355529
Two unit vectors \(\vec{A}\) and \(\vec{B}\) are inclined to each other at an angle \(45^{\circ}\). The angle made by the resultant vector w.r.to \(\vec{A}\) is?
1 \(\tan ^{-1}\left(\dfrac{1}{\sqrt{3}}\right)\)
2 \(\tan ^{-1}(\sqrt{3})\)
3 \(\tan ^{-1} \dfrac{1}{\sqrt{2}}\)
4 \(\tan ^{-1} \dfrac{1}{3}\)
Explanation:
\(\tan \alpha=\dfrac{B \sin \theta}{A+B \cos \theta}\) \(\alpha\) - Angle made by the resultant vector with \(\vec{A}\) \(\begin{gathered}\theta=60^{\circ}, A=B=1 \Rightarrow \tan \alpha=\dfrac{2 \sqrt{2}}{1+\left(\dfrac{1}{2}\right)}=\dfrac{1}{\sqrt{3}} \\\alpha=\tan ^{-1}\left(\dfrac{1}{\sqrt{3}}\right)\end{gathered}\)
PHXI06:WORK ENERGY AND POWER
355530
The angle between vector \(\vec{Q}\) and the resultant of \((2 \vec{Q}+2 \vec{P})\) and \((2 \vec{Q}-2 \vec{P})\) is
1 \(0^{\circ}\)
2 \(\tan ^{-1} \dfrac{(2 Q-2 P)}{2 Q+2 P}\)
3 \(\tan ^{-1}\left(\dfrac{2 Q}{P}\right)\)
4 \(\tan ^{-1}\left(\dfrac{P}{Q}\right)\)
Explanation:
Let the resultant be \({\vec R}\) its value is given by \(\therefore \quad \vec{R}=(2 \vec{Q}+2 \vec{P})+(2 \vec{Q}-2 \vec{P})\) or \(\vec{R}=4 \vec{Q}\) Thus, resultant \(\vec{R}\) is in the same direction along vector \(\vec{Q}\). So \(\vec{R}\) is parallel to \(\vec{Q}\). \(\therefore\) The angle between vector \(\vec{Q}\) and the resultant is zero. So, option (1) is correct.
JEE - 2024
PHXI06:WORK ENERGY AND POWER
355531
The vectors \(\vec{A}\) and \(\vec{B}\) are such that \(|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}|\) The angle between the two vectors is
1 \(75^{\circ}\)
2 \(45^{\circ}\)
3 \(90^{\circ}\)
4 \(60^{\circ}\)
Explanation:
\(|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}|\) Squaring on both sides, \(|\vec{A}+\vec{B}|^{2}=|\vec{A}-\vec{B}|^{2}\) \(\Rightarrow(\vec{A}+\vec{B}) \cdot(\vec{A}+\vec{B})=(\vec{A}-\vec{B}) \cdot(\vec{A}-\vec{B})\)\(\Rightarrow \vec{A} \cdot \vec{A}+2 \vec{A} \cdot \vec{B}+\vec{B} \cdot \vec{B}=\vec{A} \cdot \vec{A}+\vec{B} \cdot \vec{B}-2 \vec{A} \cdot \vec{B}\)\(\Rightarrow 4 \vec{A} \cdot \vec{B}=0 \Rightarrow 4 A B \cos \theta=0\)\(\Rightarrow \cos \theta=0 \Rightarrow \theta=90^{\circ}\)
PHXI06:WORK ENERGY AND POWER
355532
Vector which is perpendicular to \((a \cos \theta \hat{i}+b \sin \theta \hat{j})\) is
355529
Two unit vectors \(\vec{A}\) and \(\vec{B}\) are inclined to each other at an angle \(45^{\circ}\). The angle made by the resultant vector w.r.to \(\vec{A}\) is?
