143552
The sum of two vectors \(\vec{A}\) and \(\vec{B}\) is at right angles to their difference. Then
1 \(\mathrm{A}=\mathrm{B}\)
2 \(A=2 B\)
3 \(\mathrm{B}=2 \mathrm{~A}\)
4 \(\vec{A}\) and \(\vec{B}\) have the same direction
Explanation:
A Let \(r_{1}\) and \(r_{2}\) be the sum and difference of vectors \(\vec{A}\) and \(\vec{B}\) respectively i.e., \(\mathrm{r}_{1}=\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}\) \(\mathrm{r}_{2}=\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}\) \(r_{1}\) is perpendicular to \(r_{2}\) Taking the dot product of \(r_{1}\) and \(r_{2}\) \(r_{1} \cdot r_{2}=(\vec{A}+\vec{B}) \cdot(\vec{A}-\vec{B})\) \(0=A^{2}-B^{2}\) \(A^{2}=B^{2}\) \(A=B\)
BCECE-2008
Motion in Plane
143553
The vectors \(\vec{A}\) and \(\vec{B}\) are such that \(|\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}|=|\overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{B}}|\) The angle between the two vectors is
1 \(60^{\circ}\)
2 \(75^{\circ}\)
3 \(45^{\circ}\)
4 \(90^{\circ}\)
Explanation:
D Let angle between \(\vec{A}\) and \(\vec{B}\) be \(\theta\) The resultant of \(\vec{A}+\vec{B}\) is given by \(R=\sqrt{A^{2}+B^{2}+2 A B \cos \theta}\) The resultant of \(\vec{A}-\vec{B}\) is given by \(R^{\prime}=\sqrt{A^{2}+B^{2}-2 A B \cos \theta}\) According to the question, \(\mathrm{R}=\mathrm{R}^{\prime}\) \(\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}+2 \mathrm{AB} \cos \theta}=\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}-2 \mathrm{AB} \cos \theta}\) \(\mathrm{A}^{2}+\mathrm{B}^{2}+2 \mathrm{AB} \cos \theta=\mathrm{A}^{2}+\mathrm{B}^{2}-2 \mathrm{AB} \cos \theta\) \(4 \mathrm{AB} \cos \theta=0\) \(\quad \theta=90^{\circ}\)
WBJEE-2016
Motion in Plane
143554
The vector sum of two forces is perpendicular to their vector differences. In that case, the forces
1 are not equal to each other in magnitude
2 cannot be predicted
3 are equal to each other
4 are equal to each other in magnitude
Explanation:
D Let \(\overrightarrow{\mathrm{f}_{1}}\) and \(\overrightarrow{\mathrm{f}_{2}}\) be the two forces Then sum of forces, \(\vec{a}=\overrightarrow{f_{1}}+\overrightarrow{f_{2}}\) And difference, \(\overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{f}_{1}}-\overrightarrow{\mathrm{f}_{2}}\) The two forces are perpendicular to each other \((\vec{a} \cdot \vec{b})=0\) \(\left(\overrightarrow{\mathrm{f}_{1}}+\overrightarrow{\mathrm{f}_{2}}\right) \cdot\left(\overrightarrow{\mathrm{f}_{1}}-\overrightarrow{\mathrm{f}_{2}}\right)=0\) \(\left|\overrightarrow{\mathrm{f}}_{1}\right|^{2}-\left|\overrightarrow{\mathrm{f}_{2}}\right|^{2}=0\) \(\left|\overrightarrow{\mathrm{f}}_{1}\right|^{2}=\left|\overrightarrow{\mathrm{f}_{2}}\right|^{2}\) \(\left|\overrightarrow{\mathrm{f}_{1}}\right|=\left|\overrightarrow{\mathrm{f}_{2}}\right|\) In that case both the force are equal and have same magnitude.
