143567
If the two vectors \(\vec{A}=2 \hat{i}+3 \hat{j}+4 \hat{k} \quad\) and \(\vec{B}=\hat{i}+2 \hat{j}-n \hat{k}\) are perpendicular then the value of \(n\) is :
1 1
2 2
3 3
4 4
5 5
Explanation:
B Given, \(\vec{A}=2 \hat{i}+3 \hat{j}+4 \hat{k}\) \(\vec{B}=\hat{i}+2 \hat{j}-n \hat{k}\) If two vectors are perpendicular then their scalar product is zero. \(\therefore \overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}=0\) \((2 \hat{i}+3 \hat{j}+4 \hat{k}) \cdot(\hat{i}+2 \hat{j}-n \hat{k})=0\) \(2+6-4 n=0\) \(\mathrm{n}=2\)
Kerala CEE 2006
Motion in Plane
143568
A particle is displaced from a position \((2 \hat{i}-\hat{j}+\hat{k})\) to another position \((3 \hat{i}+2 \hat{j}-2 \hat{k})\) under the action of the force of \((2 \hat{i}+\hat{j}-\hat{k})\). The work done by the force in an arbitrary unit is:
143570
The angle between two vectors \(A\) and \(B\) is \(\theta\). Vector \(R\) is the resultant of the two vectors. If \(R\) makes an \(\frac{\theta}{2}\) with \(A\), then
1 \(\mathrm{A}=2 \mathrm{~B}\)
2 \(\mathrm{A}=\frac{\mathrm{B}}{2}\)
3 \(\mathrm{A}=\mathrm{B}\)
4 \(\mathrm{AB}=1\)
Explanation:
C The angle \(\alpha\) which the resultant \(\overrightarrow{\mathrm{R}}\) makes with \(\overrightarrow{\mathrm{A}}\) is given by- \(\tan \alpha=\frac{\mathrm{B} \sin \theta}{\mathrm{A}+\mathrm{B} \cos \theta} \quad\) Here, \(\alpha=\frac{\theta}{2}\) Hence, \(\tan \frac{\theta}{2}=\frac{\mathrm{B} \sin \theta}{\mathrm{A}+\mathrm{B} \cos \theta}\) \(\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}=\frac{2 B \sin (\theta / 2) \cos (\theta / 2)}{\mathrm{A}+\mathrm{B} \cos \theta}\) \(\mathrm{A}+\mathrm{B} \cos \theta=2 \mathrm{~B} \cos ^{2}(\theta / 2)\) \(\mathrm{A}+\mathrm{B}\left[2 \cos ^{2}(\theta / 2)-1\right]=2 \mathrm{~B} \cos ^{2}(\theta / 2)\) \(\mathrm{A}+2 \mathrm{~B} \cos ^{2}\left(\frac{\theta}{2}\right)-\mathrm{B}=2 \mathrm{~B} \cos ^{2}\left(\frac{\theta}{2}\right)\) \(\mathrm{A}-\mathrm{B}=0\) \(\therefore \mathrm{A}=\mathrm{B}\)
UPSEE - 2013
Motion in Plane
143571
The power utilised when a force of \((2 \hat{i}+3 \hat{j}+4 \hat{k}) N\) acts on a body for \(4 s\), producing a displacement of \((3 \hat{i}+4 \hat{j}+5 \hat{k}) \mathrm{m}\), is
1 \(9.5 \mathrm{~W}\)
2 \(7.5 \mathrm{~W}\)
3 \(6.5 \mathrm{~W}\)
4 \(4.5 \mathrm{~W}\)
Explanation:
A Power \(=\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{v}}\) And, \(\quad \vec{v}=\frac{\vec{d}}{t}\) \(\overrightarrow{\mathrm{v}}=\frac{3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}}{4}\) \(\therefore\) Power \(=(2 \hat{i}+3 \hat{j}+4 \hat{k}) \cdot \frac{(3 \hat{i}+4 \hat{j}+5 \hat{k})}{4}\) \(=2 \times \frac{3}{4}+3 \times \frac{4}{4}+4 \times \frac{5}{4}\) \(=\frac{6}{4}+\frac{12}{4}+\frac{20}{4}=\frac{6+12+20}{4}\) \(=\frac{38}{4}\) Power \(=9.5 \mathrm{~W}\)
AP EAMCET (21.09.2020) Shift-I
Motion in Plane
143572
\(\quad \vec{A}=4 \hat{i}+3 \hat{j}\) and \(\vec{B}=4 \hat{i}+2 \hat{j}\). Find a vector parallel to \(\vec{A}\) but has magnitude five times that of \(\overrightarrow{\mathbf{B}}\).
