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PHXI04:MOTION IN A PLANE

361785 Vector $\vec{A}$ makes equal angles with $x$ , $y$ and $z$ axis. Value of its components (in terms of magnitude of $\vec{A}$ ) will be

1

\frac{A}{\sqrt{2}}

2

\frac{A}{\sqrt{3}}

3

\frac{\sqrt{3}}{A}

4

\sqrt{3} A

Explanation:

Let the component of $\vec{A}$ makes angles $α, β a n d γ$ with $x$ , $y$ and $z$ axis respectively then $α = β = γ$
$\cos^{2} α + \cos^{2} β + \cos^{2} γ = 1$
$\Rightarrow 3 \cos^{2} α = 1 \Rightarrow \cos α = \frac{1}{\sqrt{3}}$
$∴ A_{x} = A_{y} = A_{z} = A \cos α = \frac{A}{\sqrt{3}}$

PHXI04:MOTION IN A PLANE

361786 The component of a vector $r$ along $X$ -axis will have maximum value if

1

r

is along positive

Y

-axis

2

r

is along positive

X

-axis

3

r

makes an angle of

45^{\circ}

with the

X

-axis

4

r

is along negative

Y

-axis

Explanation:

If $r$ makes an angle $θ$ with $x$ -axis, then component of $r$ along $x$ -axis. $= r \cos θ$
It will be maximum if $\cos θ = max = 1 \Rightarrow θ = 0^{0},$ i.e., $r$ is along positive $x$ -axis.

NCERT Exemplar

PHXI04:MOTION IN A PLANE

361787 The $X$ and $Y$ components of a force $F$ acting at $30^{\circ}$ to $x$ -axis are respectively

1

\frac{F}{2}, \frac{\sqrt{3}}{2} F

2

\frac{F}{\sqrt{2}}, F

3

F, \frac{F}{\sqrt{2}}

4

\frac{\sqrt{3}}{2} F, \frac{F}{2}

Explanation:

The $X$ component of force $F$ is
$F_{x} = F \cos 30^{\circ} = F \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} F$
The $Y$ component of force $F$ is
$F_{y} = F \sin 30^{\circ} = F \times \frac{1}{2} = \frac{1}{2} F$
supporting img

PHXI04:MOTION IN A PLANE

361788 If $\vec{A} = 3 \hat{i} + 5 \hat{j} - 7 \hat{k}$ , the direction of cosines of the vector $\vec{A}$ are:

1

\frac{2}{\sqrt{83}}, \frac{5}{\sqrt{83}}, \frac{7}{\sqrt{83}}

2

\frac{3}{\sqrt{83}}, \frac{5}{\sqrt{83}}, \frac{7}{\sqrt{83}}

3

\frac{1}{\sqrt{83}}, \frac{2}{\sqrt{83}}, \frac{5}{\sqrt{83}}

4

\frac{3}{\sqrt{83}}, \frac{5}{\sqrt{83}}, \frac{- 7}{\sqrt{83}}

Explanation:

$\vec{A} = 3 \hat{i} + 5 \hat{j} - 7 \hat{k}$
The direction cosines are
$\begin{aligned} \cos α = \frac{A_{x}}{| A |}, \cos β = \frac{A_{y}}{| A |}, \cos γ = \frac{A_{z}}{| A |} \\ | \vec{A} | = \sqrt{{(A_{x})}^{2} + {(A_{y})}^{2} + {(A_{z})}^{2}} \\ = \sqrt{9 + 25 + 49} = \sqrt{83} \end{aligned}$
$\cos α = \frac{3}{\sqrt{83}}, \cos β = \frac{5}{\sqrt{83}}, \cos γ = \frac{- 7}{\sqrt{83}}$

PHXI04:MOTION IN A PLANE

361785 Vector $\vec{A}$ makes equal angles with $x$ , $y$ and $z$ axis. Value of its components (in terms of magnitude of $\vec{A}$ ) will be

1

\frac{A}{\sqrt{2}}

2

\frac{A}{\sqrt{3}}

3

\frac{\sqrt{3}}{A}

4

\sqrt{3} A

Explanation:

Let the component of $\vec{A}$ makes angles $α, β a n d γ$ with $x$ , $y$ and $z$ axis respectively then $α = β = γ$
$\cos^{2} α + \cos^{2} β + \cos^{2} γ = 1$
$\Rightarrow 3 \cos^{2} α = 1 \Rightarrow \cos α = \frac{1}{\sqrt{3}}$
$∴ A_{x} = A_{y} = A_{z} = A \cos α = \frac{A}{\sqrt{3}}$

PHXI04:MOTION IN A PLANE

361786 The component of a vector $r$ along $X$ -axis will have maximum value if

1

r

is along positive

Y

-axis

2

r

is along positive

X

-axis

3

r

makes an angle of

45^{\circ}

with the

X

-axis

4

r

is along negative

Y

-axis

Explanation:

If $r$ makes an angle $θ$ with $x$ -axis, then component of $r$ along $x$ -axis. $= r \cos θ$
It will be maximum if $\cos θ = max = 1 \Rightarrow θ = 0^{0},$ i.e., $r$ is along positive $x$ -axis.

