QBManager

Motion in Plane

143544 Three forces $F_{1}, F_{2}$ and $F_{3}$ together keep a body in equilibrium. If $F_{1} = 3 N$ along the positive $x$ axis, $F_{2} = 4 N$ along the positive $y$ -axis, then the third force $F_{3}$ is

1

5 N

making an angle

θ = \tan^{- 1} (\frac{3}{4})

with negative

y

-axis

2

5 N

making an angle

θ = \tan^{- 1} (\frac{4}{3})

with negative

y

-axis

3

7 N

making an angle

θ = \tan^{- 1} (\frac{3}{4})

with negative

y

-axis

4

7 N

making an angle

θ = \tan^{- 1} (\frac{4}{3})

with negative

y

-axis

Explanation:

C $F_{1}, F_{2}, F_{3}$ keep a body in equilibrium then resultant of force,

$Σ F = 0$
$F_{1} + F_{2} + F_{3} = 0$
$3 + 4 + F_{3} = 0$
$F_{3} = - 7 N$
Magnitude of $F_{3} = 7 N$
$θ$ is angle made with- $Y$ axis
$\tan θ = \frac{3}{4}$
$θ = \tan^{- 1} 3 / 4$
$F_{3}$ make angle with negative y-axis.

J and K CET- 2010

Motion in Plane

143545 Magnitudes of four pairs of displacement vectors are given. Which pair of displacement vectors, under vector addition, fails to give a resultant vector of magnitude $3 cm$ ?

1

2 cm, 7 cm

2

1 cm, 4 cm

3

2 cm, 3 cm

4

2 cm, 4 cm

Explanation:

A The magnitude $R$ of the resultant of two vectors $A$ and $B$ depends upon the magnitudes of $A$ and $B$ and the angle $θ$ between them and is given by
$R^{2} = A^{2} + B^{2} + 2 AB \cos θ$
When $θ = 0, R$ is maximum and given by
$R_{max}^{2} = A^{2} + B^{2} + 2 AB = (A + B)^{2}$
$R_{max} = A + B$
When $θ = 180^{\circ}, R$ is minimum and given by
$R_{min}^{2} = A^{2} + B^{2} - 2 AB = (A - B)^{2}$
$R_{min} = A - B$
Thus, the magnitude of resultant will lie between A - B and $A + B$ .
Now,
$Checking option (a)$
$| A - B | = | 2 - 7 | = 5$
$| A + B | = | 2 + 7 | = 9$
So, $5 \leq R \leq 9$ and $R = 4$
Hence, the option (a) is the correct answer.

J and K CET- 2009

Motion in Plane

143546 A body is under the action of two mutually perpendicular forces of $3 N$ and $4 N$ . The resultant force acting on the body is

1

7 N

2

1 N

3

5 N

4 zero

Explanation:

C The two forces be
$A = 3 N and B = 4 N$
$A$ is mutually perpendicular to $B$ .
$∴ θ = 90^{\circ}$
$R = \sqrt{A^{2} + B^{2} + 2 AB \cos θ}$
$R = \sqrt{4^{2} + (3)^{2} + 2 AB \cos 90^{\circ}}$
$R = \sqrt{16 + 9 + 0}$
$R = 5 N$

J and K CET- 2008

Motion in Plane

143549 Two vectors are given by $\vec{A} = 3 \hat{i} + \hat{j} + 3 \hat{k}$ and $\vec{B} = 3 \hat{i} + 5 \hat{j} - 2 \hat{k}$ . Find the third vector $\vec{C}$ if $\vec{A} + 3 \vec{B} - \vec{C} = 0$

1

12 \hat{i} + 14 \hat{j} + 12 \hat{k}

2

13 \hat{i} + 17 \hat{j} + 12 \hat{k}

3

12 \hat{i} + 16 \hat{j} - 3 \hat{k}

4

15 \hat{i} + 13 \hat{j} + 4 \hat{k}

Explanation:

C Given, $\vec{A} = 3 \hat{i} + \hat{j} + 3 \hat{k}, \vec{B} = 3 \hat{i} + 5 \hat{j} - 2 \hat{k}$
$\vec{A} + 3 \vec{B} - \vec{C} = 0$
$3 \hat{i} + \hat{j} + 3 \hat{k} + 3 (3 \hat{i} + 5 \hat{j} - 2 \hat{k}) - \vec{C} = 0$
$3 \hat{i} + \hat{j} + 3 \hat{k} + 9 \hat{i} + 15 \hat{j} - 6 \hat{k} - \vec{C} = 0$
$12 \hat{i} + 16 \hat{j} - 3 \hat{k} - \vec{C} = 0$
$\vec{C} = 12 \hat{i} + 16 \hat{j} - 3 \hat{k}$

J and K CET- 2007

Motion in Plane

143550 Vector which is perpendicular to a
$(a \cos θ \hat{i} + b \sin θ \hat{j}) is$

1

b \sin θ \hat{i} - a \cos θ \hat{j}

2

\frac{1}{a} \sin θ \hat{i} - \frac{1}{b} \cos θ \hat{j}

3

5 \hat{k}

4 all of these

Explanation:

D Two vectors are perpendicular if their dot product is zero i.e., $\vec{A} \cdot \vec{B} = 0$ .
In option (a)
$(a \cos θ \hat{i} + b \sin θ \hat{j}) \cdot (b \sin θ \hat{i} - a \cos θ \hat{j})$
$= a b \cos θ \sin θ - a b \sin θ \cos θ = 0$
In option (b)
$(a \cos θ \hat{i} + b \sin θ \hat{j}) \cdot (\frac{1}{a} \sin θ \hat{i} - \frac{1}{b} \cos θ \hat{j})$
$= \sin θ \cos θ - \sin θ \cos θ = 0$
In option (c)
$(a \cos θ \hat{i} + b \sin θ \hat{j}) \cdot (5 \hat{k}) = 0$ .

J and K CET- 2006

Motion in Plane

143544 Three forces $F_{1}, F_{2}$ and $F_{3}$ together keep a body in equilibrium. If $F_{1} = 3 N$ along the positive $x$ axis, $F_{2} = 4 N$ along the positive $y$ -axis, then the third force $F_{3}$ is

1

5 N

making an angle

θ = \tan^{- 1} (\frac{3}{4})

with negative

y

-axis

2

5 N

making an angle

θ = \tan^{- 1} (\frac{4}{3})

with negative

y

-axis

3

7 N

making an angle

θ = \tan^{- 1} (\frac{3}{4})

with negative

y

-axis

4

7 N

making an angle

θ = \tan^{- 1} (\frac{4}{3})

with negative

y

-axis

Explanation:

C $F_{1}, F_{2}, F_{3}$ keep a body in equilibrium then resultant of force,

$Σ F = 0$
$F_{1} + F_{2} + F_{3} = 0$
$3 + 4 + F_{3} = 0$
$F_{3} = - 7 N$
Magnitude of $F_{3} = 7 N$
$θ$ is angle made with- $Y$ axis
$\tan θ = \frac{3}{4}$
$θ = \tan^{- 1} 3 / 4$
$F_{3}$ make angle with negative y-axis.

J and K CET- 2010

Motion in Plane

143545 Magnitudes of four pairs of displacement vectors are given. Which pair of displacement vectors, under vector addition, fails to give a resultant vector of magnitude $3 cm$ ?

1

2 cm, 7 cm

2

1 cm, 4 cm

3

2 cm, 3 cm

4

2 cm, 4 cm

Explanation:

A The magnitude $R$ of the resultant of two vectors $A$ and $B$ depends upon the magnitudes of $A$ and $B$ and the angle $θ$ between them and is given by
$R^{2} = A^{2} + B^{2} + 2 AB \cos θ$
When $θ = 0, R$ is maximum and given by
$R_{max}^{2} = A^{2} + B^{2} + 2 AB = (A + B)^{2}$
$R_{max} = A + B$
When $θ = 180^{\circ}, R$ is minimum and given by
$R_{min}^{2} = A^{2} + B^{2} - 2 AB = (A - B)^{2}$
$R_{min} = A - B$
Thus, the magnitude of resultant will lie between A - B and $A + B$ .
Now,
$Checking option (a)$
$| A - B | = | 2 - 7 | = 5$
$| A + B | = | 2 + 7 | = 9$
So, $5 \leq R \leq 9$ and $R = 4$
Hence, the option (a) is the correct answer.

