Kinematics 6 Question 1
1. Let $\left|\mathbf{A} _1\right|=3,\left|\mathbf{A} _2\right|=5$ and $\left|\mathbf{A} _1+\mathbf{A} _2\right|=5$. The value of $\left(2 \mathbf{A} _1+3 \mathbf{A} _2\right) \cdot\left(3 \mathbf{A} _1-2 \mathbf{A} _2\right)$ is
(a) -106.5
(b) -112.5
(c) $\rightarrow 9.5$
(d) -118.5
(2019 Main, 08 April II)
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Answer:
Correct Answer: 1. (d)
Solution:
- For vector $\mathbf{A} _1+\mathbf{A} _2$, we have
$$ \begin{aligned} & \left|\mathbf{A} _1+\mathbf{A} _2\right|^{2}=\left(\mathbf{A} _1+\mathbf{A} _2\right) \cdot\left(\mathbf{A} _1+\mathbf{A} _2\right) \quad\left[\because \mathbf{x} \cdot \mathbf{x}=|\mathbf{x}|^{2}\right] \\ \Rightarrow & \left|\mathbf{A} _1+\mathbf{A} _2\right|^{2}=\left|\mathbf{A} _1\right|^{2}+\left|\mathbf{A} _2\right|^{2}+2 \mathbf{A} _1 \cdot \mathbf{A} _2 \end{aligned} $$
Given, $\left|\mathbf{A} _1\right|=3,\left|\mathbf{A} _2\right|=5$ and $\left|\mathbf{A}+\mathbf{A} _2\right|=5$
So, we have
$$ \begin{aligned} & \qquad \begin{aligned} (5)^{2} & =9+25+2 \mathbf{A} _1 \cdot \mathbf{A} _2 \Rightarrow \mathbf{A} _1 \cdot \mathbf{A} _2=-\frac{9}{2} \\ \text { Now, }\left(2 \mathbf{A} _1\right. & \left.+3 \mathbf{A} _2\right) \cdot\left(3 \mathbf{A} _1-2 \mathbf{A} _2\right) \\ & =6\left|\mathbf{A} _1\right|^{2}-4 \mathbf{A} _1 \cdot \mathbf{A} _2+9 \mathbf{A} _1 \cdot \mathbf{A} _2-6\left|\mathbf{A} _2\right|^{2} \\ =6\left|\mathbf{A} _1\right|^{2} & -6\left|\mathbf{A} _2\right|^{2}+5 \mathbf{A} _1 \cdot \mathbf{A} _2 \end{aligned} \end{aligned} $$
Substituting values, we have
$$ \begin{aligned} & \left(2 \mathbf{A} _1+3 \mathbf{A} _2\right) \cdot\left(3 \mathbf{A} _1-2 \mathbf{A} _2\right) \\ = & 6(9)-6(25)+5-\frac{9}{2}=-118.5 \end{aligned} $$
Alternate Solution
As we know, $\left|\mathbf{A} _1+\mathbf{A} _2\right|$ can also be written as
$$ \left|\mathbf{A}+\mathbf{A} _2\right|=\sqrt{\left|\mathbf{A} _1\right|^{2}+\left|\mathbf{A} _2\right|^{2}+2\left|\mathbf{A} _1\right|\left|\mathbf{A} _2\right| \cos \theta} $$
Substituting the given values, we get
$$ \begin{gathered} 5=\sqrt{(3)^{2}+(5)^{2}+2 \times 3 \times 5 \cos \theta} \\ \text { or } \cos \theta=-\frac{9}{2 \times 3 \times 5}=-\frac{3}{10} \\ \text { So, }\left(2 \mathbf{A} _1+3 \mathbf{A} _2\right) \cdot\left(3 \mathbf{A} _1-2 \mathbf{A} _2\right) \\ =6\left|\mathbf{A} _1\right|^{2}-6\left|\mathbf{A} _2\right|^{2}+5 \mathbf{A} _1 \cdot \mathbf{A} _2 \\ =6\left|\mathbf{A} _1\right|^{2}-6\left|\mathbf{A} _2\right|^{2}+5\left|\mathbf{A} _1 | \mathbf{A} _2\right| \cos \theta \\ =6 \times 9-6 \times 25+5 \times 2 \times 3 \times \frac{-3}{10}=-118.5 \end{gathered} $$
