QBManager

Motion in Plane

143524 If $\vec{A}, \vec{B}$ are perpendicular vectors
$\vec{A} = 5 \hat{i} + 7 \hat{j} - 3 \hat{k}$
$\vec{B} = 2 \hat{i} + 2 \hat{j} - c \hat{k}$
The value of $c$ is

1 -2

2 8

3 -7

4 -8

Explanation:

D $\vec{A} = 5 \hat{i} + 7 \hat{j} - 3 \hat{k}$
$\vec{B} = 2 \hat{i} + 2 \hat{j} - c \hat{k}$
For perpendicular vectors
$\vec{A} \cdot \vec{B} = 0$
$(5 \hat{i} + 7 \hat{j} - 3 \hat{k}) \cdot (2 \hat{i} + 2 \hat{j} - ck) = 0$
$10 + 14 + 3 c = 0$
$24 = - 3 c$
$c = - 8$

EAMCET-1991

Motion in Plane

143525 The resultant of the vectors $A$ and $B$ depends also on the angle $θ$ between them. The magnitude of the resultant is always given by

1

A + B + 2 AB \cos θ

2

\sqrt{(A + B + 2 AB \cos θ)}

3

\sqrt{A^{2} + B^{2} + 2 AB \cos θ}

4

(A^{2} + B^{2} + 2 AB \cos θ)

Explanation:

C
Resultant vector, $\vec{R}$
From $△ DOF$ ,
$OD)^{2} = (OF)^{2} + (DF)^{2}$
$(OD)^{2} = (OE + EF)^{2} + (DF)^{2}$
${\vec{R}}^{2} = (\vec{A} + \vec{B} \cos θ)^{2} + (\vec{B} \sin θ)^{2}$
$= A^{2} + B^{2} \cos^{2} θ + 2 AB \cos θ + B^{2} \sin^{2} θ$
$= A^{2} + B^{2} (\cos^{2} θ + \sin^{2} θ) + 2 AB \cos θ$
${\vec{R}}^{2} = A^{2} + B^{2} + 2 AB \cos θ$
$| \vec{R} | = \sqrt{A^{2} + B^{2} + 2 AB \cos θ}$

EAMCET-1992

Motion in Plane

143526 $\vec{A}$ and $\vec{B}$ are vectors such that $| \vec{A} + \vec{B} | =$ $| \vec{A} - \vec{B} |$ . Then, the angle between them is

1

90^{\circ}

2

60^{\circ}

3

45^{\circ}

4

0^{\circ}

Explanation:

A Given,
$| \vec{A} + \vec{B} | = | \vec{A} - \vec{B} |$
Squaring both sides,
$| \vec{A} + \vec{B} |^{2} = | \vec{A} - \vec{B} |^{2}$
${\vec{A}}^{2} + {\vec{B}}^{2} + 2 \vec{A} \cdot \vec{B} = {\vec{A}}^{2} + {\vec{B}}^{2} - 2 \vec{A} \cdot \vec{B}$
$4 \vec{A} \cdot \vec{B} = 0$
$\vec{A} \cdot \vec{B} = 0$
$| \vec{A} ‖ \vec{B} | \cos θ = 0$
$\cos θ = 0$
$θ = 90^{\circ}$

EAMCET-1993

Motion in Plane

143527 When two vectors $\vec{A}$ and $\vec{B}$ of magnitude a and $b$ are added, the magnitude of the resultant vector is always

1 equal to

(a + b)

2 less than

(a + b)

3 greater than

(a + b)

4 not greater than

(a + b)

Explanation:

D Given,
$| \vec{A} | = a, | \vec{B} | = b$
$| \vec{A} + \vec{B} | = \sqrt{a^{2} + b^{2} + 2 ab \cos θ}$
$| \vec{A} + \vec{B} |_{max} = \sqrt{a^{2} + b^{2} + 2 ab} [For max, θ = 0]$
$| \vec{A} + \vec{B} |_{max} = (a + b)$
Hence, magnitude of resultant vector is not greater than $(a + b)$

