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PHXI06:WORK ENERGY AND POWER

355551 The angle between the two vectors $\vec{A} = 3 \hat{i} + 4 \hat{j} + 5 \hat{k}$ and $\vec{B} = 3 \hat{i} + 4 \hat{j} + 5 \hat{k}$ is

1

60^{\circ}

2 Zero

3

90^{\circ}

4 None of these

PHXI06:WORK ENERGY AND POWER

355552 If $\hat{i}, \hat{j}$ and $\hat{k}$ represent unit vectors along the $x, y$ and $z$ - axes respectively, then the angle $θ$ between the vectors $\hat{i} + \hat{j} + \hat{k}$ and $\hat{i} + \hat{j}$ is equal to

1

\sin^{- 1} (\frac{1}{\sqrt{3}})

2

\sin^{- 1} (\sqrt{\frac{2}{3}})

3

\cos^{- 1} (\frac{1}{\sqrt{3}})

4

90^{\circ}

Explanation:

Let the given vectors be
$\begin{matrix} \vec{A} = \hat{i} + \hat{j} + \hat{k}; \vec{B} = \hat{i} + \hat{j} \\ ∴ A = \sqrt{1^{2} + 1^{2} + 1^{2}} = \sqrt{3}, B = \sqrt{1^{2} + 1^{2}} = \sqrt{2} \\ \vec{A} \cdot \vec{B} = 2 \\ \cos θ = \frac{\vec{A} \cdot \vec{B}}{A B} = \frac{2}{\sqrt{3} \sqrt{2}} = (\sqrt{\frac{2}{3}}) \\ \sin θ = \sqrt{1 - \cos^{2} θ} = \sqrt{1 - \frac{2}{3}} = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}} \\ θ = \sin^{- 1} (\frac{1}{\sqrt{3}}) \end{matrix}$

PHXI06:WORK ENERGY AND POWER

355553 Vectors $\vec{A}$ and $\vec{B}$ include an angle $θ$ between them.If $(\vec{A} + \vec{B})$ and $(\vec{A} - \vec{B})$ respectively subtend angles $α$ and $β$ with $A$ , then $(\tan α + \tan β)$ is

1

\frac{(A B \sin θ)}{(A^{2} + B^{2} \cos^{2} θ)}

2

\frac{(2 A B \sin θ)}{(A^{2} - B^{2} \cos^{2} θ)}

3

\frac{(A^{2} \sin^{2} θ)}{(A^{2} + B^{2} \cos^{2} θ)}

4

\frac{(B^{2} \sin^{2} θ)}{(A^{2} - B^{2} \cos^{2} θ)}

Explanation:

The vector diagram is
supporting img
$\tan α = (\frac{B \sin θ}{A + B \cos θ}) (1)$
Where $α$ is the angle made by the vector $(\vec{A} + \vec{B})$ with $\vec{A}$ .
Similarly, $\tan β = \frac{B \sin θ}{A - B \cos θ} (2)$
Where $β$ is the angle made by the vector $(\vec{A} - \vec{B})$ with $\vec{A}$ .
Note that the angle between $\vec{A}$ and $(- \vec{B})$ is $(180^{\circ} - θ) .$ Adding (1) and (2), we get
$\tan α + \tan β = \frac{B \sin θ}{A + B \cos θ} + \frac{B \sin θ}{A - B \cos θ}$
$= \frac{2 A B \sin θ}{(A^{2} - B^{2} \cos^{2} θ)}$

PHXI06:WORK ENERGY AND POWER

355554 $\vec{A} = 3 \hat{i} - \hat{j} + 7 \hat{k}$ and $\vec{B} = 5 \hat{i} - \hat{j} + 9 \hat{k}$ . The direction cosine of the vector $\vec{A} + \vec{B}$ with $x$ -axis is

1

\frac{3}{\sqrt{31}}

2

\frac{5}{\sqrt{324}}

3 5

4

\frac{8}{\sqrt{324}}

Explanation:

Let $\vec{C} = \vec{A} + \vec{B} = 3 \hat{i} - \hat{j} + 7 \hat{k} + 5 \hat{i} - \hat{j} + 9 \hat{k}$
$\vec{C} = \vec{A} + \vec{B} = 8 \hat{i} - 2 \hat{j} + 16 \hat{k}$
The direction cosine is $\cos θ = \frac{(8 \hat{i} - 2 \hat{j} + 16 \hat{k}) \cdot \hat{i}}{\sqrt{8^{2} + {(2)}^{2} + {(16)}^{2}}}$
$= \frac{8}{\sqrt{324}}$