NCERT Exemplar

PHXI04:MOTION IN A PLANE

361787 The $X$ and $Y$ components of a force $F$ acting at $30^{\circ}$ to $x$ -axis are respectively

1

\frac{F}{2}, \frac{\sqrt{3}}{2} F

2

\frac{F}{\sqrt{2}}, F

3

F, \frac{F}{\sqrt{2}}

4

\frac{\sqrt{3}}{2} F, \frac{F}{2}

Explanation:

The $X$ component of force $F$ is
$F_{x} = F \cos 30^{\circ} = F \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} F$
The $Y$ component of force $F$ is
$F_{y} = F \sin 30^{\circ} = F \times \frac{1}{2} = \frac{1}{2} F$
supporting img

PHXI04:MOTION IN A PLANE

361788 If $\vec{A} = 3 \hat{i} + 5 \hat{j} - 7 \hat{k}$ , the direction of cosines of the vector $\vec{A}$ are:

1

\frac{2}{\sqrt{83}}, \frac{5}{\sqrt{83}}, \frac{7}{\sqrt{83}}

2

\frac{3}{\sqrt{83}}, \frac{5}{\sqrt{83}}, \frac{7}{\sqrt{83}}

3

\frac{1}{\sqrt{83}}, \frac{2}{\sqrt{83}}, \frac{5}{\sqrt{83}}

4

\frac{3}{\sqrt{83}}, \frac{5}{\sqrt{83}}, \frac{- 7}{\sqrt{83}}

Explanation:

$\vec{A} = 3 \hat{i} + 5 \hat{j} - 7 \hat{k}$
The direction cosines are
$\begin{aligned} \cos α = \frac{A_{x}}{| A |}, \cos β = \frac{A_{y}}{| A |}, \cos γ = \frac{A_{z}}{| A |} \\ | \vec{A} | = \sqrt{{(A_{x})}^{2} + {(A_{y})}^{2} + {(A_{z})}^{2}} \\ = \sqrt{9 + 25 + 49} = \sqrt{83} \end{aligned}$
$\cos α = \frac{3}{\sqrt{83}}, \cos β = \frac{5}{\sqrt{83}}, \cos γ = \frac{- 7}{\sqrt{83}}$

PHXI04:MOTION IN A PLANE

361785 Vector $\vec{A}$ makes equal angles with $x$ , $y$ and $z$ axis. Value of its components (in terms of magnitude of $\vec{A}$ ) will be

1

\frac{A}{\sqrt{2}}

2

\frac{A}{\sqrt{3}}

3

\frac{\sqrt{3}}{A}

4

\sqrt{3} A

Explanation:

Let the component of $\vec{A}$ makes angles $α, β a n d γ$ with $x$ , $y$ and $z$ axis respectively then $α = β = γ$
$\cos^{2} α + \cos^{2} β + \cos^{2} γ = 1$
$\Rightarrow 3 \cos^{2} α = 1 \Rightarrow \cos α = \frac{1}{\sqrt{3}}$
$∴ A_{x} = A_{y} = A_{z} = A \cos α = \frac{A}{\sqrt{3}}$

PHXI04:MOTION IN A PLANE

361786 The component of a vector $r$ along $X$ -axis will have maximum value if

1

r

is along positive

Y

-axis

2

r

is along positive

X

-axis

3

r

makes an angle of

45^{\circ}

with the

X

-axis

4

r

is along negative

Y

-axis

Explanation:

If $r$ makes an angle $θ$ with $x$ -axis, then component of $r$ along $x$ -axis. $= r \cos θ$
It will be maximum if $\cos θ = max = 1 \Rightarrow θ = 0^{0},$ i.e., $r$ is along positive $x$ -axis.

NCERT Exemplar

PHXI04:MOTION IN A PLANE

361787 The $X$ and $Y$ components of a force $F$ acting at $30^{\circ}$ to $x$ -axis are respectively

1

\frac{F}{2}, \frac{\sqrt{3}}{2} F

2

\frac{F}{\sqrt{2}}, F

3

F, \frac{F}{\sqrt{2}}

4

\frac{\sqrt{3}}{2} F, \frac{F}{2}

Explanation:

The $X$ component of force $F$ is
$F_{x} = F \cos 30^{\circ} = F \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} F$
The $Y$ component of force $F$ is
$F_{y} = F \sin 30^{\circ} = F \times \frac{1}{2} = \frac{1}{2} F$
supporting img

