355551
The angle between the two vectors \(\vec{A}=3 \hat{i}+4 \hat{j}+5 \hat{k}\) and \(\vec{B}=3 \hat{i}+4 \hat{j}+5 \hat{k}\) is
1 \(60^{\circ}\)
2 Zero
3 \(90^{\circ}\)
4 None of these
Explanation:
As two vectors are same so the angle between them is zero.
PHXI06:WORK ENERGY AND POWER
355552
If \(\hat{i}, \hat{j}\) and \(\hat{k}\) represent unit vectors along the \(x, y\) and \(z\) - axes respectively, then the angle \(\theta\) between the vectors \(\hat{i}+\hat{j}+\hat{k}\) and \(\hat{i}+\hat{j}\) is equal to
1 \(\sin ^{-1}\left(\dfrac{1}{\sqrt{3}}\right)\)
2 \(\sin ^{-1}\left(\sqrt{\dfrac{2}{3}}\right)\)
3 \(\cos ^{-1}\left(\dfrac{1}{\sqrt{3}}\right)\)
4 \(90^{\circ}\)
Explanation:
Let the given vectors be \(\begin{gathered}\vec{A}=\hat{i}+\hat{j}+\hat{k} ; \vec{B}=\hat{i}+\hat{j} \\\therefore A=\sqrt{1^{2}+1^{2}+1^{2}}=\sqrt{3}, B=\sqrt{1^{2}+1^{2}}=\sqrt{2} \\\vec{A} \cdot \vec{B}=2 \\\cos \theta=\dfrac{\vec{A} \cdot \vec{B}}{A B}=\dfrac{2}{\sqrt{3} \sqrt{2}}=\left(\sqrt{\dfrac{2}{3}}\right) \\\sin \theta=\sqrt{1-\cos ^{2} \theta}=\sqrt{1-\dfrac{2}{3}}=\sqrt{\dfrac{1}{3}}=\dfrac{1}{\sqrt{3}} \\\theta=\sin ^{-1}\left(\dfrac{1}{\sqrt{3}}\right)\end{gathered}\)
PHXI06:WORK ENERGY AND POWER
355553
Vectors \(\vec{A}\) and \(\vec{B}\) include an angle \(\theta\) between them.If \((\vec{A}+\vec{B})\) and \((\vec{A}-\vec{B})\) respectively subtend angles \(\alpha\) and \(\beta\) with \(\mathrm{A}\), then \((\tan \alpha+\tan \beta)\) is
1 \(\dfrac{(A B \sin \theta)}{\left(A^{2}+B^{2} \cos ^{2} \theta\right)}\)
2 \(\dfrac{(2 A B \sin \theta)}{\left(A^{2}-B^{2} \cos ^{2} \theta\right)}\)
The vector diagram is \(\tan \alpha = \left( {\frac{{B\sin \theta }}{{A + B\cos \theta }}} \right)\,\,\,\,\,\,\,(1)\) Where \(\alpha \) is the angle made by the vector \((\vec A + \vec B)\) with \({\vec A}\). Similarly, \(\tan \beta = \frac{{B\sin \theta }}{{A - B\cos \theta }}\,\,\,\,\,\,\,\,\,\,\,(2)\) Where \(\beta \) is the angle made by the vector \((\vec A - \vec B)\) with \({\vec A}\). Note that the angle between \({\vec A}\) and \(( - \vec B)\) is \((180^\circ - \theta ).\) Adding (1) and (2), we get \(\tan \alpha + \tan \beta = \frac{{B\sin \theta }}{{A + B\cos \theta }} + \frac{{B\sin \theta }}{{A - B\cos \theta }}\) \( = \frac{{2AB\sin \theta }}{{({A^2} - {B^2}{{\cos }^2}\theta )}}\)