AIPMT 2003
Motion in Plane
143555
The sum of two vectors \(\vec{A}\) and \(\vec{B}\) is at right angles to their difference. This is possible if
1 \(\mathrm{A}=2 \mathrm{~B}\)
2 \(\mathrm{A}=\mathrm{B}\)
3 \(\mathrm{A}=3 \mathrm{~B}\)
4 \(\mathrm{B}=2 \mathrm{~A}\)
Explanation:
B Let, \(\overrightarrow{\mathrm{P}}_{1}\) and \(\overrightarrow{\mathrm{P}}_{2}\) sum and difference of vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\), \(\overrightarrow{\mathrm{P}_{1}}=(\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}})\) \(\overrightarrow{\mathrm{P}_{2}}=(\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}})\) \(\overrightarrow{\mathrm{P}_{1}} \cdot \overrightarrow{\mathrm{P}_{2}}=(\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}) \cdot(\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}})\) \(0=\mathrm{A}^{2}-\mathrm{B}^{2}\) \(\mathrm{A}^{2}=\mathrm{B}^{2}\) \(\mathrm{A}=\mathrm{B}\)
143552
The sum of two vectors \(\vec{A}\) and \(\vec{B}\) is at right angles to their difference. Then
1 \(\mathrm{A}=\mathrm{B}\)
2 \(A=2 B\)
3 \(\mathrm{B}=2 \mathrm{~A}\)
4 \(\vec{A}\) and \(\vec{B}\) have the same direction
Explanation:
A Let \(r_{1}\) and \(r_{2}\) be the sum and difference of vectors \(\vec{A}\) and \(\vec{B}\) respectively i.e., \(\mathrm{r}_{1}=\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}\) \(\mathrm{r}_{2}=\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}\) \(r_{1}\) is perpendicular to \(r_{2}\) Taking the dot product of \(r_{1}\) and \(r_{2}\) \(r_{1} \cdot r_{2}=(\vec{A}+\vec{B}) \cdot(\vec{A}-\vec{B})\) \(0=A^{2}-B^{2}\) \(A^{2}=B^{2}\) \(A=B\)
BCECE-2008
Motion in Plane
143553
The vectors \(\vec{A}\) and \(\vec{B}\) are such that \(|\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}|=|\overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{B}}|\) The angle between the two vectors is
1 \(60^{\circ}\)
2 \(75^{\circ}\)
3 \(45^{\circ}\)
4 \(90^{\circ}\)
Explanation:
D Let angle between \(\vec{A}\) and \(\vec{B}\) be \(\theta\) The resultant of \(\vec{A}+\vec{B}\) is given by \(R=\sqrt{A^{2}+B^{2}+2 A B \cos \theta}\) The resultant of \(\vec{A}-\vec{B}\) is given by \(R^{\prime}=\sqrt{A^{2}+B^{2}-2 A B \cos \theta}\) According to the question, \(\mathrm{R}=\mathrm{R}^{\prime}\) \(\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}+2 \mathrm{AB} \cos \theta}=\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}-2 \mathrm{AB} \cos \theta}\) \(\mathrm{A}^{2}+\mathrm{B}^{2}+2 \mathrm{AB} \cos \theta=\mathrm{A}^{2}+\mathrm{B}^{2}-2 \mathrm{AB} \cos \theta\) \(4 \mathrm{AB} \cos \theta=0\) \(\quad \theta=90^{\circ}\)
WBJEE-2016
Motion in Plane
143554
The vector sum of two forces is perpendicular to their vector differences. In that case, the forces
1 are not equal to each other in magnitude
2 cannot be predicted
3 are equal to each other
4 are equal to each other in magnitude
Explanation:
D Let \(\overrightarrow{\mathrm{f}_{1}}\) and \(\overrightarrow{\mathrm{f}_{2}}\) be the two forces Then sum of forces, \(\vec{a}=\overrightarrow{f_{1}}+\overrightarrow{f_{2}}\) And difference, \(\overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{f}_{1}}-\overrightarrow{\mathrm{f}_{2}}\) The two forces are perpendicular to each other \((\vec{a} \cdot \vec{b})=0\) \(\left(\overrightarrow{\mathrm{f}_{1}}+\overrightarrow{\mathrm{f}_{2}}\right) \cdot\left(\overrightarrow{\mathrm{f}_{1}}-\overrightarrow{\mathrm{f}_{2}}\right)=0\) \(\left|\overrightarrow{\mathrm{f}}_{1}\right|^{2}-\left|\overrightarrow{\mathrm{f}_{2}}\right|^{2}=0\) \(\left|\overrightarrow{\mathrm{f}}_{1}\right|^{2}=\left|\overrightarrow{\mathrm{f}_{2}}\right|^{2}\) \(\left|\overrightarrow{\mathrm{f}_{1}}\right|=\left|\overrightarrow{\mathrm{f}_{2}}\right|\) In that case both the force are equal and have same magnitude.
