NEET Test Series from KOTA - 10 Papers In MS WORD
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Motion in Plane
143631
If a unit vector is represented by \(0.5 \hat{i}+0.8 \hat{j}+c \hat{k}\), then the value of \(c\) is
1 1
2 \(\sqrt{0.11}\)
3 \(\sqrt{0.01}\)
4 0.39
Explanation:
B Given, \(\overrightarrow{\mathrm{A}}=0.5 \hat{\mathrm{i}}+0.8 \hat{\mathrm{j}}+\mathrm{c} \hat{\mathrm{k}}\) It is unit vector so it has magnitude \(|\overrightarrow{\mathrm{A}}|=1\) \(\sqrt{(0.5)^{2}+(0.8)^{2}+\mathrm{c}^{2}}=1\) \(\mathrm{c}^{2}=0.11\) \(\mathrm{c}=\sqrt{0.11}\)
TS EAMCET (Medical)-2017
Motion in Plane
143633
The angle between the two vectors \(A=3 \hat{i}+4 \hat{j}+5 \hat{k}\) and \(B=3 \hat{i}+4 \hat{j}-5 \hat{k}\) will be
143634
The angle between \(\vec{A}\) and \(\vec{B}\) is \(\theta\). The value of the triple product \(\overrightarrow{\mathbf{A}} \cdot(\overrightarrow{\mathbf{B}} \times \overrightarrow{\mathbf{A}})\) is
1 \(A^{2} B\)
2 zero
3 \(\mathrm{A}^{2} \mathrm{~B} \sin \theta\)
4 \(\mathrm{A}^{2} \mathrm{~B} \cos \theta\)
Explanation:
B Let \(\vec{A} \cdot(\vec{B} \times \vec{A})=\vec{A} \cdot \vec{C}\) Here \(\vec{C}=\vec{B} \times \vec{A}\) which is perpendicular to both vector \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\). \(\therefore \quad \overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{C}}=0\) \(\because \quad \overrightarrow{\mathrm{C}}\) is perpendicular to \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) \(\therefore\) Angle between \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{C}}\) is \(90^{\circ}\) \(\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{C}}=\mathrm{AC} \cos \theta\) \(\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{C}}=0\) \(\therefore \quad \overrightarrow{\mathrm{A}} \cdot(\overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{A}})=0\)
JIPMER-2007
Motion in Plane
143635
\(\overrightarrow{\mathbf{A}}\) is a vector quantity such that \(|\overrightarrow{\mathbf{A}}|=\) non-zero constant. Which of the following expression is true for \(\overrightarrow{\mathbf{A}}\) ?
143631
If a unit vector is represented by \(0.5 \hat{i}+0.8 \hat{j}+c \hat{k}\), then the value of \(c\) is
1 1
2 \(\sqrt{0.11}\)
3 \(\sqrt{0.01}\)
4 0.39
Explanation:
B Given, \(\overrightarrow{\mathrm{A}}=0.5 \hat{\mathrm{i}}+0.8 \hat{\mathrm{j}}+\mathrm{c} \hat{\mathrm{k}}\) It is unit vector so it has magnitude \(|\overrightarrow{\mathrm{A}}|=1\) \(\sqrt{(0.5)^{2}+(0.8)^{2}+\mathrm{c}^{2}}=1\) \(\mathrm{c}^{2}=0.11\) \(\mathrm{c}=\sqrt{0.11}\)
TS EAMCET (Medical)-2017
Motion in Plane
143633
The angle between the two vectors \(A=3 \hat{i}+4 \hat{j}+5 \hat{k}\) and \(B=3 \hat{i}+4 \hat{j}-5 \hat{k}\) will be
143634
The angle between \(\vec{A}\) and \(\vec{B}\) is \(\theta\). The value of the triple product \(\overrightarrow{\mathbf{A}} \cdot(\overrightarrow{\mathbf{B}} \times \overrightarrow{\mathbf{A}})\) is
1 \(A^{2} B\)
2 zero
3 \(\mathrm{A}^{2} \mathrm{~B} \sin \theta\)
4 \(\mathrm{A}^{2} \mathrm{~B} \cos \theta\)
Explanation:
B Let \(\vec{A} \cdot(\vec{B} \times \vec{A})=\vec{A} \cdot \vec{C}\) Here \(\vec{C}=\vec{B} \times \vec{A}\) which is perpendicular to both vector \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\). \(\therefore \quad \overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{C}}=0\) \(\because \quad \overrightarrow{\mathrm{C}}\) is perpendicular to \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) \(\therefore\) Angle between \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{C}}\) is \(90^{\circ}\) \(\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{C}}=\mathrm{AC} \cos \theta\) \(\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{C}}=0\) \(\therefore \quad \overrightarrow{\mathrm{A}} \cdot(\overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{A}})=0\)
JIPMER-2007
Motion in Plane
143635
\(\overrightarrow{\mathbf{A}}\) is a vector quantity such that \(|\overrightarrow{\mathbf{A}}|=\) non-zero constant. Which of the following expression is true for \(\overrightarrow{\mathbf{A}}\) ?