1 \(\tan ^{-1}\left(\dfrac{1}{\sqrt{3}}\right)\)
2 \(\tan ^{-1}(\sqrt{3})\)
3 \(\tan ^{-1} \dfrac{1}{\sqrt{2}}\)
4 \(\tan ^{-1} \dfrac{1}{3}\)
Explanation:
\(\tan \alpha=\dfrac{B \sin \theta}{A+B \cos \theta}\) \(\alpha\) - Angle made by the resultant vector with \(\vec{A}\) \(\begin{gathered}\theta=60^{\circ}, A=B=1 \Rightarrow \tan \alpha=\dfrac{2 \sqrt{2}}{1+\left(\dfrac{1}{2}\right)}=\dfrac{1}{\sqrt{3}} \\\alpha=\tan ^{-1}\left(\dfrac{1}{\sqrt{3}}\right)\end{gathered}\)
PHXI06:WORK ENERGY AND POWER
355530
The angle between vector \(\vec{Q}\) and the resultant of \((2 \vec{Q}+2 \vec{P})\) and \((2 \vec{Q}-2 \vec{P})\) is
1 \(0^{\circ}\)
2 \(\tan ^{-1} \dfrac{(2 Q-2 P)}{2 Q+2 P}\)
3 \(\tan ^{-1}\left(\dfrac{2 Q}{P}\right)\)
4 \(\tan ^{-1}\left(\dfrac{P}{Q}\right)\)
Explanation:
Let the resultant be \({\vec R}\) its value is given by \(\therefore \quad \vec{R}=(2 \vec{Q}+2 \vec{P})+(2 \vec{Q}-2 \vec{P})\) or \(\vec{R}=4 \vec{Q}\) Thus, resultant \(\vec{R}\) is in the same direction along vector \(\vec{Q}\). So \(\vec{R}\) is parallel to \(\vec{Q}\). \(\therefore\) The angle between vector \(\vec{Q}\) and the resultant is zero. So, option (1) is correct.
JEE - 2024
PHXI06:WORK ENERGY AND POWER
355531
The vectors \(\vec{A}\) and \(\vec{B}\) are such that \(|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}|\) The angle between the two vectors is
1 \(75^{\circ}\)
2 \(45^{\circ}\)
3 \(90^{\circ}\)
4 \(60^{\circ}\)
Explanation:
\(|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}|\) Squaring on both sides, \(|\vec{A}+\vec{B}|^{2}=|\vec{A}-\vec{B}|^{2}\) \(\Rightarrow(\vec{A}+\vec{B}) \cdot(\vec{A}+\vec{B})=(\vec{A}-\vec{B}) \cdot(\vec{A}-\vec{B})\)\(\Rightarrow \vec{A} \cdot \vec{A}+2 \vec{A} \cdot \vec{B}+\vec{B} \cdot \vec{B}=\vec{A} \cdot \vec{A}+\vec{B} \cdot \vec{B}-2 \vec{A} \cdot \vec{B}\)\(\Rightarrow 4 \vec{A} \cdot \vec{B}=0 \Rightarrow 4 A B \cos \theta=0\)\(\Rightarrow \cos \theta=0 \Rightarrow \theta=90^{\circ}\)
PHXI06:WORK ENERGY AND POWER
355532
Vector which is perpendicular to \((a \cos \theta \hat{i}+b \sin \theta \hat{j})\) is
355529
Two unit vectors \(\vec{A}\) and \(\vec{B}\) are inclined to each other at an angle \(45^{\circ}\). The angle made by the resultant vector w.r.to \(\vec{A}\) is?
1 \(\tan ^{-1}\left(\dfrac{1}{\sqrt{3}}\right)\)
2 \(\tan ^{-1}(\sqrt{3})\)
3 \(\tan ^{-1} \dfrac{1}{\sqrt{2}}\)
4 \(\tan ^{-1} \dfrac{1}{3}\)
Explanation:
\(\tan \alpha=\dfrac{B \sin \theta}{A+B \cos \theta}\) \(\alpha\) - Angle made by the resultant vector with \(\vec{A}\) \(\begin{gathered}\theta=60^{\circ}, A=B=1 \Rightarrow \tan \alpha=\dfrac{2 \sqrt{2}}{1+\left(\dfrac{1}{2}\right)}=\dfrac{1}{\sqrt{3}} \\\alpha=\tan ^{-1}\left(\dfrac{1}{\sqrt{3}}\right)\end{gathered}\)
PHXI06:WORK ENERGY AND POWER
355530
The angle between vector \(\vec{Q}\) and the resultant of \((2 \vec{Q}+2 \vec{P})\) and \((2 \vec{Q}-2 \vec{P})\) is
1 \(0^{\circ}\)
2 \(\tan ^{-1} \dfrac{(2 Q-2 P)}{2 Q+2 P}\)
3 \(\tan ^{-1}\left(\dfrac{2 Q}{P}\right)\)
4 \(\tan ^{-1}\left(\dfrac{P}{Q}\right)\)
Explanation:
Let the resultant be \({\vec R}\) its value is given by \(\therefore \quad \vec{R}=(2 \vec{Q}+2 \vec{P})+(2 \vec{Q}-2 \vec{P})\) or \(\vec{R}=4 \vec{Q}\) Thus, resultant \(\vec{R}\) is in the same direction along vector \(\vec{Q}\). So \(\vec{R}\) is parallel to \(\vec{Q}\). \(\therefore\) The angle between vector \(\vec{Q}\) and the resultant is zero. So, option (1) is correct.