PHXI06:WORK ENERGY AND POWER

355551 The angle between the two vectors $\vec{A} = 3 \hat{i} + 4 \hat{j} + 5 \hat{k}$ and $\vec{B} = 3 \hat{i} + 4 \hat{j} + 5 \hat{k}$ is

1

60^{\circ}

2 Zero

3

90^{\circ}

4 None of these

PHXI06:WORK ENERGY AND POWER

355552 If $\hat{i}, \hat{j}$ and $\hat{k}$ represent unit vectors along the $x, y$ and $z$ - axes respectively, then the angle $θ$ between the vectors $\hat{i} + \hat{j} + \hat{k}$ and $\hat{i} + \hat{j}$ is equal to

1

\sin^{- 1} (\frac{1}{\sqrt{3}})

2

\sin^{- 1} (\sqrt{\frac{2}{3}})

3

\cos^{- 1} (\frac{1}{\sqrt{3}})

4

90^{\circ}

Explanation:

Let the given vectors be
$\begin{matrix} \vec{A} = \hat{i} + \hat{j} + \hat{k}; \vec{B} = \hat{i} + \hat{j} \\ ∴ A = \sqrt{1^{2} + 1^{2} + 1^{2}} = \sqrt{3}, B = \sqrt{1^{2} + 1^{2}} = \sqrt{2} \\ \vec{A} \cdot \vec{B} = 2 \\ \cos θ = \frac{\vec{A} \cdot \vec{B}}{A B} = \frac{2}{\sqrt{3} \sqrt{2}} = (\sqrt{\frac{2}{3}}) \\ \sin θ = \sqrt{1 - \cos^{2} θ} = \sqrt{1 - \frac{2}{3}} = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}} \\ θ = \sin^{- 1} (\frac{1}{\sqrt{3}}) \end{matrix}$

PHXI06:WORK ENERGY AND POWER

355553 Vectors $\vec{A}$ and $\vec{B}$ include an angle $θ$ between them.If $(\vec{A} + \vec{B})$ and $(\vec{A} - \vec{B})$ respectively subtend angles $α$ and $β$ with $A$ , then $(\tan α + \tan β)$ is

1

\frac{(A B \sin θ)}{(A^{2} + B^{2} \cos^{2} θ)}

2

\frac{(2 A B \sin θ)}{(A^{2} - B^{2} \cos^{2} θ)}

3

\frac{(A^{2} \sin^{2} θ)}{(A^{2} + B^{2} \cos^{2} θ)}

4

\frac{(B^{2} \sin^{2} θ)}{(A^{2} - B^{2} \cos^{2} θ)}

Explanation:

The vector diagram is
supporting img
$\tan α = (\frac{B \sin θ}{A + B \cos θ}) (1)$
Where $α$ is the angle made by the vector $(\vec{A} + \vec{B})$ with $\vec{A}$ .
Similarly, $\tan β = \frac{B \sin θ}{A - B \cos θ} (2)$
Where $β$ is the angle made by the vector $(\vec{A} - \vec{B})$ with $\vec{A}$ .
Note that the angle between $\vec{A}$ and $(- \vec{B})$ is $(180^{\circ} - θ) .$ Adding (1) and (2), we get
$\tan α + \tan β = \frac{B \sin θ}{A + B \cos θ} + \frac{B \sin θ}{A - B \cos θ}$
$= \frac{2 A B \sin θ}{(A^{2} - B^{2} \cos^{2} θ)}$

PHXI06:WORK ENERGY AND POWER

355554 $\vec{A} = 3 \hat{i} - \hat{j} + 7 \hat{k}$ and $\vec{B} = 5 \hat{i} - \hat{j} + 9 \hat{k}$ . The direction cosine of the vector $\vec{A} + \vec{B}$ with $x$ -axis is

1

\frac{3}{\sqrt{31}}

2

\frac{5}{\sqrt{324}}

3 5

4

\frac{8}{\sqrt{324}}

Explanation:

Let $\vec{C} = \vec{A} + \vec{B} = 3 \hat{i} - \hat{j} + 7 \hat{k} + 5 \hat{i} - \hat{j} + 9 \hat{k}$
$\vec{C} = \vec{A} + \vec{B} = 8 \hat{i} - 2 \hat{j} + 16 \hat{k}$
The direction cosine is $\cos θ = \frac{(8 \hat{i} - 2 \hat{j} + 16 \hat{k}) \cdot \hat{i}}{\sqrt{8^{2} + {(2)}^{2} + {(16)}^{2}}}$
$= \frac{8}{\sqrt{324}}$