143567
If the two vectors \(\vec{A}=2 \hat{i}+3 \hat{j}+4 \hat{k} \quad\) and \(\vec{B}=\hat{i}+2 \hat{j}-n \hat{k}\) are perpendicular then the value of \(n\) is :
1 1
2 2
3 3
4 4
5 5
Explanation:
B Given, \(\vec{A}=2 \hat{i}+3 \hat{j}+4 \hat{k}\) \(\vec{B}=\hat{i}+2 \hat{j}-n \hat{k}\) If two vectors are perpendicular then their scalar product is zero. \(\therefore \overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}=0\) \((2 \hat{i}+3 \hat{j}+4 \hat{k}) \cdot(\hat{i}+2 \hat{j}-n \hat{k})=0\) \(2+6-4 n=0\) \(\mathrm{n}=2\)
Kerala CEE 2006
Motion in Plane
143568
A particle is displaced from a position \((2 \hat{i}-\hat{j}+\hat{k})\) to another position \((3 \hat{i}+2 \hat{j}-2 \hat{k})\) under the action of the force of \((2 \hat{i}+\hat{j}-\hat{k})\). The work done by the force in an arbitrary unit is:
143570
The angle between two vectors \(A\) and \(B\) is \(\theta\). Vector \(R\) is the resultant of the two vectors. If \(R\) makes an \(\frac{\theta}{2}\) with \(A\), then
1 \(\mathrm{A}=2 \mathrm{~B}\)
2 \(\mathrm{A}=\frac{\mathrm{B}}{2}\)
3 \(\mathrm{A}=\mathrm{B}\)
4 \(\mathrm{AB}=1\)
Explanation:
C The angle \(\alpha\) which the resultant \(\overrightarrow{\mathrm{R}}\) makes with \(\overrightarrow{\mathrm{A}}\) is given by- \(\tan \alpha=\frac{\mathrm{B} \sin \theta}{\mathrm{A}+\mathrm{B} \cos \theta} \quad\) Here, \(\alpha=\frac{\theta}{2}\) Hence, \(\tan \frac{\theta}{2}=\frac{\mathrm{B} \sin \theta}{\mathrm{A}+\mathrm{B} \cos \theta}\) \(\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}=\frac{2 B \sin (\theta / 2) \cos (\theta / 2)}{\mathrm{A}+\mathrm{B} \cos \theta}\) \(\mathrm{A}+\mathrm{B} \cos \theta=2 \mathrm{~B} \cos ^{2}(\theta / 2)\) \(\mathrm{A}+\mathrm{B}\left[2 \cos ^{2}(\theta / 2)-1\right]=2 \mathrm{~B} \cos ^{2}(\theta / 2)\) \(\mathrm{A}+2 \mathrm{~B} \cos ^{2}\left(\frac{\theta}{2}\right)-\mathrm{B}=2 \mathrm{~B} \cos ^{2}\left(\frac{\theta}{2}\right)\) \(\mathrm{A}-\mathrm{B}=0\) \(\therefore \mathrm{A}=\mathrm{B}\)
UPSEE - 2013
Motion in Plane
143571
The power utilised when a force of \((2 \hat{i}+3 \hat{j}+4 \hat{k}) N\) acts on a body for \(4 s\), producing a displacement of \((3 \hat{i}+4 \hat{j}+5 \hat{k}) \mathrm{m}\), is
1 \(9.5 \mathrm{~W}\)
2 \(7.5 \mathrm{~W}\)
3 \(6.5 \mathrm{~W}\)
4 \(4.5 \mathrm{~W}\)
