QBManager

Motion in Plane

143598 The vectors are given by $\vec{A} = \hat{i} + 2 \hat{j} + 2 \hat{k}$ and $\vec{B} = 3 \hat{i} + 6 \hat{j} + 2 \hat{k}$ . Another vector $C$ has the same magnitude as $B$ but has the same direction as $A$ . Then which of the following vectors represents $C$ ?

1

\frac{7}{3} (\hat{i} + 2 \hat{j} + 2 \hat{k})

2

\frac{3}{7} (\hat{i} - 2 \hat{j} + 2 \hat{k})

3

\frac{7}{9} (\hat{i} - 2 \hat{j} + 2 \hat{k})

4

\frac{9}{7} (\hat{i} + 2 \hat{j} + 2 \hat{k})

Explanation:

A Given that, $\vec{A} = \hat{i} + 2 \hat{j} + 2 \hat{k}$
Magnitude of $| \vec{A} | = \sqrt{1^{2} + 2^{2} + 2^{2}} = 3$
Unit vector along $\vec{A}$ .
$\hat{A} = \frac{\vec{A}}{| \vec{A} |} = \frac{\hat{i} + 2 \hat{j} + 2 \hat{k}}{3}$
$\vec{B} = 3 \hat{i} + 6 \hat{j} + 2 \hat{k}$
Magnitude of
$| \vec{B} | = \sqrt{3^{2} + 6^{2} + 2^{2}}$
$= \sqrt{9 + 36 + 4}$
$| \vec{B} | = 7$
So, $C = | \vec{B} | \hat{A}$
$C = 7 \times \frac{\hat{i} + 2 \hat{j} + 2 \hat{k}}{3}$
$C = \frac{7}{3} (\hat{i} + 2 \hat{j} + 2 \hat{k})$

WB JEE 2013

Motion in Plane

143599 The (x,y,z) coordinates of two points $A$ and $B$ are given respectively as $(0, 3, - 1)$ and $(- 2, 6, 4)$ . The displacement vector form $A$ to $B$ may be given by :

1

- 2 \hat{i} + 6 \hat{j} + 4 \hat{k}

2

- 2 \hat{i} + 3 \hat{j} + 3 \hat{k}

3

- 2 \hat{i} + 3 \hat{j} + 5 \hat{k}

4

2 \hat{i} - 3 \hat{j} - 5 \hat{k}

Explanation:

C Given,
$r_{A} = 3 \hat{j} - \hat{k}, r_{B} = - 2 \hat{i} + 6 \hat{j} + 4 \hat{k}$
Displacement vector $(r_{A B}) = r_{B} - r_{A}$ .
$r_{A B} = (- 2 \hat{i} + 6 \hat{j} + 4 \hat{k}) - (3 \hat{j} - \hat{k})$
$r_{A B} = - 2 \hat{i} + 3 \hat{j} + 5 \hat{k}$
Therefore, the displacement vector from A to B is $- 2 \hat{i} + 3 \hat{j} + 5 \hat{k}$

BCECE-2006

Motion in Plane

143600 The resultant of two forces $P$ and $Q$ is of magnitude $P$ . If $P$ be doubled, the resultant will be inclined to $Q$ at an angle.

1

0^{0}

2

30^{\circ}

3

60^{\circ}

4

90^{\circ}

Explanation:

D Let the angle between $P$ and $Q$ be $θ$ .
Since, the resultant of $P$ and $Q$ is $P$
$∴ P^{2} = P^{2} + Q^{2} + 2 PQ \cos θ$
$- Q^{2} = 2 PQ \cos θ$
$Q = - 2 P \cos θ$
According to the question, given $P$ is doubled then $4 P^{2} = 4 P^{2} + Q^{2} + 4 PQ \cos θ$
Put the value of $Q$
$0 = (- 2 P \cos θ)^{2} + 4 P (- 2 P \cos θ) \cdot \cos θ$
$0 = 4 P^{2} \cos^{2} θ - 8 P^{2} \cos^{2} θ$
$0 = - 4 P^{2} \cos^{2} θ$
$\cos θ = 0$
$θ = 90^{\circ}$

