QBManager

Motion in Plane

143636 Which of the following relations is true for two unit vector $\hat{A}$ and $\hat{B}$ making an angle $θ$ to each other?

1

| \hat{A} + \hat{B} | = | \hat{A} - \hat{B} | \tan \frac{θ}{2}

2

| \hat{A} - \hat{B} | = | \hat{A} + \hat{B} | \tan \frac{θ}{2}

3

| \hat{A} + \hat{B} | = | \hat{A} - \hat{B} | \cos \frac{θ}{2}

4

| \hat{A} - \hat{B} | = | \hat{A} + \hat{B} | \cos \frac{θ}{2}

Explanation:

B Given that,
$| \hat{A} + \hat{B} | = \sqrt{| \hat{A} |^{2} + | \hat{B} |^{2} + 2 | \hat{A} | | \hat{B} | \cos θ}$
$| \hat{A} | = 1 | \hat{B} |^{2} = 1$
$| \hat{A} + \hat{B} | = \sqrt{(1)^{2} + (1)^{2} + 2 \cos θ}$
$= \sqrt{2 + 2 \cos θ}$
$= \sqrt{2 + 2 (2 \cos^{2} \frac{θ}{2} - 1)}$
$= \sqrt{2 (1 + 2 \cos^{2} \frac{θ}{2} - 1)} = 2 \cos \frac{θ}{2}$
$\hat{A} + \hat{B} = 2 \cos \frac{θ}{2}$
$| \hat{A} - \hat{B} | = \sqrt{| \hat{A} |^{2} + | \hat{B} |^{2} - 2 | \hat{A} | | \hat{B} | \cos θ}$
$| \hat{A} - \hat{B} | = \sqrt{1 + 1 - 2 \cos θ}$
$| \hat{A} - \hat{B} | = \sqrt{2 - 2 \cos θ}$
$= \sqrt{2 (1 - \cos θ)}$
$= \sqrt{2 (1 - (1 - 2 \sin^{2} \frac{θ}{2})}$
$= \sqrt{2 (1 - 1 + 2 \sin^{2} \frac{θ}{2})}$
$= \sqrt{2 (1 - 1 + 2 \sin^{2} \frac{θ}{2})} = 2 \sin \frac{θ}{2}$
$\hat{A} - \hat{B} = 2 \sin \frac{θ}{2}$
Equation (i) divide by equation (ii), we get -
$\frac{| \hat{A} - \hat{B} |}{| \hat{A} + \hat{B} |} = \frac{2 \sin \frac{θ}{2}}{2 \cos \frac{θ}{2}}$
$\frac{| \hat{A} - \hat{B} |}{| \hat{A} + \hat{B} |} = \tan \frac{θ}{2}$
$| \hat{A} - \hat{B} | = | \hat{A} + \hat{B} | \tan \frac{θ}{2}$

Shift-I]

Motion in Plane

1 (A)

\to

(iv), (B)

\to

(i), (C)

\to

(iii), (D)

\to

(ii)

2 (A)

\to

(iv), (B)

\to

(iii), (C)

\to

(i), (D)

\to

(ii)

3 (A)

\to

(iii), (B)

\to

(ii), (C)

\to

(iv), (D)

\to

(i)

4 (A)

\to

(i), (B)

\to

(iv), (C)

\to

(ii), (D)

\to

(iii)

Explanation:

B Applying triangle law of vectors to the diagram.
(i) $\vec{B} = \vec{A} + \vec{C}$
$∴ \vec{B} - \vec{A} - \vec{C} = 0$
This matches with (C).
(ii) $- \vec{C} = \vec{A} + \vec{B}$
This matches with (D).
(iii) $\vec{C} + \vec{B} = \vec{A}$
$∴ \vec{A} - \vec{C} = \vec{B}$
This matches with (B).
(iv) $\vec{A} + \vec{B} = \vec{C}$
$∴ \vec{C} - \vec{A} - \vec{B} = 0$

JEE Main-25.07.2021

Motion in Plane

143638 Two vectors $P$ and $Q$ have equal magnitudes. If the magnitude of $P + Q$ is $n$ times the magnitude of $P - Q$ , then angle between $P$ and $Q$ is