PHXI06:WORK ENERGY AND POWER

355551 The angle between the two vectors $\vec{A} = 3 \hat{i} + 4 \hat{j} + 5 \hat{k}$ and $\vec{B} = 3 \hat{i} + 4 \hat{j} + 5 \hat{k}$ is

1

60^{\circ}

2 Zero

3

90^{\circ}

4 None of these

PHXI06:WORK ENERGY AND POWER

355552 If $\hat{i}, \hat{j}$ and $\hat{k}$ represent unit vectors along the $x, y$ and $z$ - axes respectively, then the angle $θ$ between the vectors $\hat{i} + \hat{j} + \hat{k}$ and $\hat{i} + \hat{j}$ is equal to

1

\sin^{- 1} (\frac{1}{\sqrt{3}})

2

\sin^{- 1} (\sqrt{\frac{2}{3}})

3

\cos^{- 1} (\frac{1}{\sqrt{3}})

4

90^{\circ}

Explanation:

Let the given vectors be
$\begin{matrix} \vec{A} = \hat{i} + \hat{j} + \hat{k}; \vec{B} = \hat{i} + \hat{j} \\ ∴ A = \sqrt{1^{2} + 1^{2} + 1^{2}} = \sqrt{3}, B = \sqrt{1^{2} + 1^{2}} = \sqrt{2} \\ \vec{A} \cdot \vec{B} = 2 \\ \cos θ = \frac{\vec{A} \cdot \vec{B}}{A B} = \frac{2}{\sqrt{3} \sqrt{2}} = (\sqrt{\frac{2}{3}}) \\ \sin θ = \sqrt{1 - \cos^{2} θ} = \sqrt{1 - \frac{2}{3}} = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}} \\ θ = \sin^{- 1} (\frac{1}{\sqrt{3}}) \end{matrix}$

PHXI06:WORK ENERGY AND POWER

355553 Vectors $\vec{A}$ and $\vec{B}$ include an angle $θ$ between them.If $(\vec{A} + \vec{B})$ and $(\vec{A} - \vec{B})$ respectively subtend angles $α$ and $β$ with $A$ , then $(\tan α + \tan β)$ is

1

\frac{(A B \sin θ)}{(A^{2} + B^{2} \cos^{2} θ)}

2

\frac{(2 A B \sin θ)}{(A^{2} - B^{2} \cos^{2} θ)}

3

\frac{(A^{2} \sin^{2} θ)}{(A^{2} + B^{2} \cos^{2} θ)}

4

\frac{(B^{2} \sin^{2} θ)}{(A^{2} - B^{2} \cos^{2} θ)}

Explanation:

The vector diagram is
supporting img
$\tan α = (\frac{B \sin θ}{A + B \cos θ}) (1)$
Where $α$ is the angle made by the vector $(\vec{A} + \vec{B})$ with $\vec{A}$ .
Similarly, $\tan β = \frac{B \sin θ}{A - B \cos θ} (2)$
Where $β$ is the angle made by the vector $(\vec{A} - \vec{B})$ with $\vec{A}$ .
Note that the angle between $\vec{A}$ and $(- \vec{B})$ is $(180^{\circ} - θ) .$ Adding (1) and (2), we get
$\tan α + \tan β = \frac{B \sin θ}{A + B \cos θ} + \frac{B \sin θ}{A - B \cos θ}$
$= \frac{2 A B \sin θ}{(A^{2} - B^{2} \cos^{2} θ)}$

PHXI06:WORK ENERGY AND POWER

355554 $\vec{A} = 3 \hat{i} - \hat{j} + 7 \hat{k}$ and $\vec{B} = 5 \hat{i} - \hat{j} + 9 \hat{k}$ . The direction cosine of the vector $\vec{A} + \vec{B}$ with $x$ -axis is

1

\frac{3}{\sqrt{31}}

2

\frac{5}{\sqrt{324}}

3 5

4

\frac{8}{\sqrt{324}}

Explanation:

Let $\vec{C} = \vec{A} + \vec{B} = 3 \hat{i} - \hat{j} + 7 \hat{k} + 5 \hat{i} - \hat{j} + 9 \hat{k}$
$\vec{C} = \vec{A} + \vec{B} = 8 \hat{i} - 2 \hat{j} + 16 \hat{k}$
The direction cosine is $\cos θ = \frac{(8 \hat{i} - 2 \hat{j} + 16 \hat{k}) \cdot \hat{i}}{\sqrt{8^{2} + {(2)}^{2} + {(16)}^{2}}}$
$= \frac{8}{\sqrt{324}}$

