
If ${{\left| a*\left. b \right| \right.}^{2}}+{{\left| a.\left. b \right| \right.}^{2}}=144$and $\left| a \right|=4$then $\left| b \right|$ is equal to?
A. 16
B. 8
C. 3
D. 12
Answer
164.4k+ views
Hint: To solve this question where they give a cross product and a dot product. First we solve the cross product and dot product and then put the whole equation equal to the given number. Then by putting the value of $\left| a \right|$in the given equation, we get the desirable answer.
Formula used :-
${{(a.b)}^{2}}={{a}^{2}}{{b}^{2}}{{\cos }^{2}}\theta{{(a\times b)}^{2}}=({{a}^{2}}{{b}^{2}}{{\sin }^{2}}\theta )and {{(a.b)}^{2}}={{a}^{2}}{{b}^{2}}{{\cos }^{2}}\theta$
Complete Step- by- step solution:
We have given the equation $|a\times b\lvert^{2}+|a.b\lvert=144$ -----------(1)
In this question, we have given the cross product and the dot product and
We have to find out the value of b.
Suppose that there is angle between a and b is θ
Then we know the product of ${{(a\times b)}^{2}}=({{a}^{2}}{{b}^{2}}.1.{{\sin }^{2}}\theta )$
And the dot product of ${{(a.b)}^{2}}={{a}^{2}}{{b}^{2}}{{\cos }^{2}}\theta$
Put the value of cross product and dot product of ${{(a*b)}^{2}}and{{(a.b)}^{2}}$in equation (1) and we get
$\begin{align}
& ({{a}^{2}}{{b}^{2}}.1.{{\sin }^{2}}\theta )+{{a}^{2}}{{b}^{2}}{{\cos }^{2}}\theta =144 \\
& {{a}^{2}}{{b}^{2}}({{\sin }^{2}}\theta +{{\cos }^{2}}\theta )=144 \\
& {{a}^{2}}{{b}^{2}}(1)=144 \\
& {{a}^{2}}{{b}^{2}}=144 \\
\end{align}$
We have given the value of a in the question
Now put the value of $|a\lvert=4$ in above equation, we get
$16\:b^{2}=144$
Now we divide the both sides by 16, we get
$\begin{align}
& {{b}^{2}}=9 \\
& b=3 \\
& \left| b \right|=3 \\
\end{align}$or
We get the value of b=3
Option C is correct.
Note : Many students make the mistakes while opening the cross product and dot product. Remember that the distinction between dot product and cross product is the dot product is the multiplication of magnitude of the vectors and also the cos of the angle between them and the cross product is the multiplication of the magnitude of the vector and also the sine of the angle between them. Remember it while solving the questions.
Formula used :-
${{(a.b)}^{2}}={{a}^{2}}{{b}^{2}}{{\cos }^{2}}\theta{{(a\times b)}^{2}}=({{a}^{2}}{{b}^{2}}{{\sin }^{2}}\theta )and {{(a.b)}^{2}}={{a}^{2}}{{b}^{2}}{{\cos }^{2}}\theta$
Complete Step- by- step solution:
We have given the equation $|a\times b\lvert^{2}+|a.b\lvert=144$ -----------(1)
In this question, we have given the cross product and the dot product and
We have to find out the value of b.
Suppose that there is angle between a and b is θ
Then we know the product of ${{(a\times b)}^{2}}=({{a}^{2}}{{b}^{2}}.1.{{\sin }^{2}}\theta )$
And the dot product of ${{(a.b)}^{2}}={{a}^{2}}{{b}^{2}}{{\cos }^{2}}\theta$
Put the value of cross product and dot product of ${{(a*b)}^{2}}and{{(a.b)}^{2}}$in equation (1) and we get
$\begin{align}
& ({{a}^{2}}{{b}^{2}}.1.{{\sin }^{2}}\theta )+{{a}^{2}}{{b}^{2}}{{\cos }^{2}}\theta =144 \\
& {{a}^{2}}{{b}^{2}}({{\sin }^{2}}\theta +{{\cos }^{2}}\theta )=144 \\
& {{a}^{2}}{{b}^{2}}(1)=144 \\
& {{a}^{2}}{{b}^{2}}=144 \\
\end{align}$
We have given the value of a in the question
Now put the value of $|a\lvert=4$ in above equation, we get
$16\:b^{2}=144$
Now we divide the both sides by 16, we get
$\begin{align}
& {{b}^{2}}=9 \\
& b=3 \\
& \left| b \right|=3 \\
\end{align}$or
We get the value of b=3
Option C is correct.
Note : Many students make the mistakes while opening the cross product and dot product. Remember that the distinction between dot product and cross product is the dot product is the multiplication of magnitude of the vectors and also the cos of the angle between them and the cross product is the multiplication of the magnitude of the vector and also the sine of the angle between them. Remember it while solving the questions.
Recently Updated Pages
Environmental Chemistry Chapter for JEE Main Chemistry

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

Difference Between Natural and Whole Numbers: JEE Main 2024

Hess Law of Constant Heat Summation: Definition, Formula & Applications

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

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

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

JEE Advanced 2025 Notes