1

\sin^{- 1} (\frac{n - 1}{n + 1})

2

\cos^{- 1} (\frac{n - 1}{n + 1})

3

\sin^{- 1} (\frac{n^{2} - 1}{n^{2} + 1})

4

\cos^{- 1} (\frac{n^{2} - 1}{n^{2} + 1})

Explanation:

D Given that,
$P = Q$
Let the magnitude of $(P + Q) = R$
$R = | P + Q | = \sqrt{P^{2} + Q^{2} + 2 PQ \cos θ}$
$R = \sqrt{P^{2} + P^{2} + 2 P^{2} \cos θ}$ [from (i)]
The magnitude of $(P - Q) = R^{'}$
$R^{'} = | P - Q | = \sqrt{P^{2} + Q^{2} - 2 PQ \cos θ}$
$R^{'} = | P - Q | = \sqrt{P^{2} + P^{2} - 2 P^{2} \cos θ}$ [from (i)]
Given that
$R = {nR}^{'}$
$\sqrt{2 P^{2} + 2 P^{2} \cos θ} = n \sqrt{2 P^{2} - 2 P^{2} \cos θ}$
squaring both side
$2 p^{2} + 2 p^{2} \cos θ = n^{2} (2 p^{2} - 2 p^{2} \cos θ)$
$2 p^{2} (1 + \cos θ) = n^{2} 2 p^{2} (1 - \cos θ)$
$1 + \cos θ = n^{2} (1 - \cos θ)$
$1 + \cos θ = n^{2} - n^{2} \cos θ$
$\cos θ + n^{2} \cos θ = n^{2} - 1$
$\cos θ (1 + n^{2}) = n^{2} - 1$
$\cos θ = \frac{n^{2} - 1}{n^{2} + 1}$
$θ = \cos^{- 1} (\frac{n^{2} - 1}{n^{2} + 1})$

JEE Main-25.07.2021

Motion in Plane

143639 If $A$ and $B$ are two vectors satisfying the relation $\vec{A} \cdot \vec{B} = | \vec{A} \times \vec{B} |$ . Then, the value of $| \vec{A} - \vec{B} |$ will be

1

\sqrt{A^{2} + B^{2}}

2

\sqrt{A^{2} + B^{2} + \sqrt{2} AB}

3

\sqrt{A^{2} + B^{2} + 2 AB}

4

\sqrt{A^{2} + B^{2} - \sqrt{2} A B}

Explanation:

D Given that,
$\vec{A} \cdot \vec{B} = | \vec{A} \times \vec{B} |$
$AB \cos θ = AB \sin θ$
$\tan θ = 1$
$θ = \tan^{- 1} (1)$
$θ = 45^{\circ}$
The value of $| \vec{A} - \vec{B} |$ is
$| \vec{A} - \vec{B} | = \sqrt{A^{2} + B^{2} - 2 AB \cos 45^{\circ}}$
$| \vec{A} - \vec{B} | = \sqrt{A^{2} + B^{2} - \sqrt{2} AB}$

JEE Main-20.07.2021

Motion in Plane

143636 Which of the following relations is true for two unit vector $\hat{A}$ and $\hat{B}$ making an angle $θ$ to each other?

1

| \hat{A} + \hat{B} | = | \hat{A} - \hat{B} | \tan \frac{θ}{2}

2

| \hat{A} - \hat{B} | = | \hat{A} + \hat{B} | \tan \frac{θ}{2}

3

| \hat{A} + \hat{B} | = | \hat{A} - \hat{B} | \cos \frac{θ}{2}

4

| \hat{A} - \hat{B} | = | \hat{A} + \hat{B} | \cos \frac{θ}{2}

Explanation:

B Given that,
$| \hat{A} + \hat{B} | = \sqrt{| \hat{A} |^{2} + | \hat{B} |^{2} + 2 | \hat{A} | | \hat{B} | \cos θ}$
$| \hat{A} | = 1 | \hat{B} |^{2} = 1$
$| \hat{A} + \hat{B} | = \sqrt{(1)^{2} + (1)^{2} + 2 \cos θ}$
$= \sqrt{2 + 2 \cos θ}$
$= \sqrt{2 + 2 (2 \cos^{2} \frac{θ}{2} - 1)}$
$= \sqrt{2 (1 + 2 \cos^{2} \frac{θ}{2} - 1)} = 2 \cos \frac{θ}{2}$
$\hat{A} + \hat{B} = 2 \cos \frac{θ}{2}$
$| \hat{A} - \hat{B} | = \sqrt{| \hat{A} |^{2} + | \hat{B} |^{2} - 2 | \hat{A} | | \hat{B} | \cos θ}$
$| \hat{A} - \hat{B} | = \sqrt{1 + 1 - 2 \cos θ}$
$| \hat{A} - \hat{B} | = \sqrt{2 - 2 \cos θ}$
$= \sqrt{2 (1 - \cos θ)}$
$= \sqrt{2 (1 - (1 - 2 \sin^{2} \frac{θ}{2})}$
$= \sqrt{2 (1 - 1 + 2 \sin^{2} \frac{θ}{2})}$
$= \sqrt{2 (1 - 1 + 2 \sin^{2} \frac{θ}{2})} = 2 \sin \frac{θ}{2}$
$\hat{A} - \hat{B} = 2 \sin \frac{θ}{2}$
Equation (i) divide by equation (ii), we get -
$\frac{| \hat{A} - \hat{B} |}{| \hat{A} + \hat{B} |} = \frac{2 \sin \frac{θ}{2}}{2 \cos \frac{θ}{2}}$
$\frac{| \hat{A} - \hat{B} |}{| \hat{A} + \hat{B} |} = \tan \frac{θ}{2}$
$| \hat{A} - \hat{B} | = | \hat{A} + \hat{B} | \tan \frac{θ}{2}$

Shift-I]

Motion in Plane

1 (A)

\to

(iv), (B)

\to

(i), (C)

\to

(iii), (D)

\to

(ii)

2 (A)

\to

(iv), (B)

\to

(iii), (C)

\to

(i), (D)

\to

(ii)

3 (A)

\to

(iii), (B)

\to

(ii), (C)

\to

(iv), (D)

\to

(i)

4 (A)

\to

(i), (B)

\to

(iv), (C)

\to

(ii), (D)

\to

(iii)

Explanation:

B Applying triangle law of vectors to the diagram.
(i) $\vec{B} = \vec{A} + \vec{C}$
$∴ \vec{B} - \vec{A} - \vec{C} = 0$
This matches with (C).
(ii) $- \vec{C} = \vec{A} + \vec{B}$
This matches with (D).
(iii) $\vec{C} + \vec{B} = \vec{A}$
$∴ \vec{A} - \vec{C} = \vec{B}$
This matches with (B).
(iv) $\vec{A} + \vec{B} = \vec{C}$
$∴ \vec{C} - \vec{A} - \vec{B} = 0$

JEE Main-25.07.2021

Motion in Plane

143638 Two vectors $P$ and $Q$ have equal magnitudes. If the magnitude of $P + Q$ is $n$ times the magnitude of $P - Q$ , then angle between $P$ and $Q$ is

1

\sin^{- 1} (\frac{n - 1}{n + 1})

2

\cos^{- 1} (\frac{n - 1}{n + 1})

3

\sin^{- 1} (\frac{n^{2} - 1}{n^{2} + 1})

4

\cos^{- 1} (\frac{n^{2} - 1}{n^{2} + 1})

Explanation:

D Given that,
$P = Q$
Let the magnitude of $(P + Q) = R$
$R = | P + Q | = \sqrt{P^{2} + Q^{2} + 2 PQ \cos θ}$
$R = \sqrt{P^{2} + P^{2} + 2 P^{2} \cos θ}$ [from (i)]
The magnitude of $(P - Q) = R^{'}$
$R^{'} = | P - Q | = \sqrt{P^{2} + Q^{2} - 2 PQ \cos θ}$
$R^{'} = | P - Q | = \sqrt{P^{2} + P^{2} - 2 P^{2} \cos θ}$ [from (i)]
Given that
$R = {nR}^{'}$
$\sqrt{2 P^{2} + 2 P^{2} \cos θ} = n \sqrt{2 P^{2} - 2 P^{2} \cos θ}$
squaring both side
$2 p^{2} + 2 p^{2} \cos θ = n^{2} (2 p^{2} - 2 p^{2} \cos θ)$
$2 p^{2} (1 + \cos θ) = n^{2} 2 p^{2} (1 - \cos θ)$
$1 + \cos θ = n^{2} (1 - \cos θ)$
$1 + \cos θ = n^{2} - n^{2} \cos θ$
$\cos θ + n^{2} \cos θ = n^{2} - 1$
$\cos θ (1 + n^{2}) = n^{2} - 1$
$\cos θ = \frac{n^{2} - 1}{n^{2} + 1}$
$θ = \cos^{- 1} (\frac{n^{2} - 1}{n^{2} + 1})$

JEE Main-25.07.2021

Motion in Plane

143639 If $A$ and $B$ are two vectors satisfying the relation $\vec{A} \cdot \vec{B} = | \vec{A} \times \vec{B} |$ . Then, the value of $| \vec{A} - \vec{B} |$ will be

1

\sqrt{A^{2} + B^{2}}

2

\sqrt{A^{2} + B^{2} + \sqrt{2} AB}

3

\sqrt{A^{2} + B^{2} + 2 AB}

4

\sqrt{A^{2} + B^{2} - \sqrt{2} A B}

Explanation:

D Given that,
$\vec{A} \cdot \vec{B} = | \vec{A} \times \vec{B} |$
$AB \cos θ = AB \sin θ$
$\tan θ = 1$
$θ = \tan^{- 1} (1)$
$θ = 45^{\circ}$
The value of $| \vec{A} - \vec{B} |$ is
$| \vec{A} - \vec{B} | = \sqrt{A^{2} + B^{2} - 2 AB \cos 45^{\circ}}$
$| \vec{A} - \vec{B} | = \sqrt{A^{2} + B^{2} - \sqrt{2} AB}$

JEE Main-20.07.2021

Motion in Plane

143636 Which of the following relations is true for two unit vector $\hat{A}$ and $\hat{B}$ making an angle $θ$ to each other?

1

| \hat{A} + \hat{B} | = | \hat{A} - \hat{B} | \tan \frac{θ}{2}

2

| \hat{A} - \hat{B} | = | \hat{A} + \hat{B} | \tan \frac{θ}{2}

3

| \hat{A} + \hat{B} | = | \hat{A} - \hat{B} | \cos \frac{θ}{2}

4

| \hat{A} - \hat{B} | = | \hat{A} + \hat{B} | \cos \frac{θ}{2}

Explanation:

B Given that,
$| \hat{A} + \hat{B} | = \sqrt{| \hat{A} |^{2} + | \hat{B} |^{2} + 2 | \hat{A} | | \hat{B} | \cos θ}$
$| \hat{A} | = 1 | \hat{B} |^{2} = 1$
$| \hat{A} + \hat{B} | = \sqrt{(1)^{2} + (1)^{2} + 2 \cos θ}$
$= \sqrt{2 + 2 \cos θ}$
$= \sqrt{2 + 2 (2 \cos^{2} \frac{θ}{2} - 1)}$
$= \sqrt{2 (1 + 2 \cos^{2} \frac{θ}{2} - 1)} = 2 \cos \frac{θ}{2}$
$\hat{A} + \hat{B} = 2 \cos \frac{θ}{2}$
$| \hat{A} - \hat{B} | = \sqrt{| \hat{A} |^{2} + | \hat{B} |^{2} - 2 | \hat{A} | | \hat{B} | \cos θ}$
$| \hat{A} - \hat{B} | = \sqrt{1 + 1 - 2 \cos θ}$
$| \hat{A} - \hat{B} | = \sqrt{2 - 2 \cos θ}$
$= \sqrt{2 (1 - \cos θ)}$
$= \sqrt{2 (1 - (1 - 2 \sin^{2} \frac{θ}{2})}$
$= \sqrt{2 (1 - 1 + 2 \sin^{2} \frac{θ}{2})}$
$= \sqrt{2 (1 - 1 + 2 \sin^{2} \frac{θ}{2})} = 2 \sin \frac{θ}{2}$
$\hat{A} - \hat{B} = 2 \sin \frac{θ}{2}$
Equation (i) divide by equation (ii), we get -
$\frac{| \hat{A} - \hat{B} |}{| \hat{A} + \hat{B} |} = \frac{2 \sin \frac{θ}{2}}{2 \cos \frac{θ}{2}}$
$\frac{| \hat{A} - \hat{B} |}{| \hat{A} + \hat{B} |} = \tan \frac{θ}{2}$
$| \hat{A} - \hat{B} | = | \hat{A} + \hat{B} | \tan \frac{θ}{2}$

