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
298.5k+ 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
Letfx be a polynomial with positive degree satisfy-class-12-maths-JEE_Main

Evaluate the definite integral given as intlimits13left class 12 maths JEE_Main

The sum of squares of two parts of a number 100 is-class-12-maths-JEE_Main

Geometry of Complex Numbers Explained

JEE Main 2023 (February 1st Shift 2) Physics Question Paper with Answer Key

JEE Main 2023 (February 1st Shift 1) Maths Question Paper with Answer Key

Trending doubts
Electron Gain Enthalpy and Electron Affinity Explained

What Are Current and Potential Difference in Electricity?

Understanding Geostationary and Geosynchronous Satellites

Isoelectronic Species: Definition, Examples & Importance

Degree of Dissociation: Meaning, Formula, Calculation & Uses

Free Radical Substitution and Its Stepwise Mechanism

Other Pages
JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

JEE Advanced 2026 Notification Out with Exam Date, Registration (Extended), Syllabus and More

JEE Advanced Weightage Chapter Wise 2026 for Physics, Chemistry, and Mathematics

Loan Amortization Schedule Calculator – Free Online Tool

Electrochemistry JEE Advanced 2027 Notes - Free PDF Download (Sign-in Required)

Chemistry Question Papers for JEE Main, NEET & Boards (PDFs)