UPSEE-2016

Motion in Plane

143601 If $\vec{a} + \vec{b} = \vec{c}$ and $a + b = c$ , then the angle included between $\vec{a}$ and $\vec{b}$ is

1

90^{\circ}

2

180^{\circ}

3

120^{\circ}

4 zero

Explanation:

D Given that, $\vec{a} + \vec{b} = \vec{c}$ and $a + b = c$
$| \vec{c} | = | \vec{a} + \vec{b} |$
$c^{2} = a^{2} + b^{2} + 2 a b \cos θ$
$c = \sqrt{a^{2} + b^{2} + 2 a b \cos θ}$
$a + b = \sqrt{a^{2} + b^{2} + 2 a b \cos θ}$
$a^{2} + b^{2} + 2 a b = a^{2} + b^{2} + 2 a b \cos θ$
$\cos θ = 1$
$θ = 0^{\circ}$

WB JEE-2010

Motion in Plane

143598 The vectors are given by $\vec{A} = \hat{i} + 2 \hat{j} + 2 \hat{k}$ and $\vec{B} = 3 \hat{i} + 6 \hat{j} + 2 \hat{k}$ . Another vector $C$ has the same magnitude as $B$ but has the same direction as $A$ . Then which of the following vectors represents $C$ ?

1

\frac{7}{3} (\hat{i} + 2 \hat{j} + 2 \hat{k})

2

\frac{3}{7} (\hat{i} - 2 \hat{j} + 2 \hat{k})

3

\frac{7}{9} (\hat{i} - 2 \hat{j} + 2 \hat{k})

4

\frac{9}{7} (\hat{i} + 2 \hat{j} + 2 \hat{k})

Explanation:

A Given that, $\vec{A} = \hat{i} + 2 \hat{j} + 2 \hat{k}$
Magnitude of $| \vec{A} | = \sqrt{1^{2} + 2^{2} + 2^{2}} = 3$
Unit vector along $\vec{A}$ .
$\hat{A} = \frac{\vec{A}}{| \vec{A} |} = \frac{\hat{i} + 2 \hat{j} + 2 \hat{k}}{3}$
$\vec{B} = 3 \hat{i} + 6 \hat{j} + 2 \hat{k}$
Magnitude of
$| \vec{B} | = \sqrt{3^{2} + 6^{2} + 2^{2}}$
$= \sqrt{9 + 36 + 4}$
$| \vec{B} | = 7$
So, $C = | \vec{B} | \hat{A}$
$C = 7 \times \frac{\hat{i} + 2 \hat{j} + 2 \hat{k}}{3}$
$C = \frac{7}{3} (\hat{i} + 2 \hat{j} + 2 \hat{k})$

WB JEE 2013

Motion in Plane

143599 The (x,y,z) coordinates of two points $A$ and $B$ are given respectively as $(0, 3, - 1)$ and $(- 2, 6, 4)$ . The displacement vector form $A$ to $B$ may be given by :

1

- 2 \hat{i} + 6 \hat{j} + 4 \hat{k}

2

- 2 \hat{i} + 3 \hat{j} + 3 \hat{k}

3

- 2 \hat{i} + 3 \hat{j} + 5 \hat{k}

4

2 \hat{i} - 3 \hat{j} - 5 \hat{k}

Explanation:

C Given,
$r_{A} = 3 \hat{j} - \hat{k}, r_{B} = - 2 \hat{i} + 6 \hat{j} + 4 \hat{k}$
Displacement vector $(r_{A B}) = r_{B} - r_{A}$ .
$r_{A B} = (- 2 \hat{i} + 6 \hat{j} + 4 \hat{k}) - (3 \hat{j} - \hat{k})$
$r_{A B} = - 2 \hat{i} + 3 \hat{j} + 5 \hat{k}$
Therefore, the displacement vector from A to B is $- 2 \hat{i} + 3 \hat{j} + 5 \hat{k}$

BCECE-2006

Motion in Plane

143600 The resultant of two forces $P$ and $Q$ is of magnitude $P$ . If $P$ be doubled, the resultant will be inclined to $Q$ at an angle.

