SATHEE: Kinematics 6 Question 1

Kinematics 6 Question 1

1. Let $| A_{1} | = 3, | A_{2} | = 5$ and $| A_{1} + A_{2} | = 5$ . The value of $(2 A_{1} + 3 A_{2}) \cdot (3 A_{1} - 2 A_{2})$ is

(a) -106.5

(b) -112.5

(d) -118.5

(2019 Main, 08 April II)

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Answer:

Correct Answer: 1. (d)

Solution:

For vector $A_{1} + A_{2}$ , we have

$\begin{aligned} {| A_{1} + A_{2} |}^{2} = (A_{1} + A_{2}) \cdot (A_{1} + A_{2}) [∵ x \cdot x = | x |^{2}] \\ \Rightarrow & {| A_{1} + A_{2} |}^{2} = {| A_{1} |}^{2} + {| A_{2} |}^{2} + 2 A_{1} \cdot A_{2} \end{aligned}$

Given, $| A_{1} | = 3, | A_{2} | = 5$ and $| A + A_{2} | = 5$

So, we have

$\begin{aligned} \begin{aligned} (5)^{2} & = 9 + 25 + 2 A_{1} \cdot A_{2} \Rightarrow A_{1} \cdot A_{2} = - \frac{9}{2} \\ Now, (2 A_{1} & + 3 A_{2}) \cdot (3 A_{1} - 2 A_{2}) \\ = 6 {| A_{1} |}^{2} - 4 A_{1} \cdot A_{2} + 9 A_{1} \cdot A_{2} - 6 {| A_{2} |}^{2} \\ = 6 {| A_{1} |}^{2} & - 6 {| A_{2} |}^{2} + 5 A_{1} \cdot A_{2} \end{aligned} \end{aligned}$

Substituting values, we have

$\begin{aligned} (2 A_{1} + 3 A_{2}) \cdot (3 A_{1} - 2 A_{2}) \\ = & 6 (9) - 6 (25) + 5 - \frac{9}{2} = - 118.5 \end{aligned}$

Alternate Solution

As we know, $| A_{1} + A_{2} |$ can also be written as

$| A + A_{2} | = \sqrt{{| A_{1} |}^{2} + {| A_{2} |}^{2} + 2 | A_{1} | | A_{2} | \cos θ}$

Substituting the given values, we get

$\begin{matrix} 5 = \sqrt{(3)^{2} + (5)^{2} + 2 \times 3 \times 5 \cos θ} \\ or \cos θ = - \frac{9}{2 \times 3 \times 5} = - \frac{3}{10} \\ So, (2 A_{1} + 3 A_{2}) \cdot (3 A_{1} - 2 A_{2}) \\ = 6 {| A_{1} |}^{2} - 6 {| A_{2} |}^{2} + 5 A_{1} \cdot A_{2} \\ = 6 {| A_{1} |}^{2} - 6 {| A_{2} |}^{2} + 5 | A_{1} | A_{2} | \cos θ \\ = 6 \times 9 - 6 \times 25 + 5 \times 2 \times 3 \times \frac{- 3}{10} = - 118.5 \end{matrix}$