AIPMT 2003
Motion in Plane
143555
The sum of two vectors \(\vec{A}\) and \(\vec{B}\) is at right angles to their difference. This is possible if
1 \(\mathrm{A}=2 \mathrm{~B}\)
2 \(\mathrm{A}=\mathrm{B}\)
3 \(\mathrm{A}=3 \mathrm{~B}\)
4 \(\mathrm{B}=2 \mathrm{~A}\)
Explanation:
B Let, \(\overrightarrow{\mathrm{P}}_{1}\) and \(\overrightarrow{\mathrm{P}}_{2}\) sum and difference of vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\), \(\overrightarrow{\mathrm{P}_{1}}=(\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}})\) \(\overrightarrow{\mathrm{P}_{2}}=(\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}})\) \(\overrightarrow{\mathrm{P}_{1}} \cdot \overrightarrow{\mathrm{P}_{2}}=(\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}) \cdot(\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}})\) \(0=\mathrm{A}^{2}-\mathrm{B}^{2}\) \(\mathrm{A}^{2}=\mathrm{B}^{2}\) \(\mathrm{A}=\mathrm{B}\)
143552
The sum of two vectors \(\vec{A}\) and \(\vec{B}\) is at right angles to their difference. Then
1 \(\mathrm{A}=\mathrm{B}\)
2 \(A=2 B\)
3 \(\mathrm{B}=2 \mathrm{~A}\)
4 \(\vec{A}\) and \(\vec{B}\) have the same direction
Explanation:
A Let \(r_{1}\) and \(r_{2}\) be the sum and difference of vectors \(\vec{A}\) and \(\vec{B}\) respectively i.e., \(\mathrm{r}_{1}=\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}\) \(\mathrm{r}_{2}=\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}\) \(r_{1}\) is perpendicular to \(r_{2}\) Taking the dot product of \(r_{1}\) and \(r_{2}\) \(r_{1} \cdot r_{2}=(\vec{A}+\vec{B}) \cdot(\vec{A}-\vec{B})\) \(0=A^{2}-B^{2}\) \(A^{2}=B^{2}\) \(A=B\)
BCECE-2008
Motion in Plane
143553
The vectors \(\vec{A}\) and \(\vec{B}\) are such that \(|\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}|=|\overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{B}}|\) The angle between the two vectors is
1 \(60^{\circ}\)
2 \(75^{\circ}\)
3 \(45^{\circ}\)
4 \(90^{\circ}\)
Explanation:
D Let angle between \(\vec{A}\) and \(\vec{B}\) be \(\theta\) The resultant of \(\vec{A}+\vec{B}\) is given by \(R=\sqrt{A^{2}+B^{2}+2 A B \cos \theta}\) The resultant of \(\vec{A}-\vec{B}\) is given by \(R^{\prime}=\sqrt{A^{2}+B^{2}-2 A B \cos \theta}\) According to the question, \(\mathrm{R}=\mathrm{R}^{\prime}\) \(\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}+2 \mathrm{AB} \cos \theta}=\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}-2 \mathrm{AB} \cos \theta}\) \(\mathrm{A}^{2}+\mathrm{B}^{2}+2 \mathrm{AB} \cos \theta=\mathrm{A}^{2}+\mathrm{B}^{2}-2 \mathrm{AB} \cos \theta\) \(4 \mathrm{AB} \cos \theta=0\) \(\quad \theta=90^{\circ}\)
WBJEE-2016
Motion in Plane
143554
The vector sum of two forces is perpendicular to their vector differences. In that case, the forces
1 are not equal to each other in magnitude
2 cannot be predicted
3 are equal to each other
