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Chapter 14: Problem 5

Two vectors have modulus 10 and 12 . The angle between them is \(\frac{\pi}{3}\). Find their scalar product.

Short Answer

Expert verified
Answer: The scalar product of the two vectors is 60.

Step by step solution

01

Write the scalar product formula for magnitudes and the angle

The scalar product( also called dot product) of two vectors can be found using magnitudes and the angle between them. The formula is: $$ \textbf{A} \cdot \textbf{B} = \| \textbf{A} \| \, \| \textbf{B} \| \cos \theta $$ where \(\textbf{A}\) and \(\textbf{B}\) are the two vectors, \(\| \textbf{A} \|\) and \(\| \textbf{B} \|\) are their magnitudes, and \(\theta\) is the angle between the two vectors.
02

Plug in the given values

We are given that \(\| \textbf{A} \| = 10\) and \(\| \textbf{B} \| = 12\). The angle between them is \(\theta = \frac{\pi}{3}\). Substituting these values into the formula, we get: $$ \textbf{A} \cdot \textbf{B} = 10 \times 12 \times \cos \frac{\pi}{3} $$
03

Calculate the scalar product

The cosine of \(\frac{\pi}{3}\) is \(\frac{1}{2}\). So, our scalar product is: $$ \textbf{A} \cdot \textbf{B} = 10 \times 12 \times \frac{1}{2} $$ Multiplying, we get: $$ \textbf{A} \cdot \textbf{B} = 60 $$ In conclusion, the scalar product of the two vectors is 60.

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Most popular questions from this chapter

State the coordinates of the point \(\mathrm{P}\) if its position vector is given as (a) \(3 \boldsymbol{i}-7 \boldsymbol{j}\), (b) \(-4 \boldsymbol{i}\), (c) \(-0.5 \boldsymbol{i}+13 \boldsymbol{j}\), (d) \(a \boldsymbol{i}+b \boldsymbol{j}\).

Explain why, for arbitrary vectors, \(\boldsymbol{p} \times \boldsymbol{q}\) is not equal to \(q \times p\).

Show on a diagram three arbitrary vectors \(\boldsymbol{p}, \boldsymbol{q}\) and \(\boldsymbol{r}\). Using the triangle law of addition verify the associativity rule $$ p+(q+\boldsymbol{r})=(p+q)+\boldsymbol{r} $$

Given three vectors \(\boldsymbol{a}, \boldsymbol{b}\) and \(\boldsymbol{c}\), their triple vector product is defined to be \((\boldsymbol{a} \times \boldsymbol{b}) \times \boldsymbol{c}\). For the vectors \(\boldsymbol{a}=4 i+2 \boldsymbol{j}+\boldsymbol{k}\) \(b=2 i-j+7 k\) and \(c=2 i-2 j+3 k\) verify that $$ (a \times b) \times c=(a \cdot c) b-(b \cdot c) a $$

Find the work done by a force of magnitude 10 newtons acting in the direction of the vector \(3 \boldsymbol{i}+\boldsymbol{j}+8 \boldsymbol{k}\) if it moves a particle from the point \((1,1,1)\) to the point \((3,1,2)\).
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