Explanation:
A Power \(=\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{v}}\) And, \(\quad \vec{v}=\frac{\vec{d}}{t}\) \(\overrightarrow{\mathrm{v}}=\frac{3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}}{4}\) \(\therefore\) Power \(=(2 \hat{i}+3 \hat{j}+4 \hat{k}) \cdot \frac{(3 \hat{i}+4 \hat{j}+5 \hat{k})}{4}\) \(=2 \times \frac{3}{4}+3 \times \frac{4}{4}+4 \times \frac{5}{4}\) \(=\frac{6}{4}+\frac{12}{4}+\frac{20}{4}=\frac{6+12+20}{4}\) \(=\frac{38}{4}\) Power \(=9.5 \mathrm{~W}\)
AP EAMCET (21.09.2020) Shift-I
Motion in Plane
143572
\(\quad \vec{A}=4 \hat{i}+3 \hat{j}\) and \(\vec{B}=4 \hat{i}+2 \hat{j}\). Find a vector parallel to \(\vec{A}\) but has magnitude five times that of \(\overrightarrow{\mathbf{B}}\).
143567
If the two vectors \(\vec{A}=2 \hat{i}+3 \hat{j}+4 \hat{k} \quad\) and \(\vec{B}=\hat{i}+2 \hat{j}-n \hat{k}\) are perpendicular then the value of \(n\) is :
1 1
2 2
3 3
4 4
5 5
Explanation:
B Given, \(\vec{A}=2 \hat{i}+3 \hat{j}+4 \hat{k}\) \(\vec{B}=\hat{i}+2 \hat{j}-n \hat{k}\) If two vectors are perpendicular then their scalar product is zero. \(\therefore \overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}=0\) \((2 \hat{i}+3 \hat{j}+4 \hat{k}) \cdot(\hat{i}+2 \hat{j}-n \hat{k})=0\) \(2+6-4 n=0\) \(\mathrm{n}=2\)
Kerala CEE 2006
Motion in Plane
143568
A particle is displaced from a position \((2 \hat{i}-\hat{j}+\hat{k})\) to another position \((3 \hat{i}+2 \hat{j}-2 \hat{k})\) under the action of the force of \((2 \hat{i}+\hat{j}-\hat{k})\). The work done by the force in an arbitrary unit is:
143570
The angle between two vectors \(A\) and \(B\) is \(\theta\). Vector \(R\) is the resultant of the two vectors. If \(R\) makes an \(\frac{\theta}{2}\) with \(A\), then
1 \(\mathrm{A}=2 \mathrm{~B}\)
2 \(\mathrm{A}=\frac{\mathrm{B}}{2}\)
3 \(\mathrm{A}=\mathrm{B}\)
4 \(\mathrm{AB}=1\)
Explanation:
C The angle \(\alpha\) which the resultant \(\overrightarrow{\mathrm{R}}\) makes with \(\overrightarrow{\mathrm{A}}\) is given by- \(\tan \alpha=\frac{\mathrm{B} \sin \theta}{\mathrm{A}+\mathrm{B} \cos \theta} \quad\) Here, \(\alpha=\frac{\theta}{2}\) Hence, \(\tan \frac{\theta}{2}=\frac{\mathrm{B} \sin \theta}{\mathrm{A}+\mathrm{B} \cos \theta}\) \(\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}=\frac{2 B \sin (\theta / 2) \cos (\theta / 2)}{\mathrm{A}+\mathrm{B} \cos \theta}\) \(\mathrm{A}+\mathrm{B} \cos \theta=2 \mathrm{~B} \cos ^{2}(\theta / 2)\) \(\mathrm{A}+\mathrm{B}\left[2 \cos ^{2}(\theta / 2)-1\right]=2 \mathrm{~B} \cos ^{2}(\theta / 2)\) \(\mathrm{A}+2 \mathrm{~B} \cos ^{2}\left(\frac{\theta}{2}\right)-\mathrm{B}=2 \mathrm{~B} \cos ^{2}\left(\frac{\theta}{2}\right)\) \(\mathrm{A}-\mathrm{B}=0\) \(\therefore \mathrm{A}=\mathrm{B}\)
UPSEE - 2013
Motion in Plane
143571
The power utilised when a force of \((2 \hat{i}+3 \hat{j}+4 \hat{k}) N\) acts on a body for \(4 s\), producing a displacement of \((3 \hat{i}+4 \hat{j}+5 \hat{k}) \mathrm{m}\), is
1 \(9.5 \mathrm{~W}\)
2 \(7.5 \mathrm{~W}\)
3 \(6.5 \mathrm{~W}\)
4 \(4.5 \mathrm{~W}\)
Explanation:
A Power \(=\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{v}}\) And, \(\quad \vec{v}=\frac{\vec{d}}{t}\) \(\overrightarrow{\mathrm{v}}=\frac{3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}}{4}\) \(\therefore\) Power \(=(2 \hat{i}+3 \hat{j}+4 \hat{k}) \cdot \frac{(3 \hat{i}+4 \hat{j}+5 \hat{k})}{4}\) \(=2 \times \frac{3}{4}+3 \times \frac{4}{4}+4 \times \frac{5}{4}\) \(=\frac{6}{4}+\frac{12}{4}+\frac{20}{4}=\frac{6+12+20}{4}\) \(=\frac{38}{4}\) Power \(=9.5 \mathrm{~W}\)
AP EAMCET (21.09.2020) Shift-I
Motion in Plane
143572
\(\quad \vec{A}=4 \hat{i}+3 \hat{j}\) and \(\vec{B}=4 \hat{i}+2 \hat{j}\). Find a vector parallel to \(\vec{A}\) but has magnitude five times that of \(\overrightarrow{\mathbf{B}}\).
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Motion in Plane
143567
If the two vectors \(\vec{A}=2 \hat{i}+3 \hat{j}+4 \hat{k} \quad\) and \(\vec{B}=\hat{i}+2 \hat{j}-n \hat{k}\) are perpendicular then the value of \(n\) is :
1 1
2 2
3 3
4 4
5 5
Explanation:
B Given, \(\vec{A}=2 \hat{i}+3 \hat{j}+4 \hat{k}\) \(\vec{B}=\hat{i}+2 \hat{j}-n \hat{k}\) If two vectors are perpendicular then their scalar product is zero. \(\therefore \overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}=0\) \((2 \hat{i}+3 \hat{j}+4 \hat{k}) \cdot(\hat{i}+2 \hat{j}-n \hat{k})=0\) \(2+6-4 n=0\) \(\mathrm{n}=2\)
Kerala CEE 2006
Motion in Plane
143568
A particle is displaced from a position \((2 \hat{i}-\hat{j}+\hat{k})\) to another position \((3 \hat{i}+2 \hat{j}-2 \hat{k})\) under the action of the force of \((2 \hat{i}+\hat{j}-\hat{k})\). The work done by the force in an arbitrary unit is:
143570
The angle between two vectors \(A\) and \(B\) is \(\theta\). Vector \(R\) is the resultant of the two vectors. If \(R\) makes an \(\frac{\theta}{2}\) with \(A\), then
1 \(\mathrm{A}=2 \mathrm{~B}\)
2 \(\mathrm{A}=\frac{\mathrm{B}}{2}\)
3 \(\mathrm{A}=\mathrm{B}\)
4 \(\mathrm{AB}=1\)
Explanation:
C The angle \(\alpha\) which the resultant \(\overrightarrow{\mathrm{R}}\) makes with \(\overrightarrow{\mathrm{A}}\) is given by- \(\tan \alpha=\frac{\mathrm{B} \sin \theta}{\mathrm{A}+\mathrm{B} \cos \theta} \quad\) Here, \(\alpha=\frac{\theta}{2}\) Hence, \(\tan \frac{\theta}{2}=\frac{\mathrm{B} \sin \theta}{\mathrm{A}+\mathrm{B} \cos \theta}\) \(\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}=\frac{2 B \sin (\theta / 2) \cos (\theta / 2)}{\mathrm{A}+\mathrm{B} \cos \theta}\) \(\mathrm{A}+\mathrm{B} \cos \theta=2 \mathrm{~B} \cos ^{2}(\theta / 2)\) \(\mathrm{A}+\mathrm{B}\left[2 \cos ^{2}(\theta / 2)-1\right]=2 \mathrm{~B} \cos ^{2}(\theta / 2)\) \(\mathrm{A}+2 \mathrm{~B} \cos ^{2}\left(\frac{\theta}{2}\right)-\mathrm{B}=2 \mathrm{~B} \cos ^{2}\left(\frac{\theta}{2}\right)\) \(\mathrm{A}-\mathrm{B}=0\) \(\therefore \mathrm{A}=\mathrm{B}\)
UPSEE - 2013
Motion in Plane
143571
The power utilised when a force of \((2 \hat{i}+3 \hat{j}+4 \hat{k}) N\) acts on a body for \(4 s\), producing a displacement of \((3 \hat{i}+4 \hat{j}+5 \hat{k}) \mathrm{m}\), is
1 \(9.5 \mathrm{~W}\)
2 \(7.5 \mathrm{~W}\)
3 \(6.5 \mathrm{~W}\)
4 \(4.5 \mathrm{~W}\)
Explanation:
A Power \(=\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{v}}\) And, \(\quad \vec{v}=\frac{\vec{d}}{t}\) \(\overrightarrow{\mathrm{v}}=\frac{3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}}{4}\) \(\therefore\) Power \(=(2 \hat{i}+3 \hat{j}+4 \hat{k}) \cdot \frac{(3 \hat{i}+4 \hat{j}+5 \hat{k})}{4}\) \(=2 \times \frac{3}{4}+3 \times \frac{4}{4}+4 \times \frac{5}{4}\) \(=\frac{6}{4}+\frac{12}{4}+\frac{20}{4}=\frac{6+12+20}{4}\) \(=\frac{38}{4}\) Power \(=9.5 \mathrm{~W}\)
AP EAMCET (21.09.2020) Shift-I
Motion in Plane
143572
\(\quad \vec{A}=4 \hat{i}+3 \hat{j}\) and \(\vec{B}=4 \hat{i}+2 \hat{j}\). Find a vector parallel to \(\vec{A}\) but has magnitude five times that of \(\overrightarrow{\mathbf{B}}\).