PHXI06:WORK ENERGY AND POWER

355551 The angle between the two vectors $\vec{A} = 3 \hat{i} + 4 \hat{j} + 5 \hat{k}$ and $\vec{B} = 3 \hat{i} + 4 \hat{j} + 5 \hat{k}$ is

1

60^{\circ}

2 Zero

3

90^{\circ}

4 None of these

PHXI06:WORK ENERGY AND POWER

355552 If $\hat{i}, \hat{j}$ and $\hat{k}$ represent unit vectors along the $x, y$ and $z$ - axes respectively, then the angle $θ$ between the vectors $\hat{i} + \hat{j} + \hat{k}$ and $\hat{i} + \hat{j}$ is equal to

1

\sin^{- 1} (\frac{1}{\sqrt{3}})

2

\sin^{- 1} (\sqrt{\frac{2}{3}})

3

\cos^{- 1} (\frac{1}{\sqrt{3}})

4

90^{\circ}

Explanation:

Let the given vectors be
$\begin{matrix} \vec{A} = \hat{i} + \hat{j} + \hat{k}; \vec{B} = \hat{i} + \hat{j} \\ ∴ A = \sqrt{1^{2} + 1^{2} + 1^{2}} = \sqrt{3}, B = \sqrt{1^{2} + 1^{2}} = \sqrt{2} \\ \vec{A} \cdot \vec{B} = 2 \\ \cos θ = \frac{\vec{A} \cdot \vec{B}}{A B} = \frac{2}{\sqrt{3} \sqrt{2}} = (\sqrt{\frac{2}{3}}) \\ \sin θ = \sqrt{1 - \cos^{2} θ} = \sqrt{1 - \frac{2}{3}} = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}} \\ θ = \sin^{- 1} (\frac{1}{\sqrt{3}}) \end{matrix}$

PHXI06:WORK ENERGY AND POWER

355553 Vectors $\vec{A}$ and $\vec{B}$ include an angle $θ$ between them.If $(\vec{A} + \vec{B})$ and $(\vec{A} - \vec{B})$ respectively subtend angles $α$ and $β$ with $A$ , then $(\tan α + \tan β)$ is

1

\frac{(A B \sin θ)}{(A^{2} + B^{2} \cos^{2} θ)}

2

\frac{(2 A B \sin θ)}{(A^{2} - B^{2} \cos^{2} θ)}

3

\frac{(A^{2} \sin^{2} θ)}{(A^{2} + B^{2} \cos^{2} θ)}

4

\frac{(B^{2} \sin^{2} θ)}{(A^{2} - B^{2} \cos^{2} θ)}

Explanation:

The vector diagram is
supporting img
$\tan α = (\frac{B \sin θ}{A + B \cos θ}) (1)$
Where $α$ is the angle made by the vector $(\vec{A} + \vec{B})$ with $\vec{A}$ .
Similarly, $\tan β = \frac{B \sin θ}{A - B \cos θ} (2)$
Where $β$ is the angle made by the vector $(\vec{A} - \vec{B})$ with $\vec{A}$ .
Note that the angle between $\vec{A}$ and $(- \vec{B})$ is $(180^{\circ} - θ) .$ Adding (1) and (2), we get
$\tan α + \tan β = \frac{B \sin θ}{A + B \cos θ} + \frac{B \sin θ}{A - B \cos θ}$
$= \frac{2 A B \sin θ}{(A^{2} - B^{2} \cos^{2} θ)}$

PHXI06:WORK ENERGY AND POWER

355554 $\vec{A} = 3 \hat{i} - \hat{j} + 7 \hat{k}$ and $\vec{B} = 5 \hat{i} - \hat{j} + 9 \hat{k}$ . The direction cosine of the vector $\vec{A} + \vec{B}$ with $x$ -axis is

1

\frac{3}{\sqrt{31}}

2

\frac{5}{\sqrt{324}}

3 5

4

\frac{8}{\sqrt{324}}

Explanation:

Let $\vec{C} = \vec{A} + \vec{B} = 3 \hat{i} - \hat{j} + 7 \hat{k} + 5 \hat{i} - \hat{j} + 9 \hat{k}$
$\vec{C} = \vec{A} + \vec{B} = 8 \hat{i} - 2 \hat{j} + 16 \hat{k}$
The direction cosine is $\cos θ = \frac{(8 \hat{i} - 2 \hat{j} + 16 \hat{k}) \cdot \hat{i}}{\sqrt{8^{2} + {(2)}^{2} + {(16)}^{2}}}$
$= \frac{8}{\sqrt{324}}$