
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
161.4k+ 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.
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