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 \[\vec a\]?

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 , \[\vec a\], \[\vec b\]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.