Let \[\vec a = \hat i + \hat j + \sqrt 2 \hat k\], \[\vec b = {b_1}\hat i + {b_2}\hat j + \sqrt 2 \hat k\] and \[\vec c = 5\hat i + \hat j + \sqrt 2 \hat k\] be three vectors such that the projection vector of \[\vec b\] on \[\vec a\] is \[\vec a\]. If \[\vec a + \vec b\] is perpendicular to \[\vec c\], then \[\left| {\vec b} \right|\] is equal to

Hint: First we find out the projection of \[\vec b\] on \[\vec a\] by the rule and equal with \[\vec a\] and we get a equation of \[{b_1}\] and \[{b_2}\]. Also we find out \[\vec a + \vec b\] and then using the formula that \[\vec a + \vec b\] perpendicular to \[\vec c\] and we get another equation of \[{b_1}\] and \[{b_2}\]. Now solving two equations we get the value of \[{b_1}\] and \[{b_2}\], hence we find \[\left| {\vec b} \right|\] by using the formula.



Formula Used:The projection vector of \[\vec b\] on \[\vec a\] is \[\left( {\dfrac{{\vec a \cdot \vec b}}{{{{\left| {\vec a} \right|}^2}}}} \right) \cdot \vec a\]. Also if \[\vec a\] is perpendicular to \[\vec b\] then \[\vec a \cdot \vec b = 0\].



Complete step by step solution:\[\vec a \cdot \vec b\]
\[ = \left( {\hat i + \hat j + \sqrt 2 \hat k} \right) \cdot \left( {{b_1}\hat i + {b_2}\hat j + \sqrt 2 \hat k} \right)\]
\[ = {b_1} + {b_2} + 2\]
and \[\left| {\vec a} \right|\]
\[ = \sqrt {{1^2} + {1^2} + {{\left( {\sqrt 2 } \right)}^2}} \]
\[ = \sqrt {1 + 1 + 2} \]
\[ = \sqrt 4 \]
\[ = 2\].
Hence \[{\left| {\vec a} \right|^2} = 4\]. Also
\[\vec a + \vec b\]
\[ = \left( {\hat i + \hat j + \sqrt 2 \hat k} \right) + \left( {{b_1}\hat i + {b_2}\hat j + \sqrt 2 \hat k} \right)\]
\[ = \left( {{b_1} + 1} \right)\hat i + \left( {{b_2} + 1} \right)\hat j + 2\sqrt 2 \hat k\].
Now \[\vec a + \vec b\] is perpendicular to \[\vec c\], hence
\[\left( {\vec a + \vec b} \right) \cdot \vec c = 0\]
\[ \Rightarrow \left\{ {\left( {{b_1} + 1} \right)\hat i + \left( {{b_2} + 1} \right)\hat j + 2\sqrt 2 \hat k} \right\} \cdot \left( {5\hat i + \hat j + \sqrt 2 \hat k} \right) = 0\]
\[ \Rightarrow 5\left( {{b_1} + 1} \right) + \left( {{b_2} + 1} \right) + 4 = 0\]
\[ \Rightarrow 5{b_1} + 5 + {b_2} + 1 + 4 = 0\]
\[ \Rightarrow 5{b_1} + {b_2} = - 10\]-----(1).
Also the projection of \[\vec b\] on \[\vec a\] is \[\vec a\], hence
\[\left( {\dfrac{{\vec a \cdot \vec b}}{{{{\left| {\vec a} \right|}^2}}}} \right) \cdot \vec a = \vec a\]
\[ \Rightarrow \left( {\dfrac{{\vec a \cdot \vec b}}{{{{\left| {\vec a} \right|}^2}}}} \right) = 1\]
\[ \Rightarrow \dfrac{{{b_1} + {b_2} + 2}}{4} = 1\]
\[ \Rightarrow {b_1} + {b_2} + 2 = 4\]
\[ \Rightarrow {b_1} + {b_2} = 4 - 2\]
\[ \Rightarrow {b_1} + {b_2} = 2\]-----(2).
Solving (1) and (2), we get \[{b_1} = - 3\] and \[{b_2} = 5\]. Hence
\[\left| {\vec b} \right|\]
\[ = \sqrt {{b_1}^2 + {b_2}^2 + {{\left( {\sqrt 2 } \right)}^2}} \]
\[ = \sqrt {{{\left( { - 3} \right)}^2} + {5^2} + 2} \]
\[ = \sqrt {9 + 25 + 2} \]
\[ = \sqrt {36} \]
\[ = 6\].
Hence \[\left| {\vec b} \right| = 6\].





Note: It is to be noted that the formula of the projection of \[\vec b\] on \[\vec a\] is \[\left( {\dfrac{{\vec a \cdot \vec b}}{{{{\left| {\vec a} \right|}^2}}}} \right)\] but not \[\left( {\dfrac{{\vec a \cdot \vec b}}{{\left| {\vec a} \right|}}} \right)\]. Also the formula of \[\left( {\vec a + \vec b} \right)\]perpendicular to \[\vec c\] is \[\left( {\vec a + \vec b} \right) \cdot \vec c\] but not \[\left( {\vec a + \vec b} \right) \times \vec c\].