Find a vector perpendicular to ${{\hat i + 2 \hat j}}$ magnitude of ${{3}}\sqrt {{5}}$.
A) $3 \hat i + 6 \hat j$
B) $6 \hat i - 3 \hat j$
C) $4 \hat i - 2 \hat j$
D) $ \hat i - 2 \hat j$
Answer
Verified
117.3k+ views
Hint: In order to find a vector, first of all consider a general vector. Let us consider the given vector to be ${{\overrightarrow r = x \hat i + y \hat j}}$. We know that if two vectors are mutually perpendicular to each other then their dot product is zero. Now, find the dot product of both the vectors. Relate the obtained equation and the given equation to find the components of the vector. And the components of the vector are found by substitution method.
Complete step by step solution:
Let us consider that the required vector is ${{\overrightarrow r = x \hat i + y \hat j}}$
According to the question, ${{x \hat i + y \hat j}}$ is perpendicular to ${{\overrightarrow A = \hat i + 2 \hat j}}$.
When one vector is perpendicular to the other vector then the dot product of both the vectors is zero.
Thus, the dot product of vector ${{\overrightarrow r = x \hat i + y \hat j}}$ and vector ${{\overrightarrow A = \hat i + 2 \hat j}}$ must be zero.
Now, finding the dot product of both the vectors
$
\Rightarrow {{\overrightarrow r}}{{. \overrightarrow A = (x \hat i + y \hat j)}}{{.(\hat i + 2 \hat j)}} \\
\Rightarrow {{x + 2 y = 0}} \\
\Rightarrow {{x = - 2 y}}...{{(i)}} $
Given that the magnitude of vector r is ${{3}}\sqrt {{5}} $
So, ${{|r| = }}\sqrt {{{{x}}^{{2}}}{{ + }}{{{y}}^{{2}}}} {{ = 3}}\sqrt 5 $
Squaring both sides, we get
$
\Rightarrow {\left( {\sqrt {{{{x}}^{{2}}}{{ + }}{{{y}}^{{2}}}} } \right)^2}{{ = }}{\left( {{{3}}\sqrt 5 } \right)^2} \\
\Rightarrow {{{x}}^{{2}}}{{ + }}{{{y}}^{{2}}}{{ }} = {{ }}45$
Substituting the value of x from (i), we get
$
\Rightarrow {{{( - 2 y)}}^{{2}}}{{ + }}{{{y}}^{{2}}}{{ = 45}} \\
\Rightarrow {{5 }}{{{y}}^2}{{ }} = {{ }}45 \\
\therefore {{y = 3}} $
Now substituting this value of y in equation (i), we get
$
\Rightarrow {{x = - 2 y = - 2 (3)}} \\
\Rightarrow {{x = - 6}}$
Hence, the required vector becomes
$\Rightarrow {{\overrightarrow r = 6 \hat i - 3 \hat j}}$
Therefore, option (B) is the correct choice.
Note: In a two-dimensional coordinate system, any vector can be broken into x - component and y - component. Let us consider the general vector, ${{\overrightarrow r = x \hat i + y \hat j + \hat z k}}$. But in the question, we have assumed the general vector to be ${{\overrightarrow r = x \hat i + y \hat j}}$. This is because of the fact that the other vector which is provided in the question stem i.e. ${{\hat i + 2 \hat j}}$ does not involve a vector of z - component. Each part of the two-dimensional vector is also known as a component. The components of a given vector depicts the influence of that vector in a given direction.
Complete step by step solution:
Let us consider that the required vector is ${{\overrightarrow r = x \hat i + y \hat j}}$
According to the question, ${{x \hat i + y \hat j}}$ is perpendicular to ${{\overrightarrow A = \hat i + 2 \hat j}}$.
When one vector is perpendicular to the other vector then the dot product of both the vectors is zero.
Thus, the dot product of vector ${{\overrightarrow r = x \hat i + y \hat j}}$ and vector ${{\overrightarrow A = \hat i + 2 \hat j}}$ must be zero.
Now, finding the dot product of both the vectors
$
\Rightarrow {{\overrightarrow r}}{{. \overrightarrow A = (x \hat i + y \hat j)}}{{.(\hat i + 2 \hat j)}} \\
\Rightarrow {{x + 2 y = 0}} \\
\Rightarrow {{x = - 2 y}}...{{(i)}} $
Given that the magnitude of vector r is ${{3}}\sqrt {{5}} $
So, ${{|r| = }}\sqrt {{{{x}}^{{2}}}{{ + }}{{{y}}^{{2}}}} {{ = 3}}\sqrt 5 $
Squaring both sides, we get
$
\Rightarrow {\left( {\sqrt {{{{x}}^{{2}}}{{ + }}{{{y}}^{{2}}}} } \right)^2}{{ = }}{\left( {{{3}}\sqrt 5 } \right)^2} \\
\Rightarrow {{{x}}^{{2}}}{{ + }}{{{y}}^{{2}}}{{ }} = {{ }}45$
Substituting the value of x from (i), we get
$
\Rightarrow {{{( - 2 y)}}^{{2}}}{{ + }}{{{y}}^{{2}}}{{ = 45}} \\
\Rightarrow {{5 }}{{{y}}^2}{{ }} = {{ }}45 \\
\therefore {{y = 3}} $
Now substituting this value of y in equation (i), we get
$
\Rightarrow {{x = - 2 y = - 2 (3)}} \\
\Rightarrow {{x = - 6}}$
Hence, the required vector becomes
$\Rightarrow {{\overrightarrow r = 6 \hat i - 3 \hat j}}$
Therefore, option (B) is the correct choice.
Note: In a two-dimensional coordinate system, any vector can be broken into x - component and y - component. Let us consider the general vector, ${{\overrightarrow r = x \hat i + y \hat j + \hat z k}}$. But in the question, we have assumed the general vector to be ${{\overrightarrow r = x \hat i + y \hat j}}$. This is because of the fact that the other vector which is provided in the question stem i.e. ${{\hat i + 2 \hat j}}$ does not involve a vector of z - component. Each part of the two-dimensional vector is also known as a component. The components of a given vector depicts the influence of that vector in a given direction.
Recently Updated Pages
Uniform Acceleration - Definition, Equation, Examples, and FAQs
How to find Oxidation Number - Important Concepts for JEE
How Electromagnetic Waves are Formed - Important Concepts for JEE
Electrical Resistance - Important Concepts and Tips for JEE
Average Atomic Mass - Important Concepts and Tips for JEE
Chemical Equation - Important Concepts and Tips for JEE
Trending doubts
JEE Main 2025: Application Form (Out), Exam Dates (Released), Eligibility & More
JEE Main Login 2045: Step-by-Step Instructions and Details
Class 11 JEE Main Physics Mock Test 2025
JEE Main Chemistry Question Paper with Answer Keys and Solutions
Learn About Angle Of Deviation In Prism: JEE Main Physics 2025
JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics
Other Pages
NCERT Solutions for Class 11 Physics Chapter 7 Gravitation
NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements
NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Fluids
Units and Measurements Class 11 Notes - CBSE Physics Chapter 1
NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line
NCERT Solutions for Class 11 Physics Chapter 8 Mechanical Properties of Solids