
If \[\left| {\overrightarrow A } \right| = 2\] and \[\left| {\overrightarrow B } \right| = 5\] and \[\overrightarrow A \times \overrightarrow B = 8\], then find the value of \[\overrightarrow A .\overrightarrow B \] .
A. 6
B. 2
C. 20
D. 8
Answer
164.7k+ views
Hint:
Firstly we find the value of sine using \[\overrightarrow A \times \overrightarrow B \] by expanding the formula and then using trigonometric formulas to find the value of cosine . And then solve the given expression \[\overrightarrow A .\overrightarrow B \] to find the required solution.
Formula Used:
Cross product of vectors \[\overrightarrow p \] and \[\overrightarrow q \] is \[\overrightarrow p \times \overrightarrow q = \left| {\overrightarrow p } \right|\left| {\overrightarrow q } \right|\sin \theta \]
Dot product of vectors \[\overrightarrow p \] and \[\overrightarrow q \] is \[\overrightarrow p \cdot \overrightarrow q = \left| {\overrightarrow p } \right|\left| {\overrightarrow q } \right|\cos \theta \]
Trigonometric formula: \[{\sin ^2}\theta + {\cos ^2}\theta = 1\]
Complete step-by-step answer:
Given that \[\left| {\overrightarrow A } \right| = 2\] and \[\left| {\overrightarrow B } \right| = 5\] and \[\overrightarrow A \times \overrightarrow B = 8\]
From the formula of cross product, we get
\[\overrightarrow A \times \overrightarrow B = \left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right|\sin \theta \]
Substitute all the values and we get
\[ \Rightarrow 8 = 2 \times 5\sin \theta \]
Simplifying and we get
\[ \Rightarrow 8 = 10\sin \theta \]
\[ \Rightarrow \sin \theta = \dfrac{8}{{10}}\]
\[ \Rightarrow \sin \theta = \dfrac{4}{5}\]
Now we use trigonometric formula \[{\sin ^2}\theta + {\cos ^2}\theta = 1\], to find the value of \[\cos \theta \]
\[{\left( {\dfrac{4}{5}} \right)^2} + {\cos ^2}\theta = 1\]
Squaring and we get
\[ \Rightarrow \dfrac{{16}}{{25}} + {\cos ^2}\theta = 1\]
Simplifying and we get
\[ \Rightarrow {\cos ^2}\theta = 1 - \dfrac{{16}}{{25}}\]
\[ \Rightarrow {\cos ^2}\theta = \dfrac{{25 - 16}}{{25}}\]
\[ \Rightarrow {\cos ^2}\theta = \dfrac{9}{{25}}\]
Taking square root in both sides of the above equation and we get
\[ \Rightarrow \sqrt {{{\cos }^2}\theta } = \sqrt {\dfrac{9}{{25}}} \]
\[ \Rightarrow \cos \theta = \dfrac{3}{5}\]
Now we find the value of \[\overrightarrow A .\overrightarrow B \] by using the formula of dot product of vector
\[\overrightarrow A .\overrightarrow B = \left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right|\cos \theta \]
Substitute all the values of \[\left| {\overrightarrow A } \right|,\left| {\overrightarrow B } \right|\] and \[\cos \theta \], we get
\[ \Rightarrow \overrightarrow A .\overrightarrow B = 2 \times 5 \times \dfrac{3}{5}\]
\[ \Rightarrow \overrightarrow A .\overrightarrow B = 2 \times 3\]
\[ \Rightarrow \overrightarrow A .\overrightarrow B = 6\]
Therefore, the value of \[\overrightarrow A .\overrightarrow B \] is 6.
Hence, the correct option is option A.
Note:
A dot product of two vectors is also called the scalar product. It is the product of the magnitude of the two vectors and the cosine of the angle that they form with each other.
A cross product of two vectors is also called the vector product. It is the product of the magnitude of the two vectors and the sine of the angle that they form with each other.