JEE - 2024
PHXI06:WORK ENERGY AND POWER
355531
The vectors \(\vec{A}\) and \(\vec{B}\) are such that \(|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}|\) The angle between the two vectors is
1 \(75^{\circ}\)
2 \(45^{\circ}\)
3 \(90^{\circ}\)
4 \(60^{\circ}\)
Explanation:
\(|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}|\) Squaring on both sides, \(|\vec{A}+\vec{B}|^{2}=|\vec{A}-\vec{B}|^{2}\) \(\Rightarrow(\vec{A}+\vec{B}) \cdot(\vec{A}+\vec{B})=(\vec{A}-\vec{B}) \cdot(\vec{A}-\vec{B})\)\(\Rightarrow \vec{A} \cdot \vec{A}+2 \vec{A} \cdot \vec{B}+\vec{B} \cdot \vec{B}=\vec{A} \cdot \vec{A}+\vec{B} \cdot \vec{B}-2 \vec{A} \cdot \vec{B}\)\(\Rightarrow 4 \vec{A} \cdot \vec{B}=0 \Rightarrow 4 A B \cos \theta=0\)\(\Rightarrow \cos \theta=0 \Rightarrow \theta=90^{\circ}\)
PHXI06:WORK ENERGY AND POWER
355532
Vector which is perpendicular to \((a \cos \theta \hat{i}+b \sin \theta \hat{j})\) is
355529
Two unit vectors \(\vec{A}\) and \(\vec{B}\) are inclined to each other at an angle \(45^{\circ}\). The angle made by the resultant vector w.r.to \(\vec{A}\) is?
1 \(\tan ^{-1}\left(\dfrac{1}{\sqrt{3}}\right)\)
2 \(\tan ^{-1}(\sqrt{3})\)
3 \(\tan ^{-1} \dfrac{1}{\sqrt{2}}\)
4 \(\tan ^{-1} \dfrac{1}{3}\)
Explanation:
\(\tan \alpha=\dfrac{B \sin \theta}{A+B \cos \theta}\) \(\alpha\) - Angle made by the resultant vector with \(\vec{A}\) \(\begin{gathered}\theta=60^{\circ}, A=B=1 \Rightarrow \tan \alpha=\dfrac{2 \sqrt{2}}{1+\left(\dfrac{1}{2}\right)}=\dfrac{1}{\sqrt{3}} \\\alpha=\tan ^{-1}\left(\dfrac{1}{\sqrt{3}}\right)\end{gathered}\)
PHXI06:WORK ENERGY AND POWER
355530
The angle between vector \(\vec{Q}\) and the resultant of \((2 \vec{Q}+2 \vec{P})\) and \((2 \vec{Q}-2 \vec{P})\) is
1 \(0^{\circ}\)
2 \(\tan ^{-1} \dfrac{(2 Q-2 P)}{2 Q+2 P}\)
3 \(\tan ^{-1}\left(\dfrac{2 Q}{P}\right)\)
4 \(\tan ^{-1}\left(\dfrac{P}{Q}\right)\)
Explanation:
Let the resultant be \({\vec R}\) its value is given by \(\therefore \quad \vec{R}=(2 \vec{Q}+2 \vec{P})+(2 \vec{Q}-2 \vec{P})\) or \(\vec{R}=4 \vec{Q}\) Thus, resultant \(\vec{R}\) is in the same direction along vector \(\vec{Q}\). So \(\vec{R}\) is parallel to \(\vec{Q}\). \(\therefore\) The angle between vector \(\vec{Q}\) and the resultant is zero. So, option (1) is correct.
