
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
Answer
163.2k+ views
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 \[\left( {\vec a + \vec b} \right) \times \vec c\].
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 \[\left( {\vec a + \vec b} \right) \times \vec c\].
Recently Updated Pages
Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Main 2025 Session 2: Exam Date, Admit Card, Syllabus, & More

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

Trending doubts
Degree of Dissociation and Its Formula With Solved Example for JEE

Instantaneous Velocity - Formula based Examples for JEE

JEE Main Chemistry Question Paper with Answer Keys and Solutions

JEE Main Reservation Criteria 2025: SC, ST, EWS, and PwD Candidates

What is Normality in Chemistry?

Chemistry Electronic Configuration of D Block Elements: JEE Main 2025

Other Pages
Total MBBS Seats in India 2025: Government College Seat Matrix

NEET Total Marks 2025: Important Information and Key Updates

Neet Cut Off 2025 for MBBS in Tamilnadu: AIQ & State Quota Analysis

Karnataka NEET Cut off 2025 - Category Wise Cut Off Marks

NEET Marks vs Rank 2024|How to Calculate?

NEET 2025: All Major Changes in Application Process, Pattern and More
