Chapter 9: Q.13 (page 228)

What is the angle $θ$ between vectors $\vec{A}$ and $\vec{B}$ in each part of $E x e r c i s e 12 ?$

Short Answer

Expert verified

a. $θ = 125^{0}$

b. $θ = 68^{0}$

Step by step solution

Given information (part a)

$\vec{A} = 3 \hat{i} + 4 \hat{j}$ and $\vec{B} = 2 \hat{i} - 6 \hat{j}$

Explanation (part a)

The angle between $\vec{A}$ and $\vec{B}$ , we have

$θ = \cos^{- 1} (\frac{\vec{A} . \vec{B}}{|\vec{A}| |\vec{B}|})$

we know that

role="math" localid="1649877276481" $\vec{A} \cdot \vec{B} = A_{x} B_{x} + A_{y} B_{y} = (3 \hat{i}) (2 \hat{i}) + (4 \hat{j}) (- 6 \hat{j}) = 6 {\hat{i}}^{2} - 24 {\hat{j}}^{2} = 6 - 24 \vec{A} \cdot \vec{B} = - 18 |\vec{A}| = \sqrt{9 + 16} = \sqrt{25} = 5 |\vec{B}| = \sqrt{4 + 36} = \sqrt{40} = 2 \sqrt{10}$

Substitute the values of $\vec{A} . \vec{B}, |\vec{A}| a n d |\vec{B}|$ in the above equation, we get

$θ = \cos^{- 1} (\frac{- 18}{5 \times 2 \sqrt{10}}) = \cos^{- 1} (\frac{- 9}{5 \sqrt{10}}) = \cos^{- 1} (- 0.569) = 125^{0}$

Given information (part b)

$\vec{A} = 3 \hat{i} - 2 \hat{j}$

$\vec{B} = 6 \hat{i} + 4 \hat{j}$

Explanation (part b)

The angle between $\vec{A}$ and $\vec{B}$ , we have

$θ = \cos^{- 1} (\frac{\vec{A} . \vec{B}}{|\vec{A}| |\vec{B}|})$

we know that

$\vec{A} \cdot \vec{B} = A_{x} B_{x} + A_{y} B_{y} = (3 \hat{i}) (6 \hat{i}) + (- 2 \hat{j}) (4 \hat{j}) = 18 {\hat{i}}^{2} - 8 {\hat{j}}^{2} = 18 - 8 \vec{A} \cdot \vec{B} = 10 |\vec{A}| = \sqrt{9 + 4} = \sqrt{13} |\vec{B}| = \sqrt{36 + 16} = \sqrt{52} = 2 \sqrt{13}$

Substitute the values of $\vec{A} . \vec{B}, |\vec{A}| a n d |\vec{B}|$ in the above equation, we get

$θ = \cos^{- 1} (\frac{10}{\sqrt{13} \times 2 \sqrt{13}}) = \cos^{- 1} (\frac{10}{26}) = \cos^{- 1} (0.384) = 68^{0}$

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What is the angle $θ$ between vectors $\vec{A}$ and $\vec{B}$ in each part of $E x e r c i s e 12 ?$

Short Answer

Step by step solution

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Explanation (part b)

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