Question

If $\left| \overline{a} \right|=1$ , $\left| \overline{b} \right|=2$ , \[\left( \overline{a},\overline{b} \right)=\dfrac{2\pi }{3}\] then ${{\left\{ \left( \overline{a}+3\overline{b} \right)\times \left( 3\overline{a}+\overline{b} \right) \right\}}^{2}}$ :
\[\begin{align}
  & \text{1}.~~~\text{425} \\
 & \text{2}.~~\text{375} \\
 & \text{3}.~~~\text{325} \\
 & \text{4}.~~~\text{3}00 \\
\end{align}\]

Answer 1

Hint: Here, for getting the value of the given question ${{\left\{ \left( \overline{a}+3\overline{b} \right)\times \left( 3\overline{a}+\overline{b} \right) \right\}}^{2}}$ , first of all we will Do cross product within the bracket. After that we will simplify the cross product with the help of cross product formula that is $\overline{a}\times \overline{b}=\overline{a}.\overline{b}.\cos \left( \overline{a},\overline{b} \right)$ . Then we will square the bracket term and put the given values according to the situation.

Complete step by step answer:
Since, the given question that we need to solve is:
$\Rightarrow {{\left\{ \left( \overline{a}+3\overline{b} \right)\times \left( 3\overline{a}+\overline{b} \right) \right\}}^{2}}$
Here, we will do cross product of the above equation as:
$\Rightarrow {{\left( \overline{a}\times 3\overline{a}+3\overline{b}\times 3\overline{a}+\overline{a}\times \overline{b}+3\overline{b}\times \overline{b} \right)}^{2}}$
We can write the multiplication of numbers in some terms as:
$\Rightarrow {{\left[ 3\left( \overline{a}\times \overline{a} \right)+9\left( \overline{b}\times \overline{a} \right)+\left( \overline{a}\times \overline{b} \right)+3\left( \overline{b}\times \overline{b} \right) \right]}^{2}}$
Now, we will expand the cross product into dot product with the use of the formula $\left( \overline{a}\times \overline{b} \right)=\left| \overline{a} \right|.\left| \overline{b} \right|\sin \left( \overline{a},\overline{b} \right)$ as:
\[\Rightarrow {{\left[ 3\left( \left| \overline{a} \right|.\left| \overline{a} \right|\sin \left( \overline{a},\overline{a} \right) \right)+9\left( \left| \overline{b} \right|.\left| \overline{a} \right|\sin \left( \overline{b},\overline{a} \right) \right)+\left( \left| \overline{a} \right|.\left| \overline{b} \right|\sin \left( \overline{a},\overline{b} \right) \right)+3\left( \left| \overline{b} \right|.\left| \overline{b} \right|\sin \left( \overline{b},\overline{b} \right) \right) \right]}^{2}}\]
Since, the angle between two vectors are given already but If the vectors are same, the angle between them is Zero and here, we can use the property of dot product \[\overline{b}.\overline{a}\sin \left( \overline{b},\overline{a} \right)=\overline{a}.\overline{b}\sin \left( \overline{a},\overline{b} \right)\] in the above step of the question as:
\[\Rightarrow {{\left[ 3\left( \left| \overline{a} \right|.\left| \overline{a} \right|\sin 0{}^\circ \right)+9\left( \left| \overline{a} \right|.\left| \overline{b} \right|\sin \left( \overline{b},\overline{a} \right) \right)+\left( \left| \overline{a} \right|.\left| \overline{b} \right|\sin \left( \overline{a},\overline{b} \right) \right)+3\left( \left| \overline{b} \right|.\left| \overline{b} \right|\sin 0{}^\circ \right) \right]}^{2}}\]
As we know that $\sin 0{}^\circ =0$ , we will have the above step as:
\[\Rightarrow {{\left[ 3.0+9\left( \left| \overline{a} \right|.\left| \overline{b} \right|\sin \dfrac{2\pi }{3} \right)+\left( \left| \overline{a} \right|.\left| \overline{b} \right|\sin \dfrac{2\pi }{3} \right)+3.0 \right]}^{2}}\]
We can write the above step below as:
\[\Rightarrow {{\left[ 10\left( \left| \overline{a} \right|.\left| \overline{b} \right|\sin \dfrac{2\pi }{3} \right) \right]}^{2}}\]
Here, we already have the given value of the two vectors and the value of \[\sin \dfrac{2\pi }{3}\] is equal to \[\dfrac{\sqrt{3}}{2}\] . So, above equation will be as:
\[\Rightarrow {{\left[ 10\left( 1\times 2\times \dfrac{\sqrt{3}}{2} \right) \right]}^{2}}\]
Since, here the equal like number will be cancel out. Then the above equation can be written as below:
\[\Rightarrow {{\left[ 10\sqrt{3} \right]}^{2}}\]
Now, we will square the above term and will get:
\[\Rightarrow 100\times 3\]
The product of the above term will be:
\[\Rightarrow 300\]

Hence, after solving the question ${{\left\{ \left( \overline{a}+3\overline{b} \right)\times \left( 3\overline{a}+\overline{b} \right) \right\}}^{2}}$ , we got \[300\]

Note: Since, the vector and dot product of two vectors are differently written. Here is formula of dot product that is $\left( \overline{a}.\overline{b} \right)=\left| \overline{a} \right|.\left| \overline{b} \right|\cos \left( \overline{a},\overline{b} \right)$ and the vector product is$\left( \overline{a}\times \overline{b} \right)=\left| \overline{a} \right|.\left| \overline{b} \right|\sin \left( \overline{a},\overline{b} \right)$ , Where, $\left| \overline{a} \right|$ and $\left| \overline{b} \right|$ are magnitude of $\overline{a}$ and $\overline{b}$ respectively and $\left( \overline{a},\overline{b} \right)$ denotes angle between $\overline{a}$ and $\overline{b}$ .