Shift-I]

Motion in Plane

1 (A)

\to

(iv), (B)

\to

(i), (C)

\to

(iii), (D)

\to

(ii)

2 (A)

\to

(iv), (B)

\to

(iii), (C)

\to

(i), (D)

\to

(ii)

3 (A)

\to

(iii), (B)

\to

(ii), (C)

\to

(iv), (D)

\to

(i)

4 (A)

\to

(i), (B)

\to

(iv), (C)

\to

(ii), (D)

\to

(iii)

Explanation:

B Applying triangle law of vectors to the diagram.
(i) $\vec{B} = \vec{A} + \vec{C}$
$∴ \vec{B} - \vec{A} - \vec{C} = 0$
This matches with (C).
(ii) $- \vec{C} = \vec{A} + \vec{B}$
This matches with (D).
(iii) $\vec{C} + \vec{B} = \vec{A}$
$∴ \vec{A} - \vec{C} = \vec{B}$
This matches with (B).
(iv) $\vec{A} + \vec{B} = \vec{C}$
$∴ \vec{C} - \vec{A} - \vec{B} = 0$

JEE Main-25.07.2021

Motion in Plane

143638 Two vectors $P$ and $Q$ have equal magnitudes. If the magnitude of $P + Q$ is $n$ times the magnitude of $P - Q$ , then angle between $P$ and $Q$ is

1

\sin^{- 1} (\frac{n - 1}{n + 1})

2

\cos^{- 1} (\frac{n - 1}{n + 1})

3

\sin^{- 1} (\frac{n^{2} - 1}{n^{2} + 1})

4

\cos^{- 1} (\frac{n^{2} - 1}{n^{2} + 1})

Explanation:

D Given that,
$P = Q$
Let the magnitude of $(P + Q) = R$
$R = | P + Q | = \sqrt{P^{2} + Q^{2} + 2 PQ \cos θ}$
$R = \sqrt{P^{2} + P^{2} + 2 P^{2} \cos θ}$ [from (i)]
The magnitude of $(P - Q) = R^{'}$
$R^{'} = | P - Q | = \sqrt{P^{2} + Q^{2} - 2 PQ \cos θ}$
$R^{'} = | P - Q | = \sqrt{P^{2} + P^{2} - 2 P^{2} \cos θ}$ [from (i)]
Given that
$R = {nR}^{'}$
$\sqrt{2 P^{2} + 2 P^{2} \cos θ} = n \sqrt{2 P^{2} - 2 P^{2} \cos θ}$
squaring both side
$2 p^{2} + 2 p^{2} \cos θ = n^{2} (2 p^{2} - 2 p^{2} \cos θ)$
$2 p^{2} (1 + \cos θ) = n^{2} 2 p^{2} (1 - \cos θ)$
$1 + \cos θ = n^{2} (1 - \cos θ)$
$1 + \cos θ = n^{2} - n^{2} \cos θ$
$\cos θ + n^{2} \cos θ = n^{2} - 1$
$\cos θ (1 + n^{2}) = n^{2} - 1$
$\cos θ = \frac{n^{2} - 1}{n^{2} + 1}$
$θ = \cos^{- 1} (\frac{n^{2} - 1}{n^{2} + 1})$

JEE Main-25.07.2021

Motion in Plane

143639 If $A$ and $B$ are two vectors satisfying the relation $\vec{A} \cdot \vec{B} = | \vec{A} \times \vec{B} |$ . Then, the value of $| \vec{A} - \vec{B} |$ will be