J and K CET- 2009

Motion in Plane

143546 A body is under the action of two mutually perpendicular forces of $3 N$ and $4 N$ . The resultant force acting on the body is

1

7 N

2

1 N

3

5 N

4 zero

Explanation:

C The two forces be
$A = 3 N and B = 4 N$
$A$ is mutually perpendicular to $B$ .
$∴ θ = 90^{\circ}$
$R = \sqrt{A^{2} + B^{2} + 2 AB \cos θ}$
$R = \sqrt{4^{2} + (3)^{2} + 2 AB \cos 90^{\circ}}$
$R = \sqrt{16 + 9 + 0}$
$R = 5 N$

J and K CET- 2008

Motion in Plane

143549 Two vectors are given by $\vec{A} = 3 \hat{i} + \hat{j} + 3 \hat{k}$ and $\vec{B} = 3 \hat{i} + 5 \hat{j} - 2 \hat{k}$ . Find the third vector $\vec{C}$ if $\vec{A} + 3 \vec{B} - \vec{C} = 0$

1

12 \hat{i} + 14 \hat{j} + 12 \hat{k}

2

13 \hat{i} + 17 \hat{j} + 12 \hat{k}

3

12 \hat{i} + 16 \hat{j} - 3 \hat{k}

4

15 \hat{i} + 13 \hat{j} + 4 \hat{k}

Explanation:

C Given, $\vec{A} = 3 \hat{i} + \hat{j} + 3 \hat{k}, \vec{B} = 3 \hat{i} + 5 \hat{j} - 2 \hat{k}$
$\vec{A} + 3 \vec{B} - \vec{C} = 0$
$3 \hat{i} + \hat{j} + 3 \hat{k} + 3 (3 \hat{i} + 5 \hat{j} - 2 \hat{k}) - \vec{C} = 0$
$3 \hat{i} + \hat{j} + 3 \hat{k} + 9 \hat{i} + 15 \hat{j} - 6 \hat{k} - \vec{C} = 0$
$12 \hat{i} + 16 \hat{j} - 3 \hat{k} - \vec{C} = 0$
$\vec{C} = 12 \hat{i} + 16 \hat{j} - 3 \hat{k}$

J and K CET- 2007

Motion in Plane

143550 Vector which is perpendicular to a
$(a \cos θ \hat{i} + b \sin θ \hat{j}) is$

1

b \sin θ \hat{i} - a \cos θ \hat{j}

2

\frac{1}{a} \sin θ \hat{i} - \frac{1}{b} \cos θ \hat{j}

3

5 \hat{k}

4 all of these

Explanation:

D Two vectors are perpendicular if their dot product is zero i.e., $\vec{A} \cdot \vec{B} = 0$ .
In option (a)
$(a \cos θ \hat{i} + b \sin θ \hat{j}) \cdot (b \sin θ \hat{i} - a \cos θ \hat{j})$
$= a b \cos θ \sin θ - a b \sin θ \cos θ = 0$
In option (b)
$(a \cos θ \hat{i} + b \sin θ \hat{j}) \cdot (\frac{1}{a} \sin θ \hat{i} - \frac{1}{b} \cos θ \hat{j})$
$= \sin θ \cos θ - \sin θ \cos θ = 0$
In option (c)
$(a \cos θ \hat{i} + b \sin θ \hat{j}) \cdot (5 \hat{k}) = 0$ .

J and K CET- 2006

Motion in Plane

143544 Three forces $F_{1}, F_{2}$ and $F_{3}$ together keep a body in equilibrium. If $F_{1} = 3 N$ along the positive $x$ axis, $F_{2} = 4 N$ along the positive $y$ -axis, then the third force $F_{3}$ is

1

5 N

making an angle

θ = \tan^{- 1} (\frac{3}{4})

with negative

y

-axis

2

5 N

making an angle

θ = \tan^{- 1} (\frac{4}{3})

with negative

y

-axis

3

7 N

making an angle

θ = \tan^{- 1} (\frac{3}{4})

with negative

y

-axis

4

7 N

making an angle

θ = \tan^{- 1} (\frac{4}{3})

with negative

y

-axis

Explanation:

C $F_{1}, F_{2}, F_{3}$ keep a body in equilibrium then resultant of force,

$Σ F = 0$
$F_{1} + F_{2} + F_{3} = 0$
$3 + 4 + F_{3} = 0$
$F_{3} = - 7 N$
Magnitude of $F_{3} = 7 N$
$θ$ is angle made with- $Y$ axis
$\tan θ = \frac{3}{4}$
$θ = \tan^{- 1} 3 / 4$
$F_{3}$ make angle with negative y-axis.