EAMCET-1993

Motion in Plane

143529 The angle made by the vector $\vec{A} = \hat{i} + \hat{j}$ with $x$ axis is

1

90^{\circ}

2

45^{\circ}

3

{22.5}^{\circ}

4

30^{\circ}

Explanation:

B Given that,
$\vec{A} = \hat{i} + \hat{j}$
$| \vec{A} | = \sqrt{1^{2} + 1^{2}} = \sqrt{2}$
$A_{x} = 1, A_{y} = 1$
If $θ$ is the angle made by the vector with $x$ -axis than,
$\cos θ = \frac{A_{x}}{| \vec{A} |} \Rightarrow \cos θ = \frac{1}{\sqrt{2}}$
$θ = 45^{\circ}$

EAMCET-1996

Motion in Plane

143524 If $\vec{A}, \vec{B}$ are perpendicular vectors
$\vec{A} = 5 \hat{i} + 7 \hat{j} - 3 \hat{k}$
$\vec{B} = 2 \hat{i} + 2 \hat{j} - c \hat{k}$
The value of $c$ is

1 -2

2 8

3 -7

4 -8

Explanation:

D $\vec{A} = 5 \hat{i} + 7 \hat{j} - 3 \hat{k}$
$\vec{B} = 2 \hat{i} + 2 \hat{j} - c \hat{k}$
For perpendicular vectors
$\vec{A} \cdot \vec{B} = 0$
$(5 \hat{i} + 7 \hat{j} - 3 \hat{k}) \cdot (2 \hat{i} + 2 \hat{j} - ck) = 0$
$10 + 14 + 3 c = 0$
$24 = - 3 c$
$c = - 8$

EAMCET-1991

Motion in Plane

143525 The resultant of the vectors $A$ and $B$ depends also on the angle $θ$ between them. The magnitude of the resultant is always given by

1

A + B + 2 AB \cos θ

2

\sqrt{(A + B + 2 AB \cos θ)}

3

\sqrt{A^{2} + B^{2} + 2 AB \cos θ}

4

(A^{2} + B^{2} + 2 AB \cos θ)

Explanation:

C
Resultant vector, $\vec{R}$
From $△ DOF$ ,
$OD)^{2} = (OF)^{2} + (DF)^{2}$
$(OD)^{2} = (OE + EF)^{2} + (DF)^{2}$
${\vec{R}}^{2} = (\vec{A} + \vec{B} \cos θ)^{2} + (\vec{B} \sin θ)^{2}$
$= A^{2} + B^{2} \cos^{2} θ + 2 AB \cos θ + B^{2} \sin^{2} θ$
$= A^{2} + B^{2} (\cos^{2} θ + \sin^{2} θ) + 2 AB \cos θ$
${\vec{R}}^{2} = A^{2} + B^{2} + 2 AB \cos θ$
$| \vec{R} | = \sqrt{A^{2} + B^{2} + 2 AB \cos θ}$

EAMCET-1992

Motion in Plane

143526 $\vec{A}$ and $\vec{B}$ are vectors such that $| \vec{A} + \vec{B} | =$ $| \vec{A} - \vec{B} |$ . Then, the angle between them is

1

90^{\circ}

2

60^{\circ}

3

45^{\circ}

4

0^{\circ}

Explanation:

A Given,
$| \vec{A} + \vec{B} | = | \vec{A} - \vec{B} |$
Squaring both sides,
$| \vec{A} + \vec{B} |^{2} = | \vec{A} - \vec{B} |^{2}$
${\vec{A}}^{2} + {\vec{B}}^{2} + 2 \vec{A} \cdot \vec{B} = {\vec{A}}^{2} + {\vec{B}}^{2} - 2 \vec{A} \cdot \vec{B}$
$4 \vec{A} \cdot \vec{B} = 0$
$\vec{A} \cdot \vec{B} = 0$
$| \vec{A} ‖ \vec{B} | \cos θ = 0$
$\cos θ = 0$
$θ = 90^{\circ}$