1

\sqrt{A^{2} + B^{2}}

2

\sqrt{A^{2} + B^{2} + \sqrt{2} AB}

3

\sqrt{A^{2} + B^{2} + 2 AB}

4

\sqrt{A^{2} + B^{2} - \sqrt{2} A B}

Explanation:

D Given that,
$\vec{A} \cdot \vec{B} = | \vec{A} \times \vec{B} |$
$AB \cos θ = AB \sin θ$
$\tan θ = 1$
$θ = \tan^{- 1} (1)$
$θ = 45^{\circ}$
The value of $| \vec{A} - \vec{B} |$ is
$| \vec{A} - \vec{B} | = \sqrt{A^{2} + B^{2} - 2 AB \cos 45^{\circ}}$
$| \vec{A} - \vec{B} | = \sqrt{A^{2} + B^{2} - \sqrt{2} AB}$

JEE Main-20.07.2021

Motion in Plane

143636 Which of the following relations is true for two unit vector $\hat{A}$ and $\hat{B}$ making an angle $θ$ to each other?

1

| \hat{A} + \hat{B} | = | \hat{A} - \hat{B} | \tan \frac{θ}{2}

2

| \hat{A} - \hat{B} | = | \hat{A} + \hat{B} | \tan \frac{θ}{2}

3

| \hat{A} + \hat{B} | = | \hat{A} - \hat{B} | \cos \frac{θ}{2}

4

| \hat{A} - \hat{B} | = | \hat{A} + \hat{B} | \cos \frac{θ}{2}

Explanation:

B Given that,
$| \hat{A} + \hat{B} | = \sqrt{| \hat{A} |^{2} + | \hat{B} |^{2} + 2 | \hat{A} | | \hat{B} | \cos θ}$
$| \hat{A} | = 1 | \hat{B} |^{2} = 1$
$| \hat{A} + \hat{B} | = \sqrt{(1)^{2} + (1)^{2} + 2 \cos θ}$
$= \sqrt{2 + 2 \cos θ}$
$= \sqrt{2 + 2 (2 \cos^{2} \frac{θ}{2} - 1)}$
$= \sqrt{2 (1 + 2 \cos^{2} \frac{θ}{2} - 1)} = 2 \cos \frac{θ}{2}$
$\hat{A} + \hat{B} = 2 \cos \frac{θ}{2}$
$| \hat{A} - \hat{B} | = \sqrt{| \hat{A} |^{2} + | \hat{B} |^{2} - 2 | \hat{A} | | \hat{B} | \cos θ}$
$| \hat{A} - \hat{B} | = \sqrt{1 + 1 - 2 \cos θ}$
$| \hat{A} - \hat{B} | = \sqrt{2 - 2 \cos θ}$
$= \sqrt{2 (1 - \cos θ)}$
$= \sqrt{2 (1 - (1 - 2 \sin^{2} \frac{θ}{2})}$
$= \sqrt{2 (1 - 1 + 2 \sin^{2} \frac{θ}{2})}$
$= \sqrt{2 (1 - 1 + 2 \sin^{2} \frac{θ}{2})} = 2 \sin \frac{θ}{2}$
$\hat{A} - \hat{B} = 2 \sin \frac{θ}{2}$
Equation (i) divide by equation (ii), we get -
$\frac{| \hat{A} - \hat{B} |}{| \hat{A} + \hat{B} |} = \frac{2 \sin \frac{θ}{2}}{2 \cos \frac{θ}{2}}$
$\frac{| \hat{A} - \hat{B} |}{| \hat{A} + \hat{B} |} = \tan \frac{θ}{2}$
$| \hat{A} - \hat{B} | = | \hat{A} + \hat{B} | \tan \frac{θ}{2}$

Shift-I]

Motion in Plane

1 (A)

\to

(iv), (B)

\to

(i), (C)

\to

(iii), (D)

\to

(ii)

2 (A)

\to

(iv), (B)

\to

(iii), (C)

\to

(i), (D)

\to

(ii)