J and K CET- 2010

Motion in Plane

143545 Magnitudes of four pairs of displacement vectors are given. Which pair of displacement vectors, under vector addition, fails to give a resultant vector of magnitude $3 cm$ ?

1

2 cm, 7 cm

2

1 cm, 4 cm

3

2 cm, 3 cm

4

2 cm, 4 cm

Explanation:

A The magnitude $R$ of the resultant of two vectors $A$ and $B$ depends upon the magnitudes of $A$ and $B$ and the angle $θ$ between them and is given by
$R^{2} = A^{2} + B^{2} + 2 AB \cos θ$
When $θ = 0, R$ is maximum and given by
$R_{max}^{2} = A^{2} + B^{2} + 2 AB = (A + B)^{2}$
$R_{max} = A + B$
When $θ = 180^{\circ}, R$ is minimum and given by
$R_{min}^{2} = A^{2} + B^{2} - 2 AB = (A - B)^{2}$
$R_{min} = A - B$
Thus, the magnitude of resultant will lie between A - B and $A + B$ .
Now,
$Checking option (a)$
$| A - B | = | 2 - 7 | = 5$
$| A + B | = | 2 + 7 | = 9$
So, $5 \leq R \leq 9$ and $R = 4$
Hence, the option (a) is the correct answer.

J and K CET- 2009

Motion in Plane

143546 A body is under the action of two mutually perpendicular forces of $3 N$ and $4 N$ . The resultant force acting on the body is

1

7 N

2

1 N

3

5 N

4 zero

Explanation:

C The two forces be
$A = 3 N and B = 4 N$
$A$ is mutually perpendicular to $B$ .
$∴ θ = 90^{\circ}$
$R = \sqrt{A^{2} + B^{2} + 2 AB \cos θ}$
$R = \sqrt{4^{2} + (3)^{2} + 2 AB \cos 90^{\circ}}$
$R = \sqrt{16 + 9 + 0}$
$R = 5 N$

J and K CET- 2008

Motion in Plane

143549 Two vectors are given by $\vec{A} = 3 \hat{i} + \hat{j} + 3 \hat{k}$ and $\vec{B} = 3 \hat{i} + 5 \hat{j} - 2 \hat{k}$ . Find the third vector $\vec{C}$ if $\vec{A} + 3 \vec{B} - \vec{C} = 0$

1

12 \hat{i} + 14 \hat{j} + 12 \hat{k}

2

13 \hat{i} + 17 \hat{j} + 12 \hat{k}

3

12 \hat{i} + 16 \hat{j} - 3 \hat{k}

4

15 \hat{i} + 13 \hat{j} + 4 \hat{k}

Explanation:

C Given, $\vec{A} = 3 \hat{i} + \hat{j} + 3 \hat{k}, \vec{B} = 3 \hat{i} + 5 \hat{j} - 2 \hat{k}$
$\vec{A} + 3 \vec{B} - \vec{C} = 0$
$3 \hat{i} + \hat{j} + 3 \hat{k} + 3 (3 \hat{i} + 5 \hat{j} - 2 \hat{k}) - \vec{C} = 0$
$3 \hat{i} + \hat{j} + 3 \hat{k} + 9 \hat{i} + 15 \hat{j} - 6 \hat{k} - \vec{C} = 0$
$12 \hat{i} + 16 \hat{j} - 3 \hat{k} - \vec{C} = 0$
$\vec{C} = 12 \hat{i} + 16 \hat{j} - 3 \hat{k}$

J and K CET- 2007

Motion in Plane

143550 Vector which is perpendicular to a
$(a \cos θ \hat{i} + b \sin θ \hat{j}) is$

1

b \sin θ \hat{i} - a \cos θ \hat{j}

2

\frac{1}{a} \sin θ \hat{i} - \frac{1}{b} \cos θ \hat{j}

3

5 \hat{k}

4 all of these

Explanation:

D Two vectors are perpendicular if their dot product is zero i.e., $\vec{A} \cdot \vec{B} = 0$ .
In option (a)
$(a \cos θ \hat{i} + b \sin θ \hat{j}) \cdot (b \sin θ \hat{i} - a \cos θ \hat{j})$
$= a b \cos θ \sin θ - a b \sin θ \cos θ = 0$
In option (b)
$(a \cos θ \hat{i} + b \sin θ \hat{j}) \cdot (\frac{1}{a} \sin θ \hat{i} - \frac{1}{b} \cos θ \hat{j})$
$= \sin θ \cos θ - \sin θ \cos θ = 0$
In option (c)
$(a \cos θ \hat{i} + b \sin θ \hat{j}) \cdot (5 \hat{k}) = 0$ .

J and K CET- 2006

Motion in Plane

143544 Three forces $F_{1}, F_{2}$ and $F_{3}$ together keep a body in equilibrium. If $F_{1} = 3 N$ along the positive $x$ axis, $F_{2} = 4 N$ along the positive $y$ -axis, then the third force $F_{3}$ is

1

5 N

making an angle

θ = \tan^{- 1} (\frac{3}{4})

with negative

y

-axis

2

5 N

making an angle

θ = \tan^{- 1} (\frac{4}{3})

with negative

y

-axis

3

7 N

making an angle

θ = \tan^{- 1} (\frac{3}{4})

with negative

y

-axis

4

7 N

making an angle

θ = \tan^{- 1} (\frac{4}{3})

with negative

y

-axis

Explanation:

C $F_{1}, F_{2}, F_{3}$ keep a body in equilibrium then resultant of force,

$Σ F = 0$
$F_{1} + F_{2} + F_{3} = 0$
$3 + 4 + F_{3} = 0$
$F_{3} = - 7 N$
Magnitude of $F_{3} = 7 N$
$θ$ is angle made with- $Y$ axis
$\tan θ = \frac{3}{4}$
$θ = \tan^{- 1} 3 / 4$
$F_{3}$ make angle with negative y-axis.

J and K CET- 2010

Motion in Plane

143545 Magnitudes of four pairs of displacement vectors are given. Which pair of displacement vectors, under vector addition, fails to give a resultant vector of magnitude $3 cm$ ?

1

2 cm, 7 cm

2

1 cm, 4 cm

3

2 cm, 3 cm

4

2 cm, 4 cm

Explanation:

A The magnitude $R$ of the resultant of two vectors $A$ and $B$ depends upon the magnitudes of $A$ and $B$ and the angle $θ$ between them and is given by
$R^{2} = A^{2} + B^{2} + 2 AB \cos θ$
When $θ = 0, R$ is maximum and given by
$R_{max}^{2} = A^{2} + B^{2} + 2 AB = (A + B)^{2}$
$R_{max} = A + B$
When $θ = 180^{\circ}, R$ is minimum and given by
$R_{min}^{2} = A^{2} + B^{2} - 2 AB = (A - B)^{2}$
$R_{min} = A - B$
Thus, the magnitude of resultant will lie between A - B and $A + B$ .
Now,
$Checking option (a)$
$| A - B | = | 2 - 7 | = 5$
$| A + B | = | 2 + 7 | = 9$
So, $5 \leq R \leq 9$ and $R = 4$
Hence, the option (a) is the correct answer.

J and K CET- 2009

Motion in Plane

143546 A body is under the action of two mutually perpendicular forces of $3 N$ and $4 N$ . The resultant force acting on the body is

1

7 N

2

1 N

3

5 N

4 zero

Explanation:

C The two forces be
$A = 3 N and B = 4 N$
$A$ is mutually perpendicular to $B$ .
$∴ θ = 90^{\circ}$
$R = \sqrt{A^{2} + B^{2} + 2 AB \cos θ}$
$R = \sqrt{4^{2} + (3)^{2} + 2 AB \cos 90^{\circ}}$
$R = \sqrt{16 + 9 + 0}$
$R = 5 N$