4 are equal to each other in magnitude
Explanation:
D Let \(\overrightarrow{\mathrm{f}_{1}}\) and \(\overrightarrow{\mathrm{f}_{2}}\) be the two forces Then sum of forces, \(\vec{a}=\overrightarrow{f_{1}}+\overrightarrow{f_{2}}\) And difference, \(\overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{f}_{1}}-\overrightarrow{\mathrm{f}_{2}}\) The two forces are perpendicular to each other \((\vec{a} \cdot \vec{b})=0\) \(\left(\overrightarrow{\mathrm{f}_{1}}+\overrightarrow{\mathrm{f}_{2}}\right) \cdot\left(\overrightarrow{\mathrm{f}_{1}}-\overrightarrow{\mathrm{f}_{2}}\right)=0\) \(\left|\overrightarrow{\mathrm{f}}_{1}\right|^{2}-\left|\overrightarrow{\mathrm{f}_{2}}\right|^{2}=0\) \(\left|\overrightarrow{\mathrm{f}}_{1}\right|^{2}=\left|\overrightarrow{\mathrm{f}_{2}}\right|^{2}\) \(\left|\overrightarrow{\mathrm{f}_{1}}\right|=\left|\overrightarrow{\mathrm{f}_{2}}\right|\) In that case both the force are equal and have same magnitude.
AIPMT 2003
Motion in Plane
143555
The sum of two vectors \(\vec{A}\) and \(\vec{B}\) is at right angles to their difference. This is possible if
1 \(\mathrm{A}=2 \mathrm{~B}\)
2 \(\mathrm{A}=\mathrm{B}\)
3 \(\mathrm{A}=3 \mathrm{~B}\)
4 \(\mathrm{B}=2 \mathrm{~A}\)
Explanation:
B Let, \(\overrightarrow{\mathrm{P}}_{1}\) and \(\overrightarrow{\mathrm{P}}_{2}\) sum and difference of vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\), \(\overrightarrow{\mathrm{P}_{1}}=(\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}})\) \(\overrightarrow{\mathrm{P}_{2}}=(\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}})\) \(\overrightarrow{\mathrm{P}_{1}} \cdot \overrightarrow{\mathrm{P}_{2}}=(\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}) \cdot(\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}})\) \(0=\mathrm{A}^{2}-\mathrm{B}^{2}\) \(\mathrm{A}^{2}=\mathrm{B}^{2}\) \(\mathrm{A}=\mathrm{B}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Motion in Plane
143552
The sum of two vectors \(\vec{A}\) and \(\vec{B}\) is at right angles to their difference. Then
1 \(\mathrm{A}=\mathrm{B}\)
2 \(A=2 B\)
3 \(\mathrm{B}=2 \mathrm{~A}\)
4 \(\vec{A}\) and \(\vec{B}\) have the same direction
Explanation:
A Let \(r_{1}\) and \(r_{2}\) be the sum and difference of vectors \(\vec{A}\) and \(\vec{B}\) respectively i.e., \(\mathrm{r}_{1}=\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}\) \(\mathrm{r}_{2}=\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}\) \(r_{1}\) is perpendicular to \(r_{2}\) Taking the dot product of \(r_{1}\) and \(r_{2}\) \(r_{1} \cdot r_{2}=(\vec{A}+\vec{B}) \cdot(\vec{A}-\vec{B})\) \(0=A^{2}-B^{2}\) \(A^{2}=B^{2}\) \(A=B\)
BCECE-2008
Motion in Plane
143553
The vectors \(\vec{A}\) and \(\vec{B}\) are such that \(|\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}|=|\overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{B}}|\) The angle between the two vectors is