143567
If the two vectors \(\vec{A}=2 \hat{i}+3 \hat{j}+4 \hat{k} \quad\) and \(\vec{B}=\hat{i}+2 \hat{j}-n \hat{k}\) are perpendicular then the value of \(n\) is :
1 1
2 2
3 3
4 4
5 5
Explanation:
B Given, \(\vec{A}=2 \hat{i}+3 \hat{j}+4 \hat{k}\) \(\vec{B}=\hat{i}+2 \hat{j}-n \hat{k}\) If two vectors are perpendicular then their scalar product is zero. \(\therefore \overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}=0\) \((2 \hat{i}+3 \hat{j}+4 \hat{k}) \cdot(\hat{i}+2 \hat{j}-n \hat{k})=0\) \(2+6-4 n=0\) \(\mathrm{n}=2\)
Kerala CEE 2006
Motion in Plane
143568
A particle is displaced from a position \((2 \hat{i}-\hat{j}+\hat{k})\) to another position \((3 \hat{i}+2 \hat{j}-2 \hat{k})\) under the action of the force of \((2 \hat{i}+\hat{j}-\hat{k})\). The work done by the force in an arbitrary unit is:
143570
The angle between two vectors \(A\) and \(B\) is \(\theta\). Vector \(R\) is the resultant of the two vectors. If \(R\) makes an \(\frac{\theta}{2}\) with \(A\), then
1 \(\mathrm{A}=2 \mathrm{~B}\)
2 \(\mathrm{A}=\frac{\mathrm{B}}{2}\)
3 \(\mathrm{A}=\mathrm{B}\)
4 \(\mathrm{AB}=1\)
Explanation:
C The angle \(\alpha\) which the resultant \(\overrightarrow{\mathrm{R}}\) makes with \(\overrightarrow{\mathrm{A}}\) is given by- \(\tan \alpha=\frac{\mathrm{B} \sin \theta}{\mathrm{A}+\mathrm{B} \cos \theta} \quad\) Here, \(\alpha=\frac{\theta}{2}\) Hence, \(\tan \frac{\theta}{2}=\frac{\mathrm{B} \sin \theta}{\mathrm{A}+\mathrm{B} \cos \theta}\) \(\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}=\frac{2 B \sin (\theta / 2) \cos (\theta / 2)}{\mathrm{A}+\mathrm{B} \cos \theta}\) \(\mathrm{A}+\mathrm{B} \cos \theta=2 \mathrm{~B} \cos ^{2}(\theta / 2)\) \(\mathrm{A}+\mathrm{B}\left[2 \cos ^{2}(\theta / 2)-1\right]=2 \mathrm{~B} \cos ^{2}(\theta / 2)\) \(\mathrm{A}+2 \mathrm{~B} \cos ^{2}\left(\frac{\theta}{2}\right)-\mathrm{B}=2 \mathrm{~B} \cos ^{2}\left(\frac{\theta}{2}\right)\) \(\mathrm{A}-\mathrm{B}=0\) \(\therefore \mathrm{A}=\mathrm{B}\)
UPSEE - 2013
Motion in Plane
143571
The power utilised when a force of \((2 \hat{i}+3 \hat{j}+4 \hat{k}) N\) acts on a body for \(4 s\), producing a displacement of \((3 \hat{i}+4 \hat{j}+5 \hat{k}) \mathrm{m}\), is
1 \(9.5 \mathrm{~W}\)
2 \(7.5 \mathrm{~W}\)
3 \(6.5 \mathrm{~W}\)
4 \(4.5 \mathrm{~W}\)
Explanation:
A Power \(=\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{v}}\) And, \(\quad \vec{v}=\frac{\vec{d}}{t}\) \(\overrightarrow{\mathrm{v}}=\frac{3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}}{4}\) \(\therefore\) Power \(=(2 \hat{i}+3 \hat{j}+4 \hat{k}) \cdot \frac{(3 \hat{i}+4 \hat{j}+5 \hat{k})}{4}\) \(=2 \times \frac{3}{4}+3 \times \frac{4}{4}+4 \times \frac{5}{4}\) \(=\frac{6}{4}+\frac{12}{4}+\frac{20}{4}=\frac{6+12+20}{4}\) \(=\frac{38}{4}\) Power \(=9.5 \mathrm{~W}\)
AP EAMCET (21.09.2020) Shift-I
Motion in Plane
143572
\(\quad \vec{A}=4 \hat{i}+3 \hat{j}\) and \(\vec{B}=4 \hat{i}+2 \hat{j}\). Find a vector parallel to \(\vec{A}\) but has magnitude five times that of \(\overrightarrow{\mathbf{B}}\).