143577
The direction of \(\vec{A}\) is vertically upward and direction of \(\vec{B}\) is in north direction. The direction of \(\vec{A} \times \vec{B}\) will be
1 Western direction
2 Eastern direction
3 At \(45^{\circ}\) upward in north
4 Vertically downward
Explanation:
A Considering vertically upward direction as z-axis and north direction as y-axis. \(\vec{A}=a \hat{k}, \vec{B}=b \hat{j}\) \(\therefore \vec{A} \times \vec{B}=a \hat{k} \times b \hat{j}=a b(-\hat{i})\) Thus, it is along negative \(\mathrm{x}\)-axis \(\therefore \overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}\) is along west.
CG PET- 2009
Motion in Plane
143578
If \(\vec{A}=\vec{B}+\vec{C}\) and the values of \(\vec{A}, \vec{B}\) and \(\vec{C}\) are 13, 12 and 5 respectively, then the angle between \(\vec{A}\) and \(\vec{C}\) will be
1 \(\cos ^{-1}(5 / 13)\)
2 \(\cos ^{-1}(13 / 12)\)
3 \(\pi / 2\)
4 \(\sin ^{-1}(5 / 12)\)
Explanation:
A Given that, \(\overrightarrow{\mathrm{A}}=13, \overrightarrow{\mathrm{B}}=12, \overrightarrow{\mathrm{C}}=5\) \(\mathrm{A}^{2}=\mathrm{B}^{2}+\mathrm{C}^{2}+2 \mathrm{BC} \cos \theta\) \((13)^{2}=(12)^{2}+(5)^{2}+2 \times 12 \times 5 \cos \theta\) \(\cos \theta=0\) \(\theta=90^{\circ}\) Hence, it is a right-angle triangle. So, angle between \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{C}}\) is, \(\cos \alpha=\frac{5}{13} \Rightarrow \alpha=\cos ^{-1}\left(\frac{5}{13}\right)\)
CG PET- 2009
Motion in Plane
143580
Let \(A_{1}+A_{2}=5 A_{3}, A_{1}-A_{2}=3 A_{3}\), \(A_{3}=2 \hat{i}+4 \hat{j} \text {, then } \frac{\left|A_{1}\right|}{\left|A_{2}\right|} \text { is }\)
143581
The magnitude of \(x\) and \(y\) components of \(A\) are 7 and 6 respectively. Also the magnitudes of \(x\) and \(y\) components of \(A+B\) are 11 and 9 respectively. Calculate the magnitude of vector B.
143577
The direction of \(\vec{A}\) is vertically upward and direction of \(\vec{B}\) is in north direction. The direction of \(\vec{A} \times \vec{B}\) will be
1 Western direction
2 Eastern direction
3 At \(45^{\circ}\) upward in north
4 Vertically downward
Explanation:
A Considering vertically upward direction as z-axis and north direction as y-axis. \(\vec{A}=a \hat{k}, \vec{B}=b \hat{j}\) \(\therefore \vec{A} \times \vec{B}=a \hat{k} \times b \hat{j}=a b(-\hat{i})\) Thus, it is along negative \(\mathrm{x}\)-axis \(\therefore \overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}\) is along west.
CG PET- 2009
Motion in Plane
143578
If \(\vec{A}=\vec{B}+\vec{C}\) and the values of \(\vec{A}, \vec{B}\) and \(\vec{C}\) are 13, 12 and 5 respectively, then the angle between \(\vec{A}\) and \(\vec{C}\) will be
1 \(\cos ^{-1}(5 / 13)\)
2 \(\cos ^{-1}(13 / 12)\)
3 \(\pi / 2\)
4 \(\sin ^{-1}(5 / 12)\)
Explanation:
A Given that, \(\overrightarrow{\mathrm{A}}=13, \overrightarrow{\mathrm{B}}=12, \overrightarrow{\mathrm{C}}=5\) \(\mathrm{A}^{2}=\mathrm{B}^{2}+\mathrm{C}^{2}+2 \mathrm{BC} \cos \theta\) \((13)^{2}=(12)^{2}+(5)^{2}+2 \times 12 \times 5 \cos \theta\) \(\cos \theta=0\) \(\theta=90^{\circ}\) Hence, it is a right-angle triangle. So, angle between \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{C}}\) is, \(\cos \alpha=\frac{5}{13} \Rightarrow \alpha=\cos ^{-1}\left(\frac{5}{13}\right)\)
CG PET- 2009
Motion in Plane
143580
Let \(A_{1}+A_{2}=5 A_{3}, A_{1}-A_{2}=3 A_{3}\), \(A_{3}=2 \hat{i}+4 \hat{j} \text {, then } \frac{\left|A_{1}\right|}{\left|A_{2}\right|} \text { is }\)
143581
The magnitude of \(x\) and \(y\) components of \(A\) are 7 and 6 respectively. Also the magnitudes of \(x\) and \(y\) components of \(A+B\) are 11 and 9 respectively. Calculate the magnitude of vector B.
