
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
162k+ 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
If there are 25 railway stations on a railway line class 11 maths JEE_Main

Minimum area of the circle which touches the parabolas class 11 maths JEE_Main

Which of the following is the empty set A x x is a class 11 maths JEE_Main

The number of ways of selecting two squares on chessboard class 11 maths JEE_Main

Find the points common to the hyperbola 25x2 9y2 2-class-11-maths-JEE_Main

A box contains 6 balls which may be all of different class 11 maths JEE_Main

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

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

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

JoSAA JEE Main & Advanced 2025 Counselling: Registration Dates, Documents, Fees, Seat Allotment & Cut‑offs

NIT Cutoff Percentile for 2025

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

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

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

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

Degree of Dissociation and Its Formula With Solved Example for JEE

Free Radical Substitution Mechanism of Alkanes for JEE Main 2025