EAMCET-1993

Motion in Plane

143527 When two vectors $\vec{A}$ and $\vec{B}$ of magnitude a and $b$ are added, the magnitude of the resultant vector is always

1 equal to

(a + b)

2 less than

(a + b)

3 greater than

(a + b)

4 not greater than

(a + b)

Explanation:

D Given,
$| \vec{A} | = a, | \vec{B} | = b$
$| \vec{A} + \vec{B} | = \sqrt{a^{2} + b^{2} + 2 ab \cos θ}$
$| \vec{A} + \vec{B} |_{max} = \sqrt{a^{2} + b^{2} + 2 ab} [For max, θ = 0]$
$| \vec{A} + \vec{B} |_{max} = (a + b)$
Hence, magnitude of resultant vector is not greater than $(a + b)$

EAMCET-1993

Motion in Plane

143529 The angle made by the vector $\vec{A} = \hat{i} + \hat{j}$ with $x$ axis is

1

90^{\circ}

2

45^{\circ}

3

{22.5}^{\circ}

4

30^{\circ}

Explanation:

B Given that,
$\vec{A} = \hat{i} + \hat{j}$
$| \vec{A} | = \sqrt{1^{2} + 1^{2}} = \sqrt{2}$
$A_{x} = 1, A_{y} = 1$
If $θ$ is the angle made by the vector with $x$ -axis than,
$\cos θ = \frac{A_{x}}{| \vec{A} |} \Rightarrow \cos θ = \frac{1}{\sqrt{2}}$
$θ = 45^{\circ}$

EAMCET-1996

Motion in Plane

143524 If $\vec{A}, \vec{B}$ are perpendicular vectors
$\vec{A} = 5 \hat{i} + 7 \hat{j} - 3 \hat{k}$
$\vec{B} = 2 \hat{i} + 2 \hat{j} - c \hat{k}$
The value of $c$ is

1 -2

2 8

3 -7

4 -8

Explanation:

D $\vec{A} = 5 \hat{i} + 7 \hat{j} - 3 \hat{k}$
$\vec{B} = 2 \hat{i} + 2 \hat{j} - c \hat{k}$
For perpendicular vectors
$\vec{A} \cdot \vec{B} = 0$
$(5 \hat{i} + 7 \hat{j} - 3 \hat{k}) \cdot (2 \hat{i} + 2 \hat{j} - ck) = 0$
$10 + 14 + 3 c = 0$
$24 = - 3 c$
$c = - 8$

EAMCET-1991

Motion in Plane

143525 The resultant of the vectors $A$ and $B$ depends also on the angle $θ$ between them. The magnitude of the resultant is always given by

1

A + B + 2 AB \cos θ

2

\sqrt{(A + B + 2 AB \cos θ)}

3

\sqrt{A^{2} + B^{2} + 2 AB \cos θ}

4

(A^{2} + B^{2} + 2 AB \cos θ)

Explanation:

C
Resultant vector, $\vec{R}$
From $△ DOF$ ,
$OD)^{2} = (OF)^{2} + (DF)^{2}$
$(OD)^{2} = (OE + EF)^{2} + (DF)^{2}$
${\vec{R}}^{2} = (\vec{A} + \vec{B} \cos θ)^{2} + (\vec{B} \sin θ)^{2}$
$= A^{2} + B^{2} \cos^{2} θ + 2 AB \cos θ + B^{2} \sin^{2} θ$
$= A^{2} + B^{2} (\cos^{2} θ + \sin^{2} θ) + 2 AB \cos θ$
${\vec{R}}^{2} = A^{2} + B^{2} + 2 AB \cos θ$
$| \vec{R} | = \sqrt{A^{2} + B^{2} + 2 AB \cos θ}$

EAMCET-1992

Motion in Plane

143526 $\vec{A}$ and $\vec{B}$ are vectors such that $| \vec{A} + \vec{B} | =$ $| \vec{A} - \vec{B} |$ . Then, the angle between them is