J and K CET- 2008

Motion in Plane

143549 Two vectors are given by $\vec{A} = 3 \hat{i} + \hat{j} + 3 \hat{k}$ and $\vec{B} = 3 \hat{i} + 5 \hat{j} - 2 \hat{k}$ . Find the third vector $\vec{C}$ if $\vec{A} + 3 \vec{B} - \vec{C} = 0$

1

12 \hat{i} + 14 \hat{j} + 12 \hat{k}

2

13 \hat{i} + 17 \hat{j} + 12 \hat{k}

3

12 \hat{i} + 16 \hat{j} - 3 \hat{k}

4

15 \hat{i} + 13 \hat{j} + 4 \hat{k}

Explanation:

C Given, $\vec{A} = 3 \hat{i} + \hat{j} + 3 \hat{k}, \vec{B} = 3 \hat{i} + 5 \hat{j} - 2 \hat{k}$
$\vec{A} + 3 \vec{B} - \vec{C} = 0$
$3 \hat{i} + \hat{j} + 3 \hat{k} + 3 (3 \hat{i} + 5 \hat{j} - 2 \hat{k}) - \vec{C} = 0$
$3 \hat{i} + \hat{j} + 3 \hat{k} + 9 \hat{i} + 15 \hat{j} - 6 \hat{k} - \vec{C} = 0$
$12 \hat{i} + 16 \hat{j} - 3 \hat{k} - \vec{C} = 0$
$\vec{C} = 12 \hat{i} + 16 \hat{j} - 3 \hat{k}$

J and K CET- 2007

Motion in Plane

143550 Vector which is perpendicular to a
$(a \cos θ \hat{i} + b \sin θ \hat{j}) is$

1

b \sin θ \hat{i} - a \cos θ \hat{j}

2

\frac{1}{a} \sin θ \hat{i} - \frac{1}{b} \cos θ \hat{j}

3

5 \hat{k}

4 all of these

Explanation:

D Two vectors are perpendicular if their dot product is zero i.e., $\vec{A} \cdot \vec{B} = 0$ .
In option (a)
$(a \cos θ \hat{i} + b \sin θ \hat{j}) \cdot (b \sin θ \hat{i} - a \cos θ \hat{j})$
$= a b \cos θ \sin θ - a b \sin θ \cos θ = 0$
In option (b)
$(a \cos θ \hat{i} + b \sin θ \hat{j}) \cdot (\frac{1}{a} \sin θ \hat{i} - \frac{1}{b} \cos θ \hat{j})$
$= \sin θ \cos θ - \sin θ \cos θ = 0$
In option (c)
$(a \cos θ \hat{i} + b \sin θ \hat{j}) \cdot (5 \hat{k}) = 0$ .

J and K CET- 2006

Motion in Plane

143544 Three forces $F_{1}, F_{2}$ and $F_{3}$ together keep a body in equilibrium. If $F_{1} = 3 N$ along the positive $x$ axis, $F_{2} = 4 N$ along the positive $y$ -axis, then the third force $F_{3}$ is

1

5 N

making an angle

θ = \tan^{- 1} (\frac{3}{4})

with negative

y

-axis

2

5 N

making an angle

θ = \tan^{- 1} (\frac{4}{3})

with negative

y

-axis

3

7 N

making an angle

θ = \tan^{- 1} (\frac{3}{4})

with negative

y

-axis

4

7 N

making an angle

θ = \tan^{- 1} (\frac{4}{3})

with negative

y

-axis

Explanation:

C $F_{1}, F_{2}, F_{3}$ keep a body in equilibrium then resultant of force,

$Σ F = 0$
$F_{1} + F_{2} + F_{3} = 0$
$3 + 4 + F_{3} = 0$
$F_{3} = - 7 N$
Magnitude of $F_{3} = 7 N$
$θ$ is angle made with- $Y$ axis
$\tan θ = \frac{3}{4}$
$θ = \tan^{- 1} 3 / 4$
$F_{3}$ make angle with negative y-axis.

J and K CET- 2010

Motion in Plane

143545 Magnitudes of four pairs of displacement vectors are given. Which pair of displacement vectors, under vector addition, fails to give a resultant vector of magnitude $3 cm$ ?