Firstly we find the value of sine using \[\overrightarrow A \times \overrightarrow B \] by expanding the formula and then using trigonometric formulas to find the value of cosine . And then solve the given expression \[\overrightarrow A .\overrightarrow B \] to find the required solution.
Formula Used:
Cross product of vectors \[\overrightarrow p \] and \[\overrightarrow q \] is \[\overrightarrow p \times \overrightarrow q = \left| {\overrightarrow p } \right|\left| {\overrightarrow q } \right|\sin \theta \]
Dot product of vectors \[\overrightarrow p \] and \[\overrightarrow q \] is \[\overrightarrow p \cdot \overrightarrow q = \left| {\overrightarrow p } \right|\left| {\overrightarrow q } \right|\cos \theta \]
Trigonometric formula: \[{\sin ^2}\theta + {\cos ^2}\theta = 1\]
Complete step-by-step answer:
Given that \[\left| {\overrightarrow A } \right| = 2\] and \[\left| {\overrightarrow B } \right| = 5\] and \[\overrightarrow A \times \overrightarrow B = 8\]
From the formula of cross product, we get
\[\overrightarrow A \times \overrightarrow B = \left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right|\sin \theta \]
Substitute all the values and we get
\[ \Rightarrow 8 = 2 \times 5\sin \theta \]
Simplifying and we get
\[ \Rightarrow 8 = 10\sin \theta \]
\[ \Rightarrow \sin \theta = \dfrac{8}{{10}}\]
\[ \Rightarrow \sin \theta = \dfrac{4}{5}\]
Now we use trigonometric formula \[{\sin ^2}\theta + {\cos ^2}\theta = 1\], to find the value of \[\cos \theta \]
\[{\left( {\dfrac{4}{5}} \right)^2} + {\cos ^2}\theta = 1\]
Squaring and we get
\[ \Rightarrow \dfrac{{16}}{{25}} + {\cos ^2}\theta = 1\]
Simplifying and we get
\[ \Rightarrow {\cos ^2}\theta = 1 - \dfrac{{16}}{{25}}\]
\[ \Rightarrow {\cos ^2}\theta = \dfrac{{25 - 16}}{{25}}\]
\[ \Rightarrow {\cos ^2}\theta = \dfrac{9}{{25}}\]
Taking square root in both sides of the above equation and we get
\[ \Rightarrow \sqrt {{{\cos }^2}\theta } = \sqrt {\dfrac{9}{{25}}} \]
\[ \Rightarrow \cos \theta = \dfrac{3}{5}\]
Now we find the value of \[\overrightarrow A .\overrightarrow B \] by using the formula of dot product of vector
\[\overrightarrow A .\overrightarrow B = \left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right|\cos \theta \]
Substitute all the values of \[\left| {\overrightarrow A } \right|,\left| {\overrightarrow B } \right|\] and \[\cos \theta \], we get
\[ \Rightarrow \overrightarrow A .\overrightarrow B = 2 \times 5 \times \dfrac{3}{5}\]
\[ \Rightarrow \overrightarrow A .\overrightarrow B = 2 \times 3\]
\[ \Rightarrow \overrightarrow A .\overrightarrow B = 6\]
Therefore, the value of \[\overrightarrow A .\overrightarrow B \] is 6.
Hence, the correct option is option A.
Note:
A dot product of two vectors is also called the scalar product. It is the product of the magnitude of the two vectors and the cosine of the angle that they form with each other.
A cross product of two vectors is also called the vector product. It is the product of the magnitude of the two vectors and the sine of the angle that they form with each other.
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 Atomic Structure and Chemical Bonding important Concepts and Tips

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

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Degree of Dissociation and Its Formula With Solved Example for JEE

Instantaneous Velocity - Formula based Examples for JEE

NCERT Solutions for Class 11 Maths In Hindi Chapter 1 Sets
