143609
The angle subtended by the vector \(A=4 \hat{i}+\) \(3 \hat{\mathbf{j}}+12 \hat{\mathrm{k}}\) with \(\mathrm{x}\)-axis is
1 \(\sin ^{-1}\left(\frac{3}{13}\right)\)
2 \(\sin ^{-1}\left(\frac{4}{13}\right)\)
3 \(\cos ^{-1}\left(\frac{4}{13}\right)\)
4 \(\cos ^{-1}\left(\frac{3}{13}\right)\)
Explanation:
C Given that, \(\overrightarrow{\mathrm{A}}=4 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+12 \hat{\mathrm{k}}\) Angle with \(\mathrm{x}\)-axis \(\cos \theta =\frac{\mathrm{A}_{\mathrm{x}}}{|\overrightarrow{\mathrm{A}}|}\) \(\cos \theta =\frac{4}{\sqrt{(4)^{2}+(3)^{2}+(12)^{2}}}\) \(\cos \theta =\frac{4}{13}\) \(\theta =\cos ^{-1}\left(\frac{4}{13}\right)\)
WB JEE 2008
Motion in Plane
143611
If \(a_{1}\) and \(a_{2}\) are two non-collinear unit vectors and if \(\left|a_{1}+a_{2}\right|=\sqrt{3}\), Then the value of \(\left(a_{1}-a_{2}\right) \cdot\left(2 a_{1}+a_{2}\right)\) is
1 2
2 \(\sqrt{3}\)
3 \(\frac{1}{2}\)
4 \(\vec{a}_{1}\), and \(\vec{a}_{2} 1\)
Explanation:
C Given that, \(\left|\mathrm{a}_{1}+\mathrm{a}_{2}\right|=\sqrt{3}\) \(\overrightarrow{\mathrm{a}}_{1}\) and \(\overrightarrow{\mathrm{a}}_{2}\) are non-collinear vector \(\therefore\left|\overrightarrow{\mathrm{a}}_{1}\right|=\left|\overrightarrow{\mathrm{a}}_{2}\right|=1\) Let the angle between \(a_{1}\) and \(a_{2}\) is \(\theta\) \(\mathrm{a}_{1}{ }^{2}+\mathrm{a}_{2}{ }^{2}+2 \mathrm{a}_{1} \mathrm{a}_{2} \cos \theta=(\sqrt{3})^{2}\) \(1+1+2 \cos \theta=3\) \(2 \cos \theta=3-2\) \(2 \cos \theta=1\) \(\cos \theta=\frac{1}{2}\) Now, \(=\left(a_{1}-a_{2}\right) \cdot\left(2 a_{1}+a_{2}\right)\) \(=2 a_{1}^{2}-2\left(a_{1} \cdot a_{2}\right)+\left(a_{1} \cdot a_{2}\right)-a_{2}^{2}\) \(=2 a_{1}^{2}-\left(a_{1} \cdot a_{2}\right)-a_{2}^{2}\) Putting the value of \(a_{1}=1\) and \(a_{2}=1\) and \(\cos \theta=1 / 2\) \(=2 a_{1}^{2}-a_{2}^{2}-a_{1} a_{2} \cos \theta\) \(=2-1-\frac{1}{2}\) \(\left(a_{1}-a_{2}\right) \cdot\left(2 a_{1}+a_{2}\right)=1-\frac{1}{2}\) \(\left(a_{1}-a_{2}\right) \cdot\left(2 a_{1}+a_{2}\right)=\frac{1}{2}\)
C We know that, \(\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}=|\overrightarrow{\mathrm{A}}||\overrightarrow{\mathrm{B}}| \cos \theta\) \(\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}=|\overrightarrow{\mathrm{A}} \| \overrightarrow{\mathrm{B}}| \cos 90^{\circ}\) \(\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}=0 \quad\left(\because \cos 90^{\circ}=0\right)\) \(\therefore \quad \overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}=0\) Hence, two vector are perpendicular if their dot product is equal to zero.