PHXI06:WORK ENERGY AND POWER
355554
\(\vec{A}=3 \hat{i}-\hat{j}+7 \hat{k}\) and \(\vec{B}=5 \hat{i}-\hat{j}+9 \hat{k}\). The direction cosine of the vector \(\vec{A}+\vec{B}\) with \(x\)-axis is
1 \(\dfrac{3}{\sqrt{31}}\)
2 \(\dfrac{5}{\sqrt{324}}\)
3 5
4 \(\dfrac{8}{\sqrt{324}}\)
Explanation:
Let \(\vec{C}=\vec{A}+\vec{B}=3 \hat{i}-\hat{j}+7 \hat{k}+5 \hat{i}-\hat{j}+9 \hat{k}\) \(\vec{C}=\vec{A}+\vec{B}=8 \hat{i}-2 \hat{j}+16 \hat{k}\) The direction cosine is \(\cos \theta = \frac{{(8\hat i - 2\hat j + 16\hat k) \cdot \hat i}}{{\sqrt {{8^2} + {{(2)}^2} + {{(16)}^2}} }}\) \( = \frac{8}{{\sqrt {324} }}\)
355551
The angle between the two vectors \(\vec{A}=3 \hat{i}+4 \hat{j}+5 \hat{k}\) and \(\vec{B}=3 \hat{i}+4 \hat{j}+5 \hat{k}\) is
1 \(60^{\circ}\)
2 Zero
3 \(90^{\circ}\)
4 None of these
Explanation:
As two vectors are same so the angle between them is zero.
PHXI06:WORK ENERGY AND POWER
355552
If \(\hat{i}, \hat{j}\) and \(\hat{k}\) represent unit vectors along the \(x, y\) and \(z\) - axes respectively, then the angle \(\theta\) between the vectors \(\hat{i}+\hat{j}+\hat{k}\) and \(\hat{i}+\hat{j}\) is equal to
1 \(\sin ^{-1}\left(\dfrac{1}{\sqrt{3}}\right)\)
2 \(\sin ^{-1}\left(\sqrt{\dfrac{2}{3}}\right)\)
3 \(\cos ^{-1}\left(\dfrac{1}{\sqrt{3}}\right)\)
4 \(90^{\circ}\)
Explanation:
Let the given vectors be \(\begin{gathered}\vec{A}=\hat{i}+\hat{j}+\hat{k} ; \vec{B}=\hat{i}+\hat{j} \\\therefore A=\sqrt{1^{2}+1^{2}+1^{2}}=\sqrt{3}, B=\sqrt{1^{2}+1^{2}}=\sqrt{2} \\\vec{A} \cdot \vec{B}=2 \\\cos \theta=\dfrac{\vec{A} \cdot \vec{B}}{A B}=\dfrac{2}{\sqrt{3} \sqrt{2}}=\left(\sqrt{\dfrac{2}{3}}\right) \\\sin \theta=\sqrt{1-\cos ^{2} \theta}=\sqrt{1-\dfrac{2}{3}}=\sqrt{\dfrac{1}{3}}=\dfrac{1}{\sqrt{3}} \\\theta=\sin ^{-1}\left(\dfrac{1}{\sqrt{3}}\right)\end{gathered}\)
PHXI06:WORK ENERGY AND POWER
355553
Vectors \(\vec{A}\) and \(\vec{B}\) include an angle \(\theta\) between them.If \((\vec{A}+\vec{B})\) and \((\vec{A}-\vec{B})\) respectively subtend angles \(\alpha\) and \(\beta\) with \(\mathrm{A}\), then \((\tan \alpha+\tan \beta)\) is
1 \(\dfrac{(A B \sin \theta)}{\left(A^{2}+B^{2} \cos ^{2} \theta\right)}\)
2 \(\dfrac{(2 A B \sin \theta)}{\left(A^{2}-B^{2} \cos ^{2} \theta\right)}\)