143577
The direction of \(\vec{A}\) is vertically upward and direction of \(\vec{B}\) is in north direction. The direction of \(\vec{A} \times \vec{B}\) will be
1 Western direction
2 Eastern direction
3 At \(45^{\circ}\) upward in north
4 Vertically downward
Explanation:
A Considering vertically upward direction as z-axis and north direction as y-axis. \(\vec{A}=a \hat{k}, \vec{B}=b \hat{j}\) \(\therefore \vec{A} \times \vec{B}=a \hat{k} \times b \hat{j}=a b(-\hat{i})\) Thus, it is along negative \(\mathrm{x}\)-axis \(\therefore \overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}\) is along west.
CG PET- 2009
Motion in Plane
143578
If \(\vec{A}=\vec{B}+\vec{C}\) and the values of \(\vec{A}, \vec{B}\) and \(\vec{C}\) are 13, 12 and 5 respectively, then the angle between \(\vec{A}\) and \(\vec{C}\) will be
1 \(\cos ^{-1}(5 / 13)\)
2 \(\cos ^{-1}(13 / 12)\)
3 \(\pi / 2\)
4 \(\sin ^{-1}(5 / 12)\)
Explanation:
A Given that, \(\overrightarrow{\mathrm{A}}=13, \overrightarrow{\mathrm{B}}=12, \overrightarrow{\mathrm{C}}=5\) \(\mathrm{A}^{2}=\mathrm{B}^{2}+\mathrm{C}^{2}+2 \mathrm{BC} \cos \theta\) \((13)^{2}=(12)^{2}+(5)^{2}+2 \times 12 \times 5 \cos \theta\) \(\cos \theta=0\) \(\theta=90^{\circ}\) Hence, it is a right-angle triangle. So, angle between \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{C}}\) is, \(\cos \alpha=\frac{5}{13} \Rightarrow \alpha=\cos ^{-1}\left(\frac{5}{13}\right)\)
CG PET- 2009
Motion in Plane
143580
Let \(A_{1}+A_{2}=5 A_{3}, A_{1}-A_{2}=3 A_{3}\), \(A_{3}=2 \hat{i}+4 \hat{j} \text {, then } \frac{\left|A_{1}\right|}{\left|A_{2}\right|} \text { is }\)
143581
The magnitude of \(x\) and \(y\) components of \(A\) are 7 and 6 respectively. Also the magnitudes of \(x\) and \(y\) components of \(A+B\) are 11 and 9 respectively. Calculate the magnitude of vector B.
143577
The direction of \(\vec{A}\) is vertically upward and direction of \(\vec{B}\) is in north direction. The direction of \(\vec{A} \times \vec{B}\) will be
1 Western direction
2 Eastern direction
3 At \(45^{\circ}\) upward in north
4 Vertically downward
Explanation:
A Considering vertically upward direction as z-axis and north direction as y-axis. \(\vec{A}=a \hat{k}, \vec{B}=b \hat{j}\) \(\therefore \vec{A} \times \vec{B}=a \hat{k} \times b \hat{j}=a b(-\hat{i})\) Thus, it is along negative \(\mathrm{x}\)-axis \(\therefore \overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}\) is along west.
CG PET- 2009
Motion in Plane
143578
If \(\vec{A}=\vec{B}+\vec{C}\) and the values of \(\vec{A}, \vec{B}\) and \(\vec{C}\) are 13, 12 and 5 respectively, then the angle between \(\vec{A}\) and \(\vec{C}\) will be
1 \(\cos ^{-1}(5 / 13)\)
2 \(\cos ^{-1}(13 / 12)\)
3 \(\pi / 2\)
4 \(\sin ^{-1}(5 / 12)\)
Explanation:
A Given that, \(\overrightarrow{\mathrm{A}}=13, \overrightarrow{\mathrm{B}}=12, \overrightarrow{\mathrm{C}}=5\) \(\mathrm{A}^{2}=\mathrm{B}^{2}+\mathrm{C}^{2}+2 \mathrm{BC} \cos \theta\) \((13)^{2}=(12)^{2}+(5)^{2}+2 \times 12 \times 5 \cos \theta\) \(\cos \theta=0\) \(\theta=90^{\circ}\) Hence, it is a right-angle triangle. So, angle between \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{C}}\) is, \(\cos \alpha=\frac{5}{13} \Rightarrow \alpha=\cos ^{-1}\left(\frac{5}{13}\right)\)
CG PET- 2009
Motion in Plane
143580
Let \(A_{1}+A_{2}=5 A_{3}, A_{1}-A_{2}=3 A_{3}\), \(A_{3}=2 \hat{i}+4 \hat{j} \text {, then } \frac{\left|A_{1}\right|}{\left|A_{2}\right|} \text { is }\)
143581
The magnitude of \(x\) and \(y\) components of \(A\) are 7 and 6 respectively. Also the magnitudes of \(x\) and \(y\) components of \(A+B\) are 11 and 9 respectively. Calculate the magnitude of vector B.