1 \(60^{\circ}\)
2 \(75^{\circ}\)
3 \(45^{\circ}\)
4 \(90^{\circ}\)
Explanation:
D Let angle between \(\vec{A}\) and \(\vec{B}\) be \(\theta\) The resultant of \(\vec{A}+\vec{B}\) is given by \(R=\sqrt{A^{2}+B^{2}+2 A B \cos \theta}\) The resultant of \(\vec{A}-\vec{B}\) is given by \(R^{\prime}=\sqrt{A^{2}+B^{2}-2 A B \cos \theta}\) According to the question, \(\mathrm{R}=\mathrm{R}^{\prime}\) \(\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}+2 \mathrm{AB} \cos \theta}=\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}-2 \mathrm{AB} \cos \theta}\) \(\mathrm{A}^{2}+\mathrm{B}^{2}+2 \mathrm{AB} \cos \theta=\mathrm{A}^{2}+\mathrm{B}^{2}-2 \mathrm{AB} \cos \theta\) \(4 \mathrm{AB} \cos \theta=0\) \(\quad \theta=90^{\circ}\)
WBJEE-2016
Motion in Plane
143554
The vector sum of two forces is perpendicular to their vector differences. In that case, the forces
1 are not equal to each other in magnitude
2 cannot be predicted
3 are equal to each other
4 are equal to each other in magnitude
Explanation:
D Let \(\overrightarrow{\mathrm{f}_{1}}\) and \(\overrightarrow{\mathrm{f}_{2}}\) be the two forces Then sum of forces, \(\vec{a}=\overrightarrow{f_{1}}+\overrightarrow{f_{2}}\) And difference, \(\overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{f}_{1}}-\overrightarrow{\mathrm{f}_{2}}\) The two forces are perpendicular to each other \((\vec{a} \cdot \vec{b})=0\) \(\left(\overrightarrow{\mathrm{f}_{1}}+\overrightarrow{\mathrm{f}_{2}}\right) \cdot\left(\overrightarrow{\mathrm{f}_{1}}-\overrightarrow{\mathrm{f}_{2}}\right)=0\) \(\left|\overrightarrow{\mathrm{f}}_{1}\right|^{2}-\left|\overrightarrow{\mathrm{f}_{2}}\right|^{2}=0\) \(\left|\overrightarrow{\mathrm{f}}_{1}\right|^{2}=\left|\overrightarrow{\mathrm{f}_{2}}\right|^{2}\) \(\left|\overrightarrow{\mathrm{f}_{1}}\right|=\left|\overrightarrow{\mathrm{f}_{2}}\right|\) In that case both the force are equal and have same magnitude.
AIPMT 2003
Motion in Plane
143555
The sum of two vectors \(\vec{A}\) and \(\vec{B}\) is at right angles to their difference. This is possible if
1 \(\mathrm{A}=2 \mathrm{~B}\)
2 \(\mathrm{A}=\mathrm{B}\)
3 \(\mathrm{A}=3 \mathrm{~B}\)
4 \(\mathrm{B}=2 \mathrm{~A}\)
Explanation:
B Let, \(\overrightarrow{\mathrm{P}}_{1}\) and \(\overrightarrow{\mathrm{P}}_{2}\) sum and difference of vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\), \(\overrightarrow{\mathrm{P}_{1}}=(\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}})\) \(\overrightarrow{\mathrm{P}_{2}}=(\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}})\) \(\overrightarrow{\mathrm{P}_{1}} \cdot \overrightarrow{\mathrm{P}_{2}}=(\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}) \cdot(\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}})\) \(0=\mathrm{A}^{2}-\mathrm{B}^{2}\) \(\mathrm{A}^{2}=\mathrm{B}^{2}\) \(\mathrm{A}=\mathrm{B}\)
