Question

Let $\left| {{{\vec{A}}}_{1}} \right|=3,\left| {{{\vec{A}}}_{2}} \right|=5$ and $\left| {{{\vec{A}}}_{1}}+{{{\vec{A}}}_{2}} \right|=5$. Calculate the value of $\left( 2{{A}_{1}}+3{{A}_{2}} \right)\cdot \left( 3{{A}_{1}}-2{{A}_{2}} \right)$?
​$\begin{aligned} & A.-112.5 \\\ & B.-106.5 \\\ & C.-118.5 \\\ & D.-99.5 \\\ \end{aligned}$

Answer 1

First of square the sum of the vectors and expand the quantity. Calculate the cosine of the angle between the vectors at first. Expand the dot product of the term mentioned in the question. Substitute the values in it. This all will help you in answering the question.

Complete step by step answer:
First of all let us mention what all are given in the question. The first vector is given as,
$\left| {{{\vec{A}}}_{1}} \right|=3$
And the second vector mentioned is,
$\left| {{{\vec{A}}}_{2}} \right|=5$
The sum of these vectors is mentioned in the question as,
$\left| {{{\vec{A}}}_{1}}+{{{\vec{A}}}_{2}} \right|=5$
Let us square this equation first. That is we can write that,
${{\left| {{{\vec{A}}}_{1}}+{{{\vec{A}}}_{2}} \right|}^{2}}={{5}^{2}}$
This equation can be expanded as,
${{\left| {{{\vec{A}}}_{1}}+{{{\vec{A}}}_{2}} \right|}^{2}}={{A}_{1}}^{2}+{{A}_{2}}^{2}+2{{A}_{1}}{{A}_{2}}\cos \alpha ={{5}^{2}}$
Let us substitute the magnitudes of the vectors after squaring them. Hence we can write that,
${{\left| {{{\vec{A}}}_{1}}+{{{\vec{A}}}_{2}} \right|}^{2}}={{3}^{2}}+{{5}^{2}}+2\times 3\times 5\cos \alpha ={{5}^{2}}$
From this, the angle between the vectors can be written as,
$\cos \alpha =-\dfrac{9}{30}=-\dfrac{3}{10}$
The dot product between the terms mentioned in the question can be written as,
$\left( 2{{A}_{1}}+3{{A}_{2}} \right)\cdot \left( 3{{A}_{1}}-2{{A}_{2}} \right)$
Let us perform the dot product now. That is,
$\left( 2{{A}_{1}}+3{{A}_{2}} \right)\cdot \left( 3{{A}_{1}}-2{{A}_{2}} \right)=6{{\left| {{A}_{1}} \right|}^{2}}-4\left| {{A}_{2}} \right|\left| {{A}_{1}} \right|\cos \alpha +9\left| {{A}_{2}} \right|\left| {{A}_{1}} \right|\cos \alpha -6{{\left| {{A}_{2}} \right|}^{2}}$
Now let us substitute the values in it. It will give,
$6\times {{3}^{2}}-4\times 5\times 3\times \dfrac{-3}{10}+9\times 5\times 3\times \dfrac{-3}{10}-6\times {{5}^{2}}$
Let us simplify the obtained equation. This can be written as,
$\left( 2{{A}_{1}}+3{{A}_{2}} \right)\cdot \left( 3{{A}_{1}}-2{{A}_{2}} \right)=54+18-40.5-150=118.5$
This is given as option C. Hence the answer has been calculated.

Note:
The dot product is defined to be the sum of the products of the corresponding terms of the two sequences of numbers. This is the explanation in the algebraic way. It is the product of the Euclidean magnitudes of the two vectors and the cosine of the angle between them when we speak about this geometrically. Euclidean spaces are sometimes explained by using vector spaces in the case of modern geometry. Dot products can be otherwise known as the scalar product.