PHXI04:MOTION IN A PLANE

361788 If $\vec{A} = 3 \hat{i} + 5 \hat{j} - 7 \hat{k}$ , the direction of cosines of the vector $\vec{A}$ are:

1

\frac{2}{\sqrt{83}}, \frac{5}{\sqrt{83}}, \frac{7}{\sqrt{83}}

2

\frac{3}{\sqrt{83}}, \frac{5}{\sqrt{83}}, \frac{7}{\sqrt{83}}

3

\frac{1}{\sqrt{83}}, \frac{2}{\sqrt{83}}, \frac{5}{\sqrt{83}}

4

\frac{3}{\sqrt{83}}, \frac{5}{\sqrt{83}}, \frac{- 7}{\sqrt{83}}

Explanation:

$\vec{A} = 3 \hat{i} + 5 \hat{j} - 7 \hat{k}$
The direction cosines are
$\begin{aligned} \cos α = \frac{A_{x}}{| A |}, \cos β = \frac{A_{y}}{| A |}, \cos γ = \frac{A_{z}}{| A |} \\ | \vec{A} | = \sqrt{{(A_{x})}^{2} + {(A_{y})}^{2} + {(A_{z})}^{2}} \\ = \sqrt{9 + 25 + 49} = \sqrt{83} \end{aligned}$
$\cos α = \frac{3}{\sqrt{83}}, \cos β = \frac{5}{\sqrt{83}}, \cos γ = \frac{- 7}{\sqrt{83}}$

PHXI04:MOTION IN A PLANE

361785 Vector $\vec{A}$ makes equal angles with $x$ , $y$ and $z$ axis. Value of its components (in terms of magnitude of $\vec{A}$ ) will be

1

\frac{A}{\sqrt{2}}

2

\frac{A}{\sqrt{3}}

3

\frac{\sqrt{3}}{A}

4

\sqrt{3} A

Explanation:

Let the component of $\vec{A}$ makes angles $α, β a n d γ$ with $x$ , $y$ and $z$ axis respectively then $α = β = γ$
$\cos^{2} α + \cos^{2} β + \cos^{2} γ = 1$
$\Rightarrow 3 \cos^{2} α = 1 \Rightarrow \cos α = \frac{1}{\sqrt{3}}$
$∴ A_{x} = A_{y} = A_{z} = A \cos α = \frac{A}{\sqrt{3}}$

PHXI04:MOTION IN A PLANE

361786 The component of a vector $r$ along $X$ -axis will have maximum value if

1

r

is along positive

Y

-axis

2

r

is along positive

X

-axis

3

r

makes an angle of

45^{\circ}

with the

X

-axis

4

r

is along negative

Y

-axis

Explanation:

If $r$ makes an angle $θ$ with $x$ -axis, then component of $r$ along $x$ -axis. $= r \cos θ$
It will be maximum if $\cos θ = max = 1 \Rightarrow θ = 0^{0},$ i.e., $r$ is along positive $x$ -axis.

NCERT Exemplar

PHXI04:MOTION IN A PLANE

361787 The $X$ and $Y$ components of a force $F$ acting at $30^{\circ}$ to $x$ -axis are respectively

1

\frac{F}{2}, \frac{\sqrt{3}}{2} F

2

\frac{F}{\sqrt{2}}, F

3

F, \frac{F}{\sqrt{2}}

4

\frac{\sqrt{3}}{2} F, \frac{F}{2}

Explanation:

The $X$ component of force $F$ is
$F_{x} = F \cos 30^{\circ} = F \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} F$
The $Y$ component of force $F$ is
$F_{y} = F \sin 30^{\circ} = F \times \frac{1}{2} = \frac{1}{2} F$
supporting img

PHXI04:MOTION IN A PLANE

361788 If $\vec{A} = 3 \hat{i} + 5 \hat{j} - 7 \hat{k}$ , the direction of cosines of the vector $\vec{A}$ are:

1

\frac{2}{\sqrt{83}}, \frac{5}{\sqrt{83}}, \frac{7}{\sqrt{83}}

2

\frac{3}{\sqrt{83}}, \frac{5}{\sqrt{83}}, \frac{7}{\sqrt{83}}

3

\frac{1}{\sqrt{83}}, \frac{2}{\sqrt{83}}, \frac{5}{\sqrt{83}}

4

\frac{3}{\sqrt{83}}, \frac{5}{\sqrt{83}}, \frac{- 7}{\sqrt{83}}

Explanation:

$\vec{A} = 3 \hat{i} + 5 \hat{j} - 7 \hat{k}$
The direction cosines are
$\begin{aligned} \cos α = \frac{A_{x}}{| A |}, \cos β = \frac{A_{y}}{| A |}, \cos γ = \frac{A_{z}}{| A |} \\ | \vec{A} | = \sqrt{{(A_{x})}^{2} + {(A_{y})}^{2} + {(A_{z})}^{2}} \\ = \sqrt{9 + 25 + 49} = \sqrt{83} \end{aligned}$
$\cos α = \frac{3}{\sqrt{83}}, \cos β = \frac{5}{\sqrt{83}}, \cos γ = \frac{- 7}{\sqrt{83}}$

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