1

90^{\circ}

2

60^{\circ}

3

45^{\circ}

4

0^{\circ}

Explanation:

A Given,
$| \vec{A} + \vec{B} | = | \vec{A} - \vec{B} |$
Squaring both sides,
$| \vec{A} + \vec{B} |^{2} = | \vec{A} - \vec{B} |^{2}$
${\vec{A}}^{2} + {\vec{B}}^{2} + 2 \vec{A} \cdot \vec{B} = {\vec{A}}^{2} + {\vec{B}}^{2} - 2 \vec{A} \cdot \vec{B}$
$4 \vec{A} \cdot \vec{B} = 0$
$\vec{A} \cdot \vec{B} = 0$
$| \vec{A} ‖ \vec{B} | \cos θ = 0$
$\cos θ = 0$
$θ = 90^{\circ}$

EAMCET-1993

Motion in Plane

143527 When two vectors $\vec{A}$ and $\vec{B}$ of magnitude a and $b$ are added, the magnitude of the resultant vector is always

1 equal to

(a + b)

2 less than

(a + b)

3 greater than

(a + b)

4 not greater than

(a + b)

Explanation:

D Given,
$| \vec{A} | = a, | \vec{B} | = b$
$| \vec{A} + \vec{B} | = \sqrt{a^{2} + b^{2} + 2 ab \cos θ}$
$| \vec{A} + \vec{B} |_{max} = \sqrt{a^{2} + b^{2} + 2 ab} [For max, θ = 0]$
$| \vec{A} + \vec{B} |_{max} = (a + b)$
Hence, magnitude of resultant vector is not greater than $(a + b)$

EAMCET-1993

Motion in Plane

143529 The angle made by the vector $\vec{A} = \hat{i} + \hat{j}$ with $x$ axis is

1

90^{\circ}

2

45^{\circ}

3

{22.5}^{\circ}

4

30^{\circ}

Explanation:

B Given that,
$\vec{A} = \hat{i} + \hat{j}$
$| \vec{A} | = \sqrt{1^{2} + 1^{2}} = \sqrt{2}$
$A_{x} = 1, A_{y} = 1$
If $θ$ is the angle made by the vector with $x$ -axis than,
$\cos θ = \frac{A_{x}}{| \vec{A} |} \Rightarrow \cos θ = \frac{1}{\sqrt{2}}$
$θ = 45^{\circ}$

EAMCET-1996

Motion in Plane

143524 If $\vec{A}, \vec{B}$ are perpendicular vectors
$\vec{A} = 5 \hat{i} + 7 \hat{j} - 3 \hat{k}$
$\vec{B} = 2 \hat{i} + 2 \hat{j} - c \hat{k}$
The value of $c$ is

1 -2

2 8

3 -7

4 -8

Explanation:

D $\vec{A} = 5 \hat{i} + 7 \hat{j} - 3 \hat{k}$
$\vec{B} = 2 \hat{i} + 2 \hat{j} - c \hat{k}$
For perpendicular vectors
$\vec{A} \cdot \vec{B} = 0$
$(5 \hat{i} + 7 \hat{j} - 3 \hat{k}) \cdot (2 \hat{i} + 2 \hat{j} - ck) = 0$
$10 + 14 + 3 c = 0$
$24 = - 3 c$
$c = - 8$

EAMCET-1991

Motion in Plane

143525 The resultant of the vectors $A$ and $B$ depends also on the angle $θ$ between them. The magnitude of the resultant is always given by

1

A + B + 2 AB \cos θ

2

\sqrt{(A + B + 2 AB \cos θ)}

3

\sqrt{A^{2} + B^{2} + 2 AB \cos θ}

4

(A^{2} + B^{2} + 2 AB \cos θ)

Explanation:

C
Resultant vector, $\vec{R}$
From $△ DOF$ ,
$OD)^{2} = (OF)^{2} + (DF)^{2}$
$(OD)^{2} = (OE + EF)^{2} + (DF)^{2}$
${\vec{R}}^{2} = (\vec{A} + \vec{B} \cos θ)^{2} + (\vec{B} \sin θ)^{2}$
$= A^{2} + B^{2} \cos^{2} θ + 2 AB \cos θ + B^{2} \sin^{2} θ$
$= A^{2} + B^{2} (\cos^{2} θ + \sin^{2} θ) + 2 AB \cos θ$
${\vec{R}}^{2} = A^{2} + B^{2} + 2 AB \cos θ$
$| \vec{R} | = \sqrt{A^{2} + B^{2} + 2 AB \cos θ}$

EAMCET-1992

Motion in Plane

143526 $\vec{A}$ and $\vec{B}$ are vectors such that $| \vec{A} + \vec{B} | =$ $| \vec{A} - \vec{B} |$ . Then, the angle between them is

1

90^{\circ}

2

60^{\circ}

3

45^{\circ}

4

0^{\circ}

Explanation:

A Given,
$| \vec{A} + \vec{B} | = | \vec{A} - \vec{B} |$
Squaring both sides,
$| \vec{A} + \vec{B} |^{2} = | \vec{A} - \vec{B} |^{2}$
${\vec{A}}^{2} + {\vec{B}}^{2} + 2 \vec{A} \cdot \vec{B} = {\vec{A}}^{2} + {\vec{B}}^{2} - 2 \vec{A} \cdot \vec{B}$
$4 \vec{A} \cdot \vec{B} = 0$
$\vec{A} \cdot \vec{B} = 0$
$| \vec{A} ‖ \vec{B} | \cos θ = 0$
$\cos θ = 0$
$θ = 90^{\circ}$

EAMCET-1993

Motion in Plane

143527 When two vectors $\vec{A}$ and $\vec{B}$ of magnitude a and $b$ are added, the magnitude of the resultant vector is always

1 equal to

(a + b)

2 less than

(a + b)

3 greater than

(a + b)

4 not greater than

(a + b)

Explanation:

D Given,
$| \vec{A} | = a, | \vec{B} | = b$
$| \vec{A} + \vec{B} | = \sqrt{a^{2} + b^{2} + 2 ab \cos θ}$
$| \vec{A} + \vec{B} |_{max} = \sqrt{a^{2} + b^{2} + 2 ab} [For max, θ = 0]$
$| \vec{A} + \vec{B} |_{max} = (a + b)$
Hence, magnitude of resultant vector is not greater than $(a + b)$

EAMCET-1993

Motion in Plane

143529 The angle made by the vector $\vec{A} = \hat{i} + \hat{j}$ with $x$ axis is

1

90^{\circ}

2

45^{\circ}

3

{22.5}^{\circ}

4

30^{\circ}

Explanation:

B Given that,
$\vec{A} = \hat{i} + \hat{j}$
$| \vec{A} | = \sqrt{1^{2} + 1^{2}} = \sqrt{2}$
$A_{x} = 1, A_{y} = 1$
If $θ$ is the angle made by the vector with $x$ -axis than,
$\cos θ = \frac{A_{x}}{| \vec{A} |} \Rightarrow \cos θ = \frac{1}{\sqrt{2}}$
$θ = 45^{\circ}$

EAMCET-1996

Motion in Plane

143524 If $\vec{A}, \vec{B}$ are perpendicular vectors
$\vec{A} = 5 \hat{i} + 7 \hat{j} - 3 \hat{k}$
$\vec{B} = 2 \hat{i} + 2 \hat{j} - c \hat{k}$
The value of $c$ is

1 -2

2 8

3 -7

4 -8

Explanation:

D $\vec{A} = 5 \hat{i} + 7 \hat{j} - 3 \hat{k}$
$\vec{B} = 2 \hat{i} + 2 \hat{j} - c \hat{k}$
For perpendicular vectors
$\vec{A} \cdot \vec{B} = 0$
$(5 \hat{i} + 7 \hat{j} - 3 \hat{k}) \cdot (2 \hat{i} + 2 \hat{j} - ck) = 0$
$10 + 14 + 3 c = 0$
$24 = - 3 c$
$c = - 8$