The vector diagram is \(\tan \alpha = \left( {\frac{{B\sin \theta }}{{A + B\cos \theta }}} \right)\,\,\,\,\,\,\,(1)\) Where \(\alpha \) is the angle made by the vector \((\vec A + \vec B)\) with \({\vec A}\). Similarly, \(\tan \beta = \frac{{B\sin \theta }}{{A - B\cos \theta }}\,\,\,\,\,\,\,\,\,\,\,(2)\) Where \(\beta \) is the angle made by the vector \((\vec A - \vec B)\) with \({\vec A}\). Note that the angle between \({\vec A}\) and \(( - \vec B)\) is \((180^\circ - \theta ).\) Adding (1) and (2), we get \(\tan \alpha + \tan \beta = \frac{{B\sin \theta }}{{A + B\cos \theta }} + \frac{{B\sin \theta }}{{A - B\cos \theta }}\) \( = \frac{{2AB\sin \theta }}{{({A^2} - {B^2}{{\cos }^2}\theta )}}\)
PHXI06:WORK ENERGY AND POWER
355554
\(\vec{A}=3 \hat{i}-\hat{j}+7 \hat{k}\) and \(\vec{B}=5 \hat{i}-\hat{j}+9 \hat{k}\). The direction cosine of the vector \(\vec{A}+\vec{B}\) with \(x\)-axis is
1 \(\dfrac{3}{\sqrt{31}}\)
2 \(\dfrac{5}{\sqrt{324}}\)
3 5
4 \(\dfrac{8}{\sqrt{324}}\)
Explanation:
Let \(\vec{C}=\vec{A}+\vec{B}=3 \hat{i}-\hat{j}+7 \hat{k}+5 \hat{i}-\hat{j}+9 \hat{k}\) \(\vec{C}=\vec{A}+\vec{B}=8 \hat{i}-2 \hat{j}+16 \hat{k}\) The direction cosine is \(\cos \theta = \frac{{(8\hat i - 2\hat j + 16\hat k) \cdot \hat i}}{{\sqrt {{8^2} + {{(2)}^2} + {{(16)}^2}} }}\) \( = \frac{8}{{\sqrt {324} }}\)
355551
The angle between the two vectors \(\vec{A}=3 \hat{i}+4 \hat{j}+5 \hat{k}\) and \(\vec{B}=3 \hat{i}+4 \hat{j}+5 \hat{k}\) is
1 \(60^{\circ}\)
2 Zero
3 \(90^{\circ}\)
4 None of these
Explanation:
As two vectors are same so the angle between them is zero.
PHXI06:WORK ENERGY AND POWER
355552
If \(\hat{i}, \hat{j}\) and \(\hat{k}\) represent unit vectors along the \(x, y\) and \(z\) - axes respectively, then the angle \(\theta\) between the vectors \(\hat{i}+\hat{j}+\hat{k}\) and \(\hat{i}+\hat{j}\) is equal to
1 \(\sin ^{-1}\left(\dfrac{1}{\sqrt{3}}\right)\)
2 \(\sin ^{-1}\left(\sqrt{\dfrac{2}{3}}\right)\)
3 \(\cos ^{-1}\left(\dfrac{1}{\sqrt{3}}\right)\)
4 \(90^{\circ}\)
Explanation:
Let the given vectors be \(\begin{gathered}\vec{A}=\hat{i}+\hat{j}+\hat{k} ; \vec{B}=\hat{i}+\hat{j} \\\therefore A=\sqrt{1^{2}+1^{2}+1^{2}}=\sqrt{3}, B=\sqrt{1^{2}+1^{2}}=\sqrt{2} \\\vec{A} \cdot \vec{B}=2 \\\cos \theta=\dfrac{\vec{A} \cdot \vec{B}}{A B}=\dfrac{2}{\sqrt{3} \sqrt{2}}=\left(\sqrt{\dfrac{2}{3}}\right) \\\sin \theta=\sqrt{1-\cos ^{2} \theta}=\sqrt{1-\dfrac{2}{3}}=\sqrt{\dfrac{1}{3}}=\dfrac{1}{\sqrt{3}} \\\theta=\sin ^{-1}\left(\dfrac{1}{\sqrt{3}}\right)\end{gathered}\)
PHXI06:WORK ENERGY AND POWER
355553
Vectors \(\vec{A}\) and \(\vec{B}\) include an angle \(\theta\) between them.If \((\vec{A}+\vec{B})\) and \((\vec{A}-\vec{B})\) respectively subtend angles \(\alpha\) and \(\beta\) with \(\mathrm{A}\), then \((\tan \alpha+\tan \beta)\) is
1 \(\dfrac{(A B \sin \theta)}{\left(A^{2}+B^{2} \cos ^{2} \theta\right)}\)
2 \(\dfrac{(2 A B \sin \theta)}{\left(A^{2}-B^{2} \cos ^{2} \theta\right)}\)
The vector diagram is \(\tan \alpha = \left( {\frac{{B\sin \theta }}{{A + B\cos \theta }}} \right)\,\,\,\,\,\,\,(1)\) Where \(\alpha \) is the angle made by the vector \((\vec A + \vec B)\) with \({\vec A}\). Similarly, \(\tan \beta = \frac{{B\sin \theta }}{{A - B\cos \theta }}\,\,\,\,\,\,\,\,\,\,\,(2)\) Where \(\beta \) is the angle made by the vector \((\vec A - \vec B)\) with \({\vec A}\). Note that the angle between \({\vec A}\) and \(( - \vec B)\) is \((180^\circ - \theta ).\) Adding (1) and (2), we get \(\tan \alpha + \tan \beta = \frac{{B\sin \theta }}{{A + B\cos \theta }} + \frac{{B\sin \theta }}{{A - B\cos \theta }}\) \( = \frac{{2AB\sin \theta }}{{({A^2} - {B^2}{{\cos }^2}\theta )}}\)
PHXI06:WORK ENERGY AND POWER
355554
\(\vec{A}=3 \hat{i}-\hat{j}+7 \hat{k}\) and \(\vec{B}=5 \hat{i}-\hat{j}+9 \hat{k}\). The direction cosine of the vector \(\vec{A}+\vec{B}\) with \(x\)-axis is
1 \(\dfrac{3}{\sqrt{31}}\)
2 \(\dfrac{5}{\sqrt{324}}\)
3 5
4 \(\dfrac{8}{\sqrt{324}}\)
Explanation:
Let \(\vec{C}=\vec{A}+\vec{B}=3 \hat{i}-\hat{j}+7 \hat{k}+5 \hat{i}-\hat{j}+9 \hat{k}\) \(\vec{C}=\vec{A}+\vec{B}=8 \hat{i}-2 \hat{j}+16 \hat{k}\) The direction cosine is \(\cos \theta = \frac{{(8\hat i - 2\hat j + 16\hat k) \cdot \hat i}}{{\sqrt {{8^2} + {{(2)}^2} + {{(16)}^2}} }}\) \( = \frac{8}{{\sqrt {324} }}\)
355551
The angle between the two vectors \(\vec{A}=3 \hat{i}+4 \hat{j}+5 \hat{k}\) and \(\vec{B}=3 \hat{i}+4 \hat{j}+5 \hat{k}\) is
1 \(60^{\circ}\)
2 Zero
3 \(90^{\circ}\)
4 None of these
Explanation:
As two vectors are same so the angle between them is zero.
