QBManager

Motion in Plane

143593 If vectors $\vec{P} = a \hat{i} + a \hat{j} + 3 \hat{k}$ and $Q = a \hat{i} - 2 \hat{j} - \hat{k}$ are perpendicular to each other, then the positive value of $a$ is

1 zero

2 1

3 2

4 3

Explanation:

D If two vector are perpendicular to each other then their dot product is zero.
$\vec{P} \cdot \vec{Q} = 0$
$(a \hat{i} + a \hat{j} + 3 \hat{k}) \cdot (a \hat{i} - 2 \hat{j} - \hat{k}) = 0$
$a^{2} - 2 a - 3 = 0$
$a^{2} - 3 a + a - 3 = 0$
$a (a - 3) + 1 (a - 3) = 0$
$(a - 3) (a + 1) = 0$
$a = 3, a = - 1$
$a = 3, - 1$
So, positive value of $a$ is 3

AIIMS-2002

Motion in Plane

143594 Two equal vectors have a resultant equal to either of them, then the angle between them will be

1

110^{\circ}

2

120^{\circ}

3

60^{\circ}

4

150^{\circ}

Explanation:

B Let the two vector be A and B both at angle of $θ$ from each other with a resultant $R$
From the triangle law of vector addition,
$R^{2} = A^{2} + B^{2} + 2 A B \cos θ$
$A = B = R (given)$
$(A)^{2} = (A)^{2} + (A)^{2} + 2 \cdot A \cdot A \cdot \cos θ$
$A^{2} = 2 A^{2} (1 + \cos θ)$
$\cos θ = \frac{- 1}{2}$
$\cos θ = \cos 120^{\circ}$
$θ = 120^{\circ}$

AIIMS-2000

Motion in Plane

143596 Assertion: If $| \vec{A} + \vec{B} | = | \vec{A} - \vec{B} |$ , then the angle between $\vec{A}$ and $\vec{B}$ is $90^{\circ}$ .
Reason: $\vec{A} + \vec{B} = \vec{B} + \vec{A}$

1 If both assertion and reason are true and reason is the correct explanation of assertion.

2 If both assertion and reason are true but reason is not the correct explanation of assertion.

3 If assertion is true but reason is false.

4 If both assertion and reason are false.

Explanation:

B Given that, $| \vec{A} + \vec{B} | = | \vec{A} - \vec{B} |$
$\sqrt{A^{2} + B^{2} + 2 AB \cos θ} = \sqrt{A^{2} + B^{2} - 2 AB \cos θ}$
Squaring on both side
$A^{2} + B^{2} + 2 AB \cos θ = A^{2} + B^{2} - 2 AB \cos θ$
$A^{2} + B^{2} + 2 AB \cos θ - A^{2} - B^{2} + 2 AB \cos θ = 0$
$4 AB \cos θ = 0$
$θ = 90^{\circ}$
Also, vector addition is cumulative
Hence, $| \vec{A} + \vec{B} | = | \vec{B} + \vec{A} |$

AIIMS-26.05.2019(M) Shift-1

Motion in Plane

143597 If $A = 3 \hat{i} + 4 \hat{j}$ and $B = 7 \hat{i} + 24 \hat{j}$ then the vector having the same magnitude as $B$ and parallel to $A$ is-

1

15 \hat{i} + 20 \hat{j}

2

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

3

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

4

5 \hat{i} + 14 \hat{j}

Explanation:

A Given that,
$\vec{A} = 3 \hat{i} + 4 \hat{j} and \vec{B} = 7 \hat{i} + 24 \hat{j}$
to get direction we will find unit vector in the direction of $\vec{a}$ ,
$\hat{A} = \frac{\vec{A}}{| \vec{A} |} = \frac{3 \hat{i} + 4 \hat{j}}{\sqrt{3^{2} + 4^{2}}} = \frac{3}{5} \hat{i} + \frac{4}{5} \hat{j}$
So required vector,
$| \vec{B} | \hat{A} = \sqrt{7^{2} + 24^{2}} (\frac{3}{5} \hat{i} + \frac{4}{5} \hat{j})$
$= \sqrt{625} (\frac{3}{5} \hat{i} + \frac{4}{5} \hat{j}) = 25 \times \frac{1}{5} (3 \hat{i} + 4 \hat{j})$
$| \vec{B} | \hat{A} = 15 \hat{i} + 20 \hat{j}$