1

0^{0}

2

30^{\circ}

3

60^{\circ}

4

90^{\circ}

Explanation:

D Let the angle between $P$ and $Q$ be $θ$ .
Since, the resultant of $P$ and $Q$ is $P$
$∴ P^{2} = P^{2} + Q^{2} + 2 PQ \cos θ$
$- Q^{2} = 2 PQ \cos θ$
$Q = - 2 P \cos θ$
According to the question, given $P$ is doubled then $4 P^{2} = 4 P^{2} + Q^{2} + 4 PQ \cos θ$
Put the value of $Q$
$0 = (- 2 P \cos θ)^{2} + 4 P (- 2 P \cos θ) \cdot \cos θ$
$0 = 4 P^{2} \cos^{2} θ - 8 P^{2} \cos^{2} θ$
$0 = - 4 P^{2} \cos^{2} θ$
$\cos θ = 0$
$θ = 90^{\circ}$

UPSEE-2016

Motion in Plane

143601 If $\vec{a} + \vec{b} = \vec{c}$ and $a + b = c$ , then the angle included between $\vec{a}$ and $\vec{b}$ is

1

90^{\circ}

2

180^{\circ}

3

120^{\circ}

4 zero

Explanation:

D Given that, $\vec{a} + \vec{b} = \vec{c}$ and $a + b = c$
$| \vec{c} | = | \vec{a} + \vec{b} |$
$c^{2} = a^{2} + b^{2} + 2 a b \cos θ$
$c = \sqrt{a^{2} + b^{2} + 2 a b \cos θ}$
$a + b = \sqrt{a^{2} + b^{2} + 2 a b \cos θ}$
$a^{2} + b^{2} + 2 a b = a^{2} + b^{2} + 2 a b \cos θ$
$\cos θ = 1$
$θ = 0^{\circ}$

WB JEE-2010

Motion in Plane

143598 The vectors are given by $\vec{A} = \hat{i} + 2 \hat{j} + 2 \hat{k}$ and $\vec{B} = 3 \hat{i} + 6 \hat{j} + 2 \hat{k}$ . Another vector $C$ has the same magnitude as $B$ but has the same direction as $A$ . Then which of the following vectors represents $C$ ?

1

\frac{7}{3} (\hat{i} + 2 \hat{j} + 2 \hat{k})

2

\frac{3}{7} (\hat{i} - 2 \hat{j} + 2 \hat{k})

3

\frac{7}{9} (\hat{i} - 2 \hat{j} + 2 \hat{k})

4

\frac{9}{7} (\hat{i} + 2 \hat{j} + 2 \hat{k})

Explanation:

A Given that, $\vec{A} = \hat{i} + 2 \hat{j} + 2 \hat{k}$
Magnitude of $| \vec{A} | = \sqrt{1^{2} + 2^{2} + 2^{2}} = 3$
Unit vector along $\vec{A}$ .
$\hat{A} = \frac{\vec{A}}{| \vec{A} |} = \frac{\hat{i} + 2 \hat{j} + 2 \hat{k}}{3}$
$\vec{B} = 3 \hat{i} + 6 \hat{j} + 2 \hat{k}$
Magnitude of
$| \vec{B} | = \sqrt{3^{2} + 6^{2} + 2^{2}}$
$= \sqrt{9 + 36 + 4}$
$| \vec{B} | = 7$
So, $C = | \vec{B} | \hat{A}$
$C = 7 \times \frac{\hat{i} + 2 \hat{j} + 2 \hat{k}}{3}$
$C = \frac{7}{3} (\hat{i} + 2 \hat{j} + 2 \hat{k})$

WB JEE 2013

Motion in Plane

143599 The (x,y,z) coordinates of two points $A$ and $B$ are given respectively as $(0, 3, - 1)$ and $(- 2, 6, 4)$ . The displacement vector form $A$ to $B$ may be given by :