PHXI06:WORK ENERGY AND POWER
355552
If \(\hat{i}, \hat{j}\) and \(\hat{k}\) represent unit vectors along the \(x, y\) and \(z\) - axes respectively, then the angle \(\theta\) between the vectors \(\hat{i}+\hat{j}+\hat{k}\) and \(\hat{i}+\hat{j}\) is equal to
1 \(\sin ^{-1}\left(\dfrac{1}{\sqrt{3}}\right)\)
2 \(\sin ^{-1}\left(\sqrt{\dfrac{2}{3}}\right)\)
3 \(\cos ^{-1}\left(\dfrac{1}{\sqrt{3}}\right)\)
4 \(90^{\circ}\)
Explanation:
Let the given vectors be \(\begin{gathered}\vec{A}=\hat{i}+\hat{j}+\hat{k} ; \vec{B}=\hat{i}+\hat{j} \\\therefore A=\sqrt{1^{2}+1^{2}+1^{2}}=\sqrt{3}, B=\sqrt{1^{2}+1^{2}}=\sqrt{2} \\\vec{A} \cdot \vec{B}=2 \\\cos \theta=\dfrac{\vec{A} \cdot \vec{B}}{A B}=\dfrac{2}{\sqrt{3} \sqrt{2}}=\left(\sqrt{\dfrac{2}{3}}\right) \\\sin \theta=\sqrt{1-\cos ^{2} \theta}=\sqrt{1-\dfrac{2}{3}}=\sqrt{\dfrac{1}{3}}=\dfrac{1}{\sqrt{3}} \\\theta=\sin ^{-1}\left(\dfrac{1}{\sqrt{3}}\right)\end{gathered}\)
PHXI06:WORK ENERGY AND POWER
355553
Vectors \(\vec{A}\) and \(\vec{B}\) include an angle \(\theta\) between them.If \((\vec{A}+\vec{B})\) and \((\vec{A}-\vec{B})\) respectively subtend angles \(\alpha\) and \(\beta\) with \(\mathrm{A}\), then \((\tan \alpha+\tan \beta)\) is
1 \(\dfrac{(A B \sin \theta)}{\left(A^{2}+B^{2} \cos ^{2} \theta\right)}\)
2 \(\dfrac{(2 A B \sin \theta)}{\left(A^{2}-B^{2} \cos ^{2} \theta\right)}\)
The vector diagram is \(\tan \alpha = \left( {\frac{{B\sin \theta }}{{A + B\cos \theta }}} \right)\,\,\,\,\,\,\,(1)\) Where \(\alpha \) is the angle made by the vector \((\vec A + \vec B)\) with \({\vec A}\). Similarly, \(\tan \beta = \frac{{B\sin \theta }}{{A - B\cos \theta }}\,\,\,\,\,\,\,\,\,\,\,(2)\) Where \(\beta \) is the angle made by the vector \((\vec A - \vec B)\) with \({\vec A}\). Note that the angle between \({\vec A}\) and \(( - \vec B)\) is \((180^\circ - \theta ).\) Adding (1) and (2), we get \(\tan \alpha + \tan \beta = \frac{{B\sin \theta }}{{A + B\cos \theta }} + \frac{{B\sin \theta }}{{A - B\cos \theta }}\) \( = \frac{{2AB\sin \theta }}{{({A^2} - {B^2}{{\cos }^2}\theta )}}\)
PHXI06:WORK ENERGY AND POWER
355554
\(\vec{A}=3 \hat{i}-\hat{j}+7 \hat{k}\) and \(\vec{B}=5 \hat{i}-\hat{j}+9 \hat{k}\). The direction cosine of the vector \(\vec{A}+\vec{B}\) with \(x\)-axis is
1 \(\dfrac{3}{\sqrt{31}}\)
2 \(\dfrac{5}{\sqrt{324}}\)
3 5
4 \(\dfrac{8}{\sqrt{324}}\)
Explanation:
Let \(\vec{C}=\vec{A}+\vec{B}=3 \hat{i}-\hat{j}+7 \hat{k}+5 \hat{i}-\hat{j}+9 \hat{k}\) \(\vec{C}=\vec{A}+\vec{B}=8 \hat{i}-2 \hat{j}+16 \hat{k}\) The direction cosine is \(\cos \theta = \frac{{(8\hat i - 2\hat j + 16\hat k) \cdot \hat i}}{{\sqrt {{8^2} + {{(2)}^2} + {{(16)}^2}} }}\) \( = \frac{8}{{\sqrt {324} }}\)