BCECE-2014

Motion in Plane

143593 If vectors $\vec{P} = a \hat{i} + a \hat{j} + 3 \hat{k}$ and $Q = a \hat{i} - 2 \hat{j} - \hat{k}$ are perpendicular to each other, then the positive value of $a$ is

1 zero

2 1

3 2

4 3

Explanation:

D If two vector are perpendicular to each other then their dot product is zero.
$\vec{P} \cdot \vec{Q} = 0$
$(a \hat{i} + a \hat{j} + 3 \hat{k}) \cdot (a \hat{i} - 2 \hat{j} - \hat{k}) = 0$
$a^{2} - 2 a - 3 = 0$
$a^{2} - 3 a + a - 3 = 0$
$a (a - 3) + 1 (a - 3) = 0$
$(a - 3) (a + 1) = 0$
$a = 3, a = - 1$
$a = 3, - 1$
So, positive value of $a$ is 3

AIIMS-2002

Motion in Plane

143594 Two equal vectors have a resultant equal to either of them, then the angle between them will be

1

110^{\circ}

2

120^{\circ}

3

60^{\circ}

4

150^{\circ}

Explanation:

B Let the two vector be A and B both at angle of $θ$ from each other with a resultant $R$
From the triangle law of vector addition,
$R^{2} = A^{2} + B^{2} + 2 A B \cos θ$
$A = B = R (given)$
$(A)^{2} = (A)^{2} + (A)^{2} + 2 \cdot A \cdot A \cdot \cos θ$
$A^{2} = 2 A^{2} (1 + \cos θ)$
$\cos θ = \frac{- 1}{2}$
$\cos θ = \cos 120^{\circ}$
$θ = 120^{\circ}$

AIIMS-2000

Motion in Plane

143596 Assertion: If $| \vec{A} + \vec{B} | = | \vec{A} - \vec{B} |$ , then the angle between $\vec{A}$ and $\vec{B}$ is $90^{\circ}$ .
Reason: $\vec{A} + \vec{B} = \vec{B} + \vec{A}$

1 If both assertion and reason are true and reason is the correct explanation of assertion.

2 If both assertion and reason are true but reason is not the correct explanation of assertion.

3 If assertion is true but reason is false.

4 If both assertion and reason are false.

Explanation:

B Given that, $| \vec{A} + \vec{B} | = | \vec{A} - \vec{B} |$
$\sqrt{A^{2} + B^{2} + 2 AB \cos θ} = \sqrt{A^{2} + B^{2} - 2 AB \cos θ}$
Squaring on both side
$A^{2} + B^{2} + 2 AB \cos θ = A^{2} + B^{2} - 2 AB \cos θ$
$A^{2} + B^{2} + 2 AB \cos θ - A^{2} - B^{2} + 2 AB \cos θ = 0$
$4 AB \cos θ = 0$
$θ = 90^{\circ}$
Also, vector addition is cumulative
Hence, $| \vec{A} + \vec{B} | = | \vec{B} + \vec{A} |$

AIIMS-26.05.2019(M) Shift-1

Motion in Plane

143597 If $A = 3 \hat{i} + 4 \hat{j}$ and $B = 7 \hat{i} + 24 \hat{j}$ then the vector having the same magnitude as $B$ and parallel to $A$ is-

1

15 \hat{i} + 20 \hat{j}

2

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

3

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

4

5 \hat{i} + 14 \hat{j}

Explanation:

A Given that,
$\vec{A} = 3 \hat{i} + 4 \hat{j} and \vec{B} = 7 \hat{i} + 24 \hat{j}$
to get direction we will find unit vector in the direction of $\vec{a}$ ,
$\hat{A} = \frac{\vec{A}}{| \vec{A} |} = \frac{3 \hat{i} + 4 \hat{j}}{\sqrt{3^{2} + 4^{2}}} = \frac{3}{5} \hat{i} + \frac{4}{5} \hat{j}$
So required vector,
$| \vec{B} | \hat{A} = \sqrt{7^{2} + 24^{2}} (\frac{3}{5} \hat{i} + \frac{4}{5} \hat{j})$
$= \sqrt{625} (\frac{3}{5} \hat{i} + \frac{4}{5} \hat{j}) = 25 \times \frac{1}{5} (3 \hat{i} + 4 \hat{j})$
$| \vec{B} | \hat{A} = 15 \hat{i} + 20 \hat{j}$