JEE - 2024
PHXI06:WORK ENERGY AND POWER
355531
The vectors \(\vec{A}\) and \(\vec{B}\) are such that \(|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}|\) The angle between the two vectors is
1 \(75^{\circ}\)
2 \(45^{\circ}\)
3 \(90^{\circ}\)
4 \(60^{\circ}\)
Explanation:
\(|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}|\) Squaring on both sides, \(|\vec{A}+\vec{B}|^{2}=|\vec{A}-\vec{B}|^{2}\) \(\Rightarrow(\vec{A}+\vec{B}) \cdot(\vec{A}+\vec{B})=(\vec{A}-\vec{B}) \cdot(\vec{A}-\vec{B})\)\(\Rightarrow \vec{A} \cdot \vec{A}+2 \vec{A} \cdot \vec{B}+\vec{B} \cdot \vec{B}=\vec{A} \cdot \vec{A}+\vec{B} \cdot \vec{B}-2 \vec{A} \cdot \vec{B}\)\(\Rightarrow 4 \vec{A} \cdot \vec{B}=0 \Rightarrow 4 A B \cos \theta=0\)\(\Rightarrow \cos \theta=0 \Rightarrow \theta=90^{\circ}\)
PHXI06:WORK ENERGY AND POWER
355532
Vector which is perpendicular to \((a \cos \theta \hat{i}+b \sin \theta \hat{j})\) is
355529
Two unit vectors \(\vec{A}\) and \(\vec{B}\) are inclined to each other at an angle \(45^{\circ}\). The angle made by the resultant vector w.r.to \(\vec{A}\) is?
1 \(\tan ^{-1}\left(\dfrac{1}{\sqrt{3}}\right)\)
2 \(\tan ^{-1}(\sqrt{3})\)
3 \(\tan ^{-1} \dfrac{1}{\sqrt{2}}\)
4 \(\tan ^{-1} \dfrac{1}{3}\)
Explanation:
\(\tan \alpha=\dfrac{B \sin \theta}{A+B \cos \theta}\) \(\alpha\) - Angle made by the resultant vector with \(\vec{A}\) \(\begin{gathered}\theta=60^{\circ}, A=B=1 \Rightarrow \tan \alpha=\dfrac{2 \sqrt{2}}{1+\left(\dfrac{1}{2}\right)}=\dfrac{1}{\sqrt{3}} \\\alpha=\tan ^{-1}\left(\dfrac{1}{\sqrt{3}}\right)\end{gathered}\)
PHXI06:WORK ENERGY AND POWER
355530
The angle between vector \(\vec{Q}\) and the resultant of \((2 \vec{Q}+2 \vec{P})\) and \((2 \vec{Q}-2 \vec{P})\) is
1 \(0^{\circ}\)
2 \(\tan ^{-1} \dfrac{(2 Q-2 P)}{2 Q+2 P}\)
3 \(\tan ^{-1}\left(\dfrac{2 Q}{P}\right)\)
4 \(\tan ^{-1}\left(\dfrac{P}{Q}\right)\)
Explanation:
Let the resultant be \({\vec R}\) its value is given by \(\therefore \quad \vec{R}=(2 \vec{Q}+2 \vec{P})+(2 \vec{Q}-2 \vec{P})\) or \(\vec{R}=4 \vec{Q}\) Thus, resultant \(\vec{R}\) is in the same direction along vector \(\vec{Q}\). So \(\vec{R}\) is parallel to \(\vec{Q}\). \(\therefore\) The angle between vector \(\vec{Q}\) and the resultant is zero. So, option (1) is correct.
JEE - 2024
PHXI06:WORK ENERGY AND POWER
355531
The vectors \(\vec{A}\) and \(\vec{B}\) are such that \(|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}|\) The angle between the two vectors is
1 \(75^{\circ}\)
2 \(45^{\circ}\)
3 \(90^{\circ}\)
4 \(60^{\circ}\)
Explanation:
\(|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}|\) Squaring on both sides, \(|\vec{A}+\vec{B}|^{2}=|\vec{A}-\vec{B}|^{2}\) \(\Rightarrow(\vec{A}+\vec{B}) \cdot(\vec{A}+\vec{B})=(\vec{A}-\vec{B}) \cdot(\vec{A}-\vec{B})\)\(\Rightarrow \vec{A} \cdot \vec{A}+2 \vec{A} \cdot \vec{B}+\vec{B} \cdot \vec{B}=\vec{A} \cdot \vec{A}+\vec{B} \cdot \vec{B}-2 \vec{A} \cdot \vec{B}\)\(\Rightarrow 4 \vec{A} \cdot \vec{B}=0 \Rightarrow 4 A B \cos \theta=0\)\(\Rightarrow \cos \theta=0 \Rightarrow \theta=90^{\circ}\)
PHXI06:WORK ENERGY AND POWER
355532
Vector which is perpendicular to \((a \cos \theta \hat{i}+b \sin \theta \hat{j})\) is