1

- 2 \hat{i} + 6 \hat{j} + 4 \hat{k}

2

- 2 \hat{i} + 3 \hat{j} + 3 \hat{k}

3

- 2 \hat{i} + 3 \hat{j} + 5 \hat{k}

4

2 \hat{i} - 3 \hat{j} - 5 \hat{k}

Explanation:

C Given,
$r_{A} = 3 \hat{j} - \hat{k}, r_{B} = - 2 \hat{i} + 6 \hat{j} + 4 \hat{k}$
Displacement vector $(r_{A B}) = r_{B} - r_{A}$ .
$r_{A B} = (- 2 \hat{i} + 6 \hat{j} + 4 \hat{k}) - (3 \hat{j} - \hat{k})$
$r_{A B} = - 2 \hat{i} + 3 \hat{j} + 5 \hat{k}$
Therefore, the displacement vector from A to B is $- 2 \hat{i} + 3 \hat{j} + 5 \hat{k}$

BCECE-2006

Motion in Plane

143600 The resultant of two forces $P$ and $Q$ is of magnitude $P$ . If $P$ be doubled, the resultant will be inclined to $Q$ at an angle.

1

0^{0}

2

30^{\circ}

3

60^{\circ}

4

90^{\circ}

Explanation:

D Let the angle between $P$ and $Q$ be $θ$ .
Since, the resultant of $P$ and $Q$ is $P$
$∴ P^{2} = P^{2} + Q^{2} + 2 PQ \cos θ$
$- Q^{2} = 2 PQ \cos θ$
$Q = - 2 P \cos θ$
According to the question, given $P$ is doubled then $4 P^{2} = 4 P^{2} + Q^{2} + 4 PQ \cos θ$
Put the value of $Q$
$0 = (- 2 P \cos θ)^{2} + 4 P (- 2 P \cos θ) \cdot \cos θ$
$0 = 4 P^{2} \cos^{2} θ - 8 P^{2} \cos^{2} θ$
$0 = - 4 P^{2} \cos^{2} θ$
$\cos θ = 0$
$θ = 90^{\circ}$

UPSEE-2016

Motion in Plane

143601 If $\vec{a} + \vec{b} = \vec{c}$ and $a + b = c$ , then the angle included between $\vec{a}$ and $\vec{b}$ is

1

90^{\circ}

2

180^{\circ}

3

120^{\circ}

4 zero

Explanation:

D Given that, $\vec{a} + \vec{b} = \vec{c}$ and $a + b = c$
$| \vec{c} | = | \vec{a} + \vec{b} |$
$c^{2} = a^{2} + b^{2} + 2 a b \cos θ$
$c = \sqrt{a^{2} + b^{2} + 2 a b \cos θ}$
$a + b = \sqrt{a^{2} + b^{2} + 2 a b \cos θ}$
$a^{2} + b^{2} + 2 a b = a^{2} + b^{2} + 2 a b \cos θ$
$\cos θ = 1$
$θ = 0^{\circ}$

WB JEE-2010

Motion in Plane

143598 The vectors are given by $\vec{A} = \hat{i} + 2 \hat{j} + 2 \hat{k}$ and $\vec{B} = 3 \hat{i} + 6 \hat{j} + 2 \hat{k}$ . Another vector $C$ has the same magnitude as $B$ but has the same direction as $A$ . Then which of the following vectors represents $C$ ?

1

\frac{7}{3} (\hat{i} + 2 \hat{j} + 2 \hat{k})

2

\frac{3}{7} (\hat{i} - 2 \hat{j} + 2 \hat{k})

3

\frac{7}{9} (\hat{i} - 2 \hat{j} + 2 \hat{k})

4

\frac{9}{7} (\hat{i} + 2 \hat{j} + 2 \hat{k})

Explanation:

