overrightarrow{b}\times \overrightarrow{a}\right)}{{\left|a\right|}^2}}$$

Complete step by step solution:
Let us first know what a unit vector is: it is such a vector whose magnitude is one.
Let us answer question A.
A. It has been given that: $$\overrightarrow{a}=5\hat{i}+4\hat{k}$$, $$\overrightarrow{b}=-2\hat{i}+2\hat{j}+3\hat{k}$$ and $$\overrightarrow{c}=4\hat{i}+3\hat{j}+2\hat{k}$$,
We have to find $$\overrightarrow{r}$$which is $$\overrightarrow{r}=\overrightarrow{a}-\overrightarrow{b}+\overrightarrow{c}$$.
Let us substitute the values of , $$\overrightarrow{a}$$, $$\overrightarrow{b}$$and $$\overrightarrow{c}$$in $$\overrightarrow{r}$$
We get : $$\overrightarrow{r}=5\hat{i}+4\hat{k}-\left(-2\hat{i}+2\hat{j}+3\hat{k}\right)+4\hat{i}+3\hat{j}+2\hat{k}$$
Or, $$\overrightarrow{r}=5\hat{i}+4\hat{k}+2\hat{i}-2\hat{j}-3\hat{k}+4\hat{i}+3\hat{j}+2\hat{k}$$
Or, $$\overrightarrow{r}=5\hat{i}+2\hat{i}+4\hat{i}-2\hat{j}+3\hat{j}-3\hat{k}+2\hat{k}+4\hat{k}$$
Or, $$\overrightarrow{r}=\left(5+2+4\right)\hat{i}+\left(3-2\right)\hat{j}+\left(-3+2+4\right)\hat{k}$$
Or, $$\overrightarrow{r}=11\hat{i}+1\hat{j}+3\hat{k}$$
Hence for unit vector notation we have to use: $$\hat{r}={\displaystyle \frac{r}{|r|}}$$
Lets find $$|r|$$.
$$|r|=\sqrt{{11}^2+1^2+3^2}$$
Or, $$|r|=\sqrt{121+1+9}$$
Or, $$|r|=\sqrt{131}$$
Hence the answer A is : $$\hat{r}={\displaystyle \frac{r}{|r|}}$$= $$\hat{r}={\displaystyle \frac{11\hat{i}+1\hat{j}+3\hat{k}}{\sqrt{131}}}$$,
B. Now we know that $$\overrightarrow{r}=11\hat{i}+1\hat{j}+3\hat{k}$$ hence the angle of this with z-axis is given by
$$\cos \theta ={\displaystyle \frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}}$$
or, $$\cos \theta ={\displaystyle \frac{\overrightarrow{r}.\hat{k}}{\left|\overrightarrow{r}\right|\left|\hat{k}\right|}}$$
Or, $$\cos \theta ={\displaystyle \frac{\left(11\hat{i}+1\hat{j}+3\hat{k}\right)\hat{k}}{\sqrt{131}\times \sqrt{1}}}$$
Or, $$\cos \theta ={\displaystyle \frac{0\hat{i}+0\hat{j}+3\hat{k}}{\sqrt{131}}}$$ $$\left(\because \hat{k}\times \hat{i}=\hat{k}\times \hat{j}=0\right)$$and $$\left(\because \hat{k}\times \hat{k}=1\right)$$
Or, $$\cos \theta ={\displaystyle \frac{3\hat{k}}{\sqrt{131}}}$$.
Hence angle $$\theta ={\cos }^{-1}{\displaystyle \frac{5}{\sqrt{131}}}$$.
C. Here the component of $$\overrightarrow{a}$$ along $$\overrightarrow{b}$$ is $${\displaystyle \frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{b}\right|}}$$
Hence projection is $${\displaystyle \frac{\left(5\hat{i}+4\hat{k}\right).\left(-2\hat{i}+2\hat{j}+3\hat{k}\right)}{\left|2\hat{i}-2\hat{j}-3\hat{k}\right|}}$$
Or, $${\displaystyle \frac{-10+0+12}{\sqrt{2^2+2^2+3^2}}}$$
Or, $${\displaystyle \frac{2}{\sqrt{4+4+9}}}$$
Or, $${\displaystyle \frac{2}{\sqrt{17}}}$$.
D. For the perpendicular to the direction of $$\overrightarrow{b}$$ but in the plane of $$\overrightarrow{b}$$ and $$\overrightarrow{a}$$:
 $$\sqrt{{\left|a\right|}^2-{\left|b\right|}^2}$$
or, $$\sqrt{5^2+4^2-{\displaystyle \frac{{\left(2\right)}^2}{{\sqrt{17}}^2}}}$$ (we have already found the value for $$\left|b\right|$$).
Or, $$\sqrt{25+16-{\displaystyle \frac{4}{17}}}$$
Or, $$\sqrt{41-{\displaystyle \frac{4}{17}}}$$
Or, $$\sqrt{{\displaystyle \frac{697-4}{17}}}$$
Or, $$\sqrt{{\displaystyle \frac{693}{17}}}$$.

Note:
In the first part, remember to check the answer, find if the answer that we have found is a unit vector or no. to check this we have to find the magnitude of $$\overrightarrow{r}$$. That is we will find $$\left|\hat{r}={\displaystyle \frac{r}{|r|}}\right|$$,
We will see that its magnitude comes out to be 1. Hence our answer is correct.
In the 2nd question, remember to use the formula for angle using $$\cos \theta$$ and not $$\sin \theta$$. We have to use this because $$\overrightarrow{a}.\overrightarrow{b}=\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \theta$$ is used to find angles between them.