Chapter 1: Q 42E (page 29)

(I)(a) Find the scalar product of the vectors $\vec{A}$ and $\vec{B}$ given in Exercise 1.38. (b) Find the angle between these two vectors.

Short Answer

Expert verified

The angle between these two vector is $1.43 °$ .

Step by step solution

Given data

$\begin{array}{l} \vec{A} = 4.00 \hat{i} + 7.00 \hat{j} \\ \vec{B} = 5.00 \hat{i} - 2.00 \hat{j} \end{array}$

The scalar product of vector

$a . b = (4.00 \hat{i} + 7.00 \hat{j}) . (5.00 \hat{i} - 7200 \hat{j}) = (4 \times 5) + (7 \times (- 2)) = 20 - 14 = 6$

The angle between two vector

The magnitude of vector is,

$\begin{array}{l} |\vec{A}| = \sqrt{4^{2} + 7^{2}} = \sqrt{65} = 8.06 \\ |\vec{B}| = \sqrt{5^{2} + 2^{2}} = \sqrt{29} = 5.3 \end{array}$

The angle between the two vector is,

$θ = \cos^{- 1} \frac{\vec{a} . \vec{b}}{|\vec{a}| |\vec{b}|} = \cos^{- 1} \frac{6}{(8.06) (5.3)} = \cos^{- 1} \frac{6}{(42.87)} = \cos^{- 1} (0.1399) θ = 1.43 °$

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(I)(a) Find the scalar product of the vectors $\vec{A}$ and $\vec{B}$ given in Exercise 1.38. (b) Find the angle between these two vectors.

Short Answer

Step by step solution

Given data

The scalar product of vector

The angle between two vector

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(I)(a) Find the scalar product of the vectorsA→ andB→given in Exercise 1.38. (b) Find the angle between these two vectors.

Short Answer

Step by step solution

Given data

The scalar product of vector

The angle between two vector

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Scientific Method Physics

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Study anywhere. Anytime. Across all devices.

(I)(a) Find the scalar product of the vectors $\vec{A}$ and $\vec{B}$ given in Exercise 1.38. (b) Find the angle between these two vectors.