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Chapter 1: Problem 47

Given \(\left|\overrightarrow{A_{1}}\right|=2,\left|\overrightarrow{A_{2}}\right|=3\) and \(\left|\overrightarrow{A_{1}}+\overrightarrow{A_{2}}\right|=3\). Find the value of \(\left(\overrightarrow{A_{1}}+2 \overrightarrow{A_{2}}\right) \cdot\left(3 \overrightarrow{A_{1}}-4 \overrightarrow{A_{2}}\right)\) (A) \(-64\) (B) 60 (C) \(-60\) (D) 64

Short Answer

Expert verified
The value of \(\left(\overrightarrow{A_{1}}+2 \overrightarrow{A_{2}}\right)\cdot\left(3 \overrightarrow{A_{1}}-4 \overrightarrow{A_{2}}\right) = -64\), which corresponds to the answer choice (A).

Step by step solution

01

Write down the information given

We are given the magnitudes of \(\overrightarrow{A_{1}}\), \(\overrightarrow{A_{2}}\), and \(\overrightarrow{A_{1}} + \overrightarrow{A_{2}}\) as follows: \(\left|\overrightarrow{A_{1}}\right| = 2\) \(\left|\overrightarrow{A_{2}}\right| = 3\) \(\left|\overrightarrow{A_{1}}+\overrightarrow{A_{2}}\right| = 3\)
02

Use the dot product property to find the dot product

Recall that \(\left|\overrightarrow{A}+\overrightarrow{B}\right|^2 = \left|\overrightarrow{A}\right|^2 + 2(\overrightarrow{A}\cdot\overrightarrow{B}) + \left|\overrightarrow{B}\right|^2\). We can plug in the given magnitudes and solve for \(\overrightarrow{A_{1}}\cdot\overrightarrow{A_{2}}\): \(3^2 = 2^2 + 2(\overrightarrow{A_{1}}\cdot\overrightarrow{A_{2}}) + 3^2\rightarrow 9=4+2(\overrightarrow{A_{1}}\cdot\overrightarrow{A_{2}}) +9\) Solving for \(\overrightarrow{A_{1}}\cdot\overrightarrow{A_{2}}\): \(2(\overrightarrow{A_{1}}\cdot\overrightarrow{A_{2}}) = -4\) \(\overrightarrow{A_{1}}\cdot\overrightarrow{A_{2}} = -2\)
03

Use the dot product of linear combinations to find the desired value

We need to find the value of \(\left(\overrightarrow{A_{1}}+2 \overrightarrow{A_{2}}\right)\cdot\left(3 \overrightarrow{A_{1}}-4 \overrightarrow{A_{2}}\right)\). Using the dot product of linear combinations property, we get: \(\left(\overrightarrow{A_{1}}+2 \overrightarrow{A_{2}}\right)\cdot\left(3 \overrightarrow{A_{1}}-4 \overrightarrow{A_{2}}\right) = (1\cdot3)\left(\overrightarrow{A_{1}}\cdot\overrightarrow{A_{1}}\right)+ (1\cdot-4)\left(\overrightarrow{A_{1}}\cdot\overrightarrow{A_{2}}\right)+(2\cdot3)\left(\overrightarrow{A_{2}}\cdot\overrightarrow{A_{1}}\right)+ (2\cdot-4)\left(\overrightarrow{A_{2}}\cdot\overrightarrow{A_{2}}\right)\) Since \(\overrightarrow{A_{1}}\cdot\overrightarrow{A_{1}} = \left|\overrightarrow{A_{1}}\right|^2 = 4\) and \(\overrightarrow{A_{2}}\cdot\overrightarrow{A_{2}} = \left|\overrightarrow{A_{2}}\right|^2 = 9\), we substitute these values, along with the value of \(\overrightarrow{A_{1}}\cdot\overrightarrow{A_{2}}\): \((1\cdot3)(4)+ (1\cdot-4)(-2)+(2\cdot3)(-2)+ (2\cdot-4)(9) = 12 + 8 - 12 - 72\) Now, calculate the final value: \(12 + 8 - 12 - 72 = -64\) Therefore, the value of \(\left(\overrightarrow{A_{1}}+2 \overrightarrow{A_{2}}\right)\cdot\left(3 \overrightarrow{A_{1}}-4 \overrightarrow{A_{2}}\right) = -64\), which corresponds to the answer choice (A).

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Most popular questions from this chapter

The pitch of a screw gauge is \(1 \mathrm{~mm}\) and there are 100 divisions on its circular scale. When nothing is put in between its jaws, the zero of the circular scale lies 4 divisions below the reference line. When a steel wire is placed between the jaws, two main scale divisions are clearly visible and 67 divisions on the circular scale are observed. The diameter of the wire is (A) \(2.71 \mathrm{~mm}\) (B) \(2.67 \mathrm{~mm}\) (C) \(2.63 \mathrm{~mm}\) (D) \(2.65 \mathrm{~mm}\)

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