EAMCET-1991

Motion in Plane

143525 The resultant of the vectors $A$ and $B$ depends also on the angle $θ$ between them. The magnitude of the resultant is always given by

1

A + B + 2 AB \cos θ

2

\sqrt{(A + B + 2 AB \cos θ)}

3

\sqrt{A^{2} + B^{2} + 2 AB \cos θ}

4

(A^{2} + B^{2} + 2 AB \cos θ)

Explanation:

C
Resultant vector, $\vec{R}$
From $△ DOF$ ,
$OD)^{2} = (OF)^{2} + (DF)^{2}$
$(OD)^{2} = (OE + EF)^{2} + (DF)^{2}$
${\vec{R}}^{2} = (\vec{A} + \vec{B} \cos θ)^{2} + (\vec{B} \sin θ)^{2}$
$= A^{2} + B^{2} \cos^{2} θ + 2 AB \cos θ + B^{2} \sin^{2} θ$
$= A^{2} + B^{2} (\cos^{2} θ + \sin^{2} θ) + 2 AB \cos θ$
${\vec{R}}^{2} = A^{2} + B^{2} + 2 AB \cos θ$
$| \vec{R} | = \sqrt{A^{2} + B^{2} + 2 AB \cos θ}$

EAMCET-1992

Motion in Plane

143526 $\vec{A}$ and $\vec{B}$ are vectors such that $| \vec{A} + \vec{B} | =$ $| \vec{A} - \vec{B} |$ . Then, the angle between them is

1

90^{\circ}

2

60^{\circ}

3

45^{\circ}

4

0^{\circ}

Explanation:

A Given,
$| \vec{A} + \vec{B} | = | \vec{A} - \vec{B} |$
Squaring both sides,
$| \vec{A} + \vec{B} |^{2} = | \vec{A} - \vec{B} |^{2}$
${\vec{A}}^{2} + {\vec{B}}^{2} + 2 \vec{A} \cdot \vec{B} = {\vec{A}}^{2} + {\vec{B}}^{2} - 2 \vec{A} \cdot \vec{B}$
$4 \vec{A} \cdot \vec{B} = 0$
$\vec{A} \cdot \vec{B} = 0$
$| \vec{A} ‖ \vec{B} | \cos θ = 0$
$\cos θ = 0$
$θ = 90^{\circ}$

EAMCET-1993

Motion in Plane

143527 When two vectors $\vec{A}$ and $\vec{B}$ of magnitude a and $b$ are added, the magnitude of the resultant vector is always

1 equal to

(a + b)

2 less than

(a + b)

3 greater than

(a + b)

4 not greater than

(a + b)

Explanation:

D Given,
$| \vec{A} | = a, | \vec{B} | = b$
$| \vec{A} + \vec{B} | = \sqrt{a^{2} + b^{2} + 2 ab \cos θ}$
$| \vec{A} + \vec{B} |_{max} = \sqrt{a^{2} + b^{2} + 2 ab} [For max, θ = 0]$
$| \vec{A} + \vec{B} |_{max} = (a + b)$
Hence, magnitude of resultant vector is not greater than $(a + b)$

EAMCET-1993

Motion in Plane

143529 The angle made by the vector $\vec{A} = \hat{i} + \hat{j}$ with $x$ axis is

1

90^{\circ}

2

45^{\circ}

3

{22.5}^{\circ}

4

30^{\circ}

Explanation:

B Given that,
$\vec{A} = \hat{i} + \hat{j}$
$| \vec{A} | = \sqrt{1^{2} + 1^{2}} = \sqrt{2}$
$A_{x} = 1, A_{y} = 1$
If $θ$ is the angle made by the vector with $x$ -axis than,
$\cos θ = \frac{A_{x}}{| \vec{A} |} \Rightarrow \cos θ = \frac{1}{\sqrt{2}}$
$θ = 45^{\circ}$

EAMCET-1996

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