Question

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 1

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.