Question

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

Answer 1

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 , $\vec a$, $\vec b$and $\vec c$in $\vec r$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= $\hat u = \dfrac{u}{{|u|}}$ where u is a vector.
Angle between vectors:
$\cos \theta = \dfrac{{\vec a.\vec b}}{{\left| {\vec a} \right|\left| {\vec b} \right|}}$ where a and b are given vectors.
For component of $\vec a$ perpendicular to $\vec b$ In the plane of $\vec a$ and $\vec b$ we use: $\dfrac{{\vec a \times \left( {\vec b \times \vec a} \right)}}{{{{\left| a \right|}^2}}}$

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

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 $\vec r$. That is we will find $\left| {\hat r = \dfrac{r}{{|r|}}} \right|$,
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 $\vec a.\vec b = \left| {\vec a} \right|\left| {\vec b} \right|\cos \theta$ is used to find angles between them.