BCECE-2014

Motion in Plane

143593 If vectors $\vec{P} = a \hat{i} + a \hat{j} + 3 \hat{k}$ and $Q = a \hat{i} - 2 \hat{j} - \hat{k}$ are perpendicular to each other, then the positive value of $a$ is

1 zero

2 1

3 2

4 3

Explanation:

D If two vector are perpendicular to each other then their dot product is zero.
$\vec{P} \cdot \vec{Q} = 0$
$(a \hat{i} + a \hat{j} + 3 \hat{k}) \cdot (a \hat{i} - 2 \hat{j} - \hat{k}) = 0$
$a^{2} - 2 a - 3 = 0$
$a^{2} - 3 a + a - 3 = 0$
$a (a - 3) + 1 (a - 3) = 0$
$(a - 3) (a + 1) = 0$
$a = 3, a = - 1$
$a = 3, - 1$
So, positive value of $a$ is 3

AIIMS-2002

Motion in Plane

143594 Two equal vectors have a resultant equal to either of them, then the angle between them will be

1

110^{\circ}

2

120^{\circ}

3

60^{\circ}

4

150^{\circ}

Explanation:

B Let the two vector be A and B both at angle of $θ$ from each other with a resultant $R$
From the triangle law of vector addition,
$R^{2} = A^{2} + B^{2} + 2 A B \cos θ$
$A = B = R (given)$
$(A)^{2} = (A)^{2} + (A)^{2} + 2 \cdot A \cdot A \cdot \cos θ$
$A^{2} = 2 A^{2} (1 + \cos θ)$
$\cos θ = \frac{- 1}{2}$
$\cos θ = \cos 120^{\circ}$
$θ = 120^{\circ}$

AIIMS-2000

Motion in Plane

143596 Assertion: If $| \vec{A} + \vec{B} | = | \vec{A} - \vec{B} |$ , then the angle between $\vec{A}$ and $\vec{B}$ is $90^{\circ}$ .
Reason: $\vec{A} + \vec{B} = \vec{B} + \vec{A}$

1 If both assertion and reason are true and reason is the correct explanation of assertion.

2 If both assertion and reason are true but reason is not the correct explanation of assertion.

3 If assertion is true but reason is false.

4 If both assertion and reason are false.

Explanation:

B Given that, $| \vec{A} + \vec{B} | = | \vec{A} - \vec{B} |$
$\sqrt{A^{2} + B^{2} + 2 AB \cos θ} = \sqrt{A^{2} + B^{2} - 2 AB \cos θ}$
Squaring on both side
$A^{2} + B^{2} + 2 AB \cos θ = A^{2} + B^{2} - 2 AB \cos θ$
$A^{2} + B^{2} + 2 AB \cos θ - A^{2} - B^{2} + 2 AB \cos θ = 0$
$4 AB \cos θ = 0$
$θ = 90^{\circ}$
Also, vector addition is cumulative
Hence, $| \vec{A} + \vec{B} | = | \vec{B} + \vec{A} |$

AIIMS-26.05.2019(M) Shift-1

Motion in Plane

143597 If $A = 3 \hat{i} + 4 \hat{j}$ and $B = 7 \hat{i} + 24 \hat{j}$ then the vector having the same magnitude as $B$ and parallel to $A$ is-

1

15 \hat{i} + 20 \hat{j}

2

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

3

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

4

5 \hat{i} + 14 \hat{j}

Explanation:

A Given that,
$\vec{A} = 3 \hat{i} + 4 \hat{j} and \vec{B} = 7 \hat{i} + 24 \hat{j}$
to get direction we will find unit vector in the direction of $\vec{a}$ ,
$\hat{A} = \frac{\vec{A}}{| \vec{A} |} = \frac{3 \hat{i} + 4 \hat{j}}{\sqrt{3^{2} + 4^{2}}} = \frac{3}{5} \hat{i} + \frac{4}{5} \hat{j}$
So required vector,
$| \vec{B} | \hat{A} = \sqrt{7^{2} + 24^{2}} (\frac{3}{5} \hat{i} + \frac{4}{5} \hat{j})$
$= \sqrt{625} (\frac{3}{5} \hat{i} + \frac{4}{5} \hat{j}) = 25 \times \frac{1}{5} (3 \hat{i} + 4 \hat{j})$
$| \vec{B} | \hat{A} = 15 \hat{i} + 20 \hat{j}$