3 (A)

\to

(iii), (B)

\to

(ii), (C)

\to

(iv), (D)

\to

(i)

4 (A)

\to

(i), (B)

\to

(iv), (C)

\to

(ii), (D)

\to

(iii)

Explanation:

B Applying triangle law of vectors to the diagram.
(i) $\vec{B} = \vec{A} + \vec{C}$
$∴ \vec{B} - \vec{A} - \vec{C} = 0$
This matches with (C).
(ii) $- \vec{C} = \vec{A} + \vec{B}$
This matches with (D).
(iii) $\vec{C} + \vec{B} = \vec{A}$
$∴ \vec{A} - \vec{C} = \vec{B}$
This matches with (B).
(iv) $\vec{A} + \vec{B} = \vec{C}$
$∴ \vec{C} - \vec{A} - \vec{B} = 0$

JEE Main-25.07.2021

Motion in Plane

143638 Two vectors $P$ and $Q$ have equal magnitudes. If the magnitude of $P + Q$ is $n$ times the magnitude of $P - Q$ , then angle between $P$ and $Q$ is

1

\sin^{- 1} (\frac{n - 1}{n + 1})

2

\cos^{- 1} (\frac{n - 1}{n + 1})

3

\sin^{- 1} (\frac{n^{2} - 1}{n^{2} + 1})

4

\cos^{- 1} (\frac{n^{2} - 1}{n^{2} + 1})

Explanation:

D Given that,
$P = Q$
Let the magnitude of $(P + Q) = R$
$R = | P + Q | = \sqrt{P^{2} + Q^{2} + 2 PQ \cos θ}$
$R = \sqrt{P^{2} + P^{2} + 2 P^{2} \cos θ}$ [from (i)]
The magnitude of $(P - Q) = R^{'}$
$R^{'} = | P - Q | = \sqrt{P^{2} + Q^{2} - 2 PQ \cos θ}$
$R^{'} = | P - Q | = \sqrt{P^{2} + P^{2} - 2 P^{2} \cos θ}$ [from (i)]
Given that
$R = {nR}^{'}$
$\sqrt{2 P^{2} + 2 P^{2} \cos θ} = n \sqrt{2 P^{2} - 2 P^{2} \cos θ}$
squaring both side
$2 p^{2} + 2 p^{2} \cos θ = n^{2} (2 p^{2} - 2 p^{2} \cos θ)$
$2 p^{2} (1 + \cos θ) = n^{2} 2 p^{2} (1 - \cos θ)$
$1 + \cos θ = n^{2} (1 - \cos θ)$
$1 + \cos θ = n^{2} - n^{2} \cos θ$
$\cos θ + n^{2} \cos θ = n^{2} - 1$
$\cos θ (1 + n^{2}) = n^{2} - 1$
$\cos θ = \frac{n^{2} - 1}{n^{2} + 1}$
$θ = \cos^{- 1} (\frac{n^{2} - 1}{n^{2} + 1})$

JEE Main-25.07.2021

Motion in Plane

143639 If $A$ and $B$ are two vectors satisfying the relation $\vec{A} \cdot \vec{B} = | \vec{A} \times \vec{B} |$ . Then, the value of $| \vec{A} - \vec{B} |$ will be

1

\sqrt{A^{2} + B^{2}}

2

\sqrt{A^{2} + B^{2} + \sqrt{2} AB}

3

\sqrt{A^{2} + B^{2} + 2 AB}

4

\sqrt{A^{2} + B^{2} - \sqrt{2} A B}

Explanation:

D Given that,
$\vec{A} \cdot \vec{B} = | \vec{A} \times \vec{B} |$
$AB \cos θ = AB \sin θ$
$\tan θ = 1$
$θ = \tan^{- 1} (1)$
$θ = 45^{\circ}$
The value of $| \vec{A} - \vec{B} |$ is
$| \vec{A} - \vec{B} | = \sqrt{A^{2} + B^{2} - 2 AB \cos 45^{\circ}}$
$| \vec{A} - \vec{B} | = \sqrt{A^{2} + B^{2} - \sqrt{2} AB}$

JEE Main-20.07.2021