1

2 cm, 7 cm

2

1 cm, 4 cm

3

2 cm, 3 cm

4

2 cm, 4 cm

Explanation:

A The magnitude $R$ of the resultant of two vectors $A$ and $B$ depends upon the magnitudes of $A$ and $B$ and the angle $θ$ between them and is given by
$R^{2} = A^{2} + B^{2} + 2 AB \cos θ$
When $θ = 0, R$ is maximum and given by
$R_{max}^{2} = A^{2} + B^{2} + 2 AB = (A + B)^{2}$
$R_{max} = A + B$
When $θ = 180^{\circ}, R$ is minimum and given by
$R_{min}^{2} = A^{2} + B^{2} - 2 AB = (A - B)^{2}$
$R_{min} = A - B$
Thus, the magnitude of resultant will lie between A - B and $A + B$ .
Now,
$Checking option (a)$
$| A - B | = | 2 - 7 | = 5$
$| A + B | = | 2 + 7 | = 9$
So, $5 \leq R \leq 9$ and $R = 4$
Hence, the option (a) is the correct answer.

J and K CET- 2009

Motion in Plane

143546 A body is under the action of two mutually perpendicular forces of $3 N$ and $4 N$ . The resultant force acting on the body is

1

7 N

2

1 N

3

5 N

4 zero

Explanation:

C The two forces be
$A = 3 N and B = 4 N$
$A$ is mutually perpendicular to $B$ .
$∴ θ = 90^{\circ}$
$R = \sqrt{A^{2} + B^{2} + 2 AB \cos θ}$
$R = \sqrt{4^{2} + (3)^{2} + 2 AB \cos 90^{\circ}}$
$R = \sqrt{16 + 9 + 0}$
$R = 5 N$

J and K CET- 2008

Motion in Plane

143549 Two vectors are given by $\vec{A} = 3 \hat{i} + \hat{j} + 3 \hat{k}$ and $\vec{B} = 3 \hat{i} + 5 \hat{j} - 2 \hat{k}$ . Find the third vector $\vec{C}$ if $\vec{A} + 3 \vec{B} - \vec{C} = 0$

1

12 \hat{i} + 14 \hat{j} + 12 \hat{k}

2

13 \hat{i} + 17 \hat{j} + 12 \hat{k}

3

12 \hat{i} + 16 \hat{j} - 3 \hat{k}

4

15 \hat{i} + 13 \hat{j} + 4 \hat{k}

Explanation:

C Given, $\vec{A} = 3 \hat{i} + \hat{j} + 3 \hat{k}, \vec{B} = 3 \hat{i} + 5 \hat{j} - 2 \hat{k}$
$\vec{A} + 3 \vec{B} - \vec{C} = 0$
$3 \hat{i} + \hat{j} + 3 \hat{k} + 3 (3 \hat{i} + 5 \hat{j} - 2 \hat{k}) - \vec{C} = 0$
$3 \hat{i} + \hat{j} + 3 \hat{k} + 9 \hat{i} + 15 \hat{j} - 6 \hat{k} - \vec{C} = 0$
$12 \hat{i} + 16 \hat{j} - 3 \hat{k} - \vec{C} = 0$
$\vec{C} = 12 \hat{i} + 16 \hat{j} - 3 \hat{k}$

J and K CET- 2007

Motion in Plane

143550 Vector which is perpendicular to a
$(a \cos θ \hat{i} + b \sin θ \hat{j}) is$

1

b \sin θ \hat{i} - a \cos θ \hat{j}

2

\frac{1}{a} \sin θ \hat{i} - \frac{1}{b} \cos θ \hat{j}

3

5 \hat{k}

4 all of these

Explanation:

D Two vectors are perpendicular if their dot product is zero i.e., $\vec{A} \cdot \vec{B} = 0$ .
In option (a)
$(a \cos θ \hat{i} + b \sin θ \hat{j}) \cdot (b \sin θ \hat{i} - a \cos θ \hat{j})$
$= a b \cos θ \sin θ - a b \sin θ \cos θ = 0$
In option (b)
$(a \cos θ \hat{i} + b \sin θ \hat{j}) \cdot (\frac{1}{a} \sin θ \hat{i} - \frac{1}{b} \cos θ \hat{j})$
$= \sin θ \cos θ - \sin θ \cos θ = 0$
In option (c)
$(a \cos θ \hat{i} + b \sin θ \hat{j}) \cdot (5 \hat{k}) = 0$ .

J and K CET- 2006