BCECE-2014

Motion in Plane

143593 If vectors $\vec{P} = a \hat{i} + a \hat{j} + 3 \hat{k}$ and $Q = a \hat{i} - 2 \hat{j} - \hat{k}$ are perpendicular to each other, then the positive value of $a$ is

1 zero

2 1

3 2

4 3

Explanation:

D If two vector are perpendicular to each other then their dot product is zero.
$\vec{P} \cdot \vec{Q} = 0$
$(a \hat{i} + a \hat{j} + 3 \hat{k}) \cdot (a \hat{i} - 2 \hat{j} - \hat{k}) = 0$
$a^{2} - 2 a - 3 = 0$
$a^{2} - 3 a + a - 3 = 0$
$a (a - 3) + 1 (a - 3) = 0$
$(a - 3) (a + 1) = 0$
$a = 3, a = - 1$
$a = 3, - 1$
So, positive value of $a$ is 3

AIIMS-2002

Motion in Plane

143594 Two equal vectors have a resultant equal to either of them, then the angle between them will be

1

110^{\circ}

2

120^{\circ}

3

60^{\circ}

4

150^{\circ}

Explanation:

B Let the two vector be A and B both at angle of $θ$ from each other with a resultant $R$
From the triangle law of vector addition,
$R^{2} = A^{2} + B^{2} + 2 A B \cos θ$
$A = B = R (given)$
$(A)^{2} = (A)^{2} + (A)^{2} + 2 \cdot A \cdot A \cdot \cos θ$
$A^{2} = 2 A^{2} (1 + \cos θ)$
$\cos θ = \frac{- 1}{2}$
$\cos θ = \cos 120^{\circ}$
$θ = 120^{\circ}$

AIIMS-2000

Motion in Plane

143596 Assertion: If $| \vec{A} + \vec{B} | = | \vec{A} - \vec{B} |$ , then the angle between $\vec{A}$ and $\vec{B}$ is $90^{\circ}$ .
Reason: $\vec{A} + \vec{B} = \vec{B} + \vec{A}$

1 If both assertion and reason are true and reason is the correct explanation of assertion.

2 If both assertion and reason are true but reason is not the correct explanation of assertion.

3 If assertion is true but reason is false.

4 If both assertion and reason are false.

Explanation:

B Given that, $| \vec{A} + \vec{B} | = | \vec{A} - \vec{B} |$
$\sqrt{A^{2} + B^{2} + 2 AB \cos θ} = \sqrt{A^{2} + B^{2} - 2 AB \cos θ}$
Squaring on both side
$A^{2} + B^{2} + 2 AB \cos θ = A^{2} + B^{2} - 2 AB \cos θ$
$A^{2} + B^{2} + 2 AB \cos θ - A^{2} - B^{2} + 2 AB \cos θ = 0$
$4 AB \cos θ = 0$
$θ = 90^{\circ}$
Also, vector addition is cumulative
Hence, $| \vec{A} + \vec{B} | = | \vec{B} + \vec{A} |$

AIIMS-26.05.2019(M) Shift-1

Motion in Plane

143597 If $A = 3 \hat{i} + 4 \hat{j}$ and $B = 7 \hat{i} + 24 \hat{j}$ then the vector having the same magnitude as $B$ and parallel to $A$ is-

1

15 \hat{i} + 20 \hat{j}

2

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

3

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

4

5 \hat{i} + 14 \hat{j}

Explanation:

A Given that,
$\vec{A} = 3 \hat{i} + 4 \hat{j} and \vec{B} = 7 \hat{i} + 24 \hat{j}$
to get direction we will find unit vector in the direction of $\vec{a}$ ,
$\hat{A} = \frac{\vec{A}}{| \vec{A} |} = \frac{3 \hat{i} + 4 \hat{j}}{\sqrt{3^{2} + 4^{2}}} = \frac{3}{5} \hat{i} + \frac{4}{5} \hat{j}$
So required vector,
$| \vec{B} | \hat{A} = \sqrt{7^{2} + 24^{2}} (\frac{3}{5} \hat{i} + \frac{4}{5} \hat{j})$
$= \sqrt{625} (\frac{3}{5} \hat{i} + \frac{4}{5} \hat{j}) = 25 \times \frac{1}{5} (3 \hat{i} + 4 \hat{j})$
$| \vec{B} | \hat{A} = 15 \hat{i} + 20 \hat{j}$

BCECE-2014

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