143631
If a unit vector is represented by \(0.5 \hat{i}+0.8 \hat{j}+c \hat{k}\), then the value of \(c\) is
1 1
2 \(\sqrt{0.11}\)
3 \(\sqrt{0.01}\)
4 0.39
Explanation:
B Given, \(\overrightarrow{\mathrm{A}}=0.5 \hat{\mathrm{i}}+0.8 \hat{\mathrm{j}}+\mathrm{c} \hat{\mathrm{k}}\) It is unit vector so it has magnitude \(|\overrightarrow{\mathrm{A}}|=1\) \(\sqrt{(0.5)^{2}+(0.8)^{2}+\mathrm{c}^{2}}=1\) \(\mathrm{c}^{2}=0.11\) \(\mathrm{c}=\sqrt{0.11}\)
TS EAMCET (Medical)-2017
Motion in Plane
143633
The angle between the two vectors \(A=3 \hat{i}+4 \hat{j}+5 \hat{k}\) and \(B=3 \hat{i}+4 \hat{j}-5 \hat{k}\) will be
143634
The angle between \(\vec{A}\) and \(\vec{B}\) is \(\theta\). The value of the triple product \(\overrightarrow{\mathbf{A}} \cdot(\overrightarrow{\mathbf{B}} \times \overrightarrow{\mathbf{A}})\) is
1 \(A^{2} B\)
2 zero
3 \(\mathrm{A}^{2} \mathrm{~B} \sin \theta\)
4 \(\mathrm{A}^{2} \mathrm{~B} \cos \theta\)
Explanation:
B Let \(\vec{A} \cdot(\vec{B} \times \vec{A})=\vec{A} \cdot \vec{C}\) Here \(\vec{C}=\vec{B} \times \vec{A}\) which is perpendicular to both vector \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\). \(\therefore \quad \overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{C}}=0\) \(\because \quad \overrightarrow{\mathrm{C}}\) is perpendicular to \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) \(\therefore\) Angle between \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{C}}\) is \(90^{\circ}\) \(\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{C}}=\mathrm{AC} \cos \theta\) \(\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{C}}=0\) \(\therefore \quad \overrightarrow{\mathrm{A}} \cdot(\overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{A}})=0\)
JIPMER-2007
Motion in Plane
143635
\(\overrightarrow{\mathbf{A}}\) is a vector quantity such that \(|\overrightarrow{\mathbf{A}}|=\) non-zero constant. Which of the following expression is true for \(\overrightarrow{\mathbf{A}}\) ?
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Motion in Plane
143631
If a unit vector is represented by \(0.5 \hat{i}+0.8 \hat{j}+c \hat{k}\), then the value of \(c\) is
1 1
2 \(\sqrt{0.11}\)
3 \(\sqrt{0.01}\)
4 0.39
Explanation:
B Given, \(\overrightarrow{\mathrm{A}}=0.5 \hat{\mathrm{i}}+0.8 \hat{\mathrm{j}}+\mathrm{c} \hat{\mathrm{k}}\) It is unit vector so it has magnitude \(|\overrightarrow{\mathrm{A}}|=1\) \(\sqrt{(0.5)^{2}+(0.8)^{2}+\mathrm{c}^{2}}=1\) \(\mathrm{c}^{2}=0.11\) \(\mathrm{c}=\sqrt{0.11}\)
TS EAMCET (Medical)-2017
Motion in Plane
143633
The angle between the two vectors \(A=3 \hat{i}+4 \hat{j}+5 \hat{k}\) and \(B=3 \hat{i}+4 \hat{j}-5 \hat{k}\) will be
143634
The angle between \(\vec{A}\) and \(\vec{B}\) is \(\theta\). The value of the triple product \(\overrightarrow{\mathbf{A}} \cdot(\overrightarrow{\mathbf{B}} \times \overrightarrow{\mathbf{A}})\) is
1 \(A^{2} B\)
2 zero
3 \(\mathrm{A}^{2} \mathrm{~B} \sin \theta\)
4 \(\mathrm{A}^{2} \mathrm{~B} \cos \theta\)
Explanation:
B Let \(\vec{A} \cdot(\vec{B} \times \vec{A})=\vec{A} \cdot \vec{C}\) Here \(\vec{C}=\vec{B} \times \vec{A}\) which is perpendicular to both vector \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\). \(\therefore \quad \overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{C}}=0\) \(\because \quad \overrightarrow{\mathrm{C}}\) is perpendicular to \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) \(\therefore\) Angle between \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{C}}\) is \(90^{\circ}\) \(\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{C}}=\mathrm{AC} \cos \theta\) \(\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{C}}=0\) \(\therefore \quad \overrightarrow{\mathrm{A}} \cdot(\overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{A}})=0\)
JIPMER-2007
Motion in Plane
143635
\(\overrightarrow{\mathbf{A}}\) is a vector quantity such that \(|\overrightarrow{\mathbf{A}}|=\) non-zero constant. Which of the following expression is true for \(\overrightarrow{\mathbf{A}}\) ?