A Given that, $\vec{A} = \hat{i} + 2 \hat{j} + 2 \hat{k}$
Magnitude of $| \vec{A} | = \sqrt{1^{2} + 2^{2} + 2^{2}} = 3$
Unit vector along $\vec{A}$ .
$\hat{A} = \frac{\vec{A}}{| \vec{A} |} = \frac{\hat{i} + 2 \hat{j} + 2 \hat{k}}{3}$
$\vec{B} = 3 \hat{i} + 6 \hat{j} + 2 \hat{k}$
Magnitude of
$| \vec{B} | = \sqrt{3^{2} + 6^{2} + 2^{2}}$
$= \sqrt{9 + 36 + 4}$
$| \vec{B} | = 7$
So, $C = | \vec{B} | \hat{A}$
$C = 7 \times \frac{\hat{i} + 2 \hat{j} + 2 \hat{k}}{3}$
$C = \frac{7}{3} (\hat{i} + 2 \hat{j} + 2 \hat{k})$

WB JEE 2013

Motion in Plane

143599 The (x,y,z) coordinates of two points $A$ and $B$ are given respectively as $(0, 3, - 1)$ and $(- 2, 6, 4)$ . The displacement vector form $A$ to $B$ may be given by :

1

- 2 \hat{i} + 6 \hat{j} + 4 \hat{k}

2

- 2 \hat{i} + 3 \hat{j} + 3 \hat{k}

3

- 2 \hat{i} + 3 \hat{j} + 5 \hat{k}

4

2 \hat{i} - 3 \hat{j} - 5 \hat{k}

Explanation:

C Given,
$r_{A} = 3 \hat{j} - \hat{k}, r_{B} = - 2 \hat{i} + 6 \hat{j} + 4 \hat{k}$
Displacement vector $(r_{A B}) = r_{B} - r_{A}$ .
$r_{A B} = (- 2 \hat{i} + 6 \hat{j} + 4 \hat{k}) - (3 \hat{j} - \hat{k})$
$r_{A B} = - 2 \hat{i} + 3 \hat{j} + 5 \hat{k}$
Therefore, the displacement vector from A to B is $- 2 \hat{i} + 3 \hat{j} + 5 \hat{k}$

BCECE-2006

Motion in Plane

143600 The resultant of two forces $P$ and $Q$ is of magnitude $P$ . If $P$ be doubled, the resultant will be inclined to $Q$ at an angle.

1

0^{0}

2

30^{\circ}

3

60^{\circ}

4

90^{\circ}

Explanation:

D Let the angle between $P$ and $Q$ be $θ$ .
Since, the resultant of $P$ and $Q$ is $P$
$∴ P^{2} = P^{2} + Q^{2} + 2 PQ \cos θ$
$- Q^{2} = 2 PQ \cos θ$
$Q = - 2 P \cos θ$
According to the question, given $P$ is doubled then $4 P^{2} = 4 P^{2} + Q^{2} + 4 PQ \cos θ$
Put the value of $Q$
$0 = (- 2 P \cos θ)^{2} + 4 P (- 2 P \cos θ) \cdot \cos θ$
$0 = 4 P^{2} \cos^{2} θ - 8 P^{2} \cos^{2} θ$
$0 = - 4 P^{2} \cos^{2} θ$
$\cos θ = 0$
$θ = 90^{\circ}$

UPSEE-2016

Motion in Plane

143601 If $\vec{a} + \vec{b} = \vec{c}$ and $a + b = c$ , then the angle included between $\vec{a}$ and $\vec{b}$ is

1

90^{\circ}

2

180^{\circ}

3

120^{\circ}

4 zero

Explanation:

D Given that, $\vec{a} + \vec{b} = \vec{c}$ and $a + b = c$
$| \vec{c} | = | \vec{a} + \vec{b} |$
$c^{2} = a^{2} + b^{2} + 2 a b \cos θ$
$c = \sqrt{a^{2} + b^{2} + 2 a b \cos θ}$
$a + b = \sqrt{a^{2} + b^{2} + 2 a b \cos θ}$
$a^{2} + b^{2} + 2 a b = a^{2} + b^{2} + 2 a b \cos θ$
$\cos θ = 1$
$θ = 0^{\circ}$

WB JEE-2010