UP CPMT-2007
Motion in Plane
143615
If force \(\mathbf{F}=\mathbf{5} \hat{\mathbf{i}}+\mathbf{3} \hat{\mathbf{j}}+4 \hat{\mathbf{k}}\) makes a displacement of \(s=6 \hat{i}-5 \hat{k}\) work done by the force is
1 10 units
2 \(122 \sqrt{5}\) units
3 \(5 \sqrt{122}\) units
4 20 units
Explanation:
A Given that, Force \((F)=5 \hat{i}+3 \hat{j}+4 \hat{k}\) Displacement \((\mathrm{s})=6 \hat{\mathrm{i}}-5 \hat{\mathrm{k}}\) The work done is given by the following relation. \(\mathrm{W}=\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{d}}\) \(\mathrm{W}=(5 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \cdot(6 \hat{\mathrm{i}}-5 \hat{\mathrm{k}})\) \(\mathrm{W}=(5 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \cdot(6 \hat{\mathrm{i}}+0 \hat{\mathrm{j}}-5 \hat{\mathrm{k}})\) \(\mathrm{W}=30+0-20\) \(\mathrm{~W}=10 \text { units }\)
143609
The angle subtended by the vector \(A=4 \hat{i}+\) \(3 \hat{\mathbf{j}}+12 \hat{\mathrm{k}}\) with \(\mathrm{x}\)-axis is
1 \(\sin ^{-1}\left(\frac{3}{13}\right)\)
2 \(\sin ^{-1}\left(\frac{4}{13}\right)\)
3 \(\cos ^{-1}\left(\frac{4}{13}\right)\)
4 \(\cos ^{-1}\left(\frac{3}{13}\right)\)
Explanation:
C Given that, \(\overrightarrow{\mathrm{A}}=4 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+12 \hat{\mathrm{k}}\) Angle with \(\mathrm{x}\)-axis \(\cos \theta =\frac{\mathrm{A}_{\mathrm{x}}}{|\overrightarrow{\mathrm{A}}|}\) \(\cos \theta =\frac{4}{\sqrt{(4)^{2}+(3)^{2}+(12)^{2}}}\) \(\cos \theta =\frac{4}{13}\) \(\theta =\cos ^{-1}\left(\frac{4}{13}\right)\)
WB JEE 2008
Motion in Plane
143611
If \(a_{1}\) and \(a_{2}\) are two non-collinear unit vectors and if \(\left|a_{1}+a_{2}\right|=\sqrt{3}\), Then the value of \(\left(a_{1}-a_{2}\right) \cdot\left(2 a_{1}+a_{2}\right)\) is
1 2
2 \(\sqrt{3}\)
3 \(\frac{1}{2}\)
4 \(\vec{a}_{1}\), and \(\vec{a}_{2} 1\)
Explanation:
C Given that, \(\left|\mathrm{a}_{1}+\mathrm{a}_{2}\right|=\sqrt{3}\) \(\overrightarrow{\mathrm{a}}_{1}\) and \(\overrightarrow{\mathrm{a}}_{2}\) are non-collinear vector \(\therefore\left|\overrightarrow{\mathrm{a}}_{1}\right|=\left|\overrightarrow{\mathrm{a}}_{2}\right|=1\) Let the angle between \(a_{1}\) and \(a_{2}\) is \(\theta\) \(\mathrm{a}_{1}{ }^{2}+\mathrm{a}_{2}{ }^{2}+2 \mathrm{a}_{1} \mathrm{a}_{2} \cos \theta=(\sqrt{3})^{2}\) \(1+1+2 \cos \theta=3\) \(2 \cos \theta=3-2\) \(2 \cos \theta=1\) \(\cos \theta=\frac{1}{2}\) Now, \(=\left(a_{1}-a_{2}\right) \cdot\left(2 a_{1}+a_{2}\right)\) \(=2 a_{1}^{2}-2\left(a_{1} \cdot a_{2}\right)+\left(a_{1} \cdot a_{2}\right)-a_{2}^{2}\) \(=2 a_{1}^{2}-\left(a_{1} \cdot a_{2}\right)-a_{2}^{2}\) Putting the value of \(a_{1}=1\) and \(a_{2}=1\) and \(\cos \theta=1 / 2\) \(=2 a_{1}^{2}-a_{2}^{2}-a_{1} a_{2} \cos \theta\) \(=2-1-\frac{1}{2}\) \(\left(a_{1}-a_{2}\right) \cdot\left(2 a_{1}+a_{2}\right)=1-\frac{1}{2}\) \(\left(a_{1}-a_{2}\right) \cdot\left(2 a_{1}+a_{2}\right)=\frac{1}{2}\)
C We know that, \(\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}=|\overrightarrow{\mathrm{A}}||\overrightarrow{\mathrm{B}}| \cos \theta\) \(\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}=|\overrightarrow{\mathrm{A}} \| \overrightarrow{\mathrm{B}}| \cos 90^{\circ}\) \(\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}=0 \quad\left(\because \cos 90^{\circ}=0\right)\) \(\therefore \quad \overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}=0\) Hence, two vector are perpendicular if their dot product is equal to zero.
UP CPMT-2007
Motion in Plane
143615
If force \(\mathbf{F}=\mathbf{5} \hat{\mathbf{i}}+\mathbf{3} \hat{\mathbf{j}}+4 \hat{\mathbf{k}}\) makes a displacement of \(s=6 \hat{i}-5 \hat{k}\) work done by the force is
1 10 units
2 \(122 \sqrt{5}\) units
3 \(5 \sqrt{122}\) units
4 20 units
Explanation:
A Given that, Force \((F)=5 \hat{i}+3 \hat{j}+4 \hat{k}\) Displacement \((\mathrm{s})=6 \hat{\mathrm{i}}-5 \hat{\mathrm{k}}\) The work done is given by the following relation. \(\mathrm{W}=\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{d}}\) \(\mathrm{W}=(5 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \cdot(6 \hat{\mathrm{i}}-5 \hat{\mathrm{k}})\) \(\mathrm{W}=(5 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \cdot(6 \hat{\mathrm{i}}+0 \hat{\mathrm{j}}-5 \hat{\mathrm{k}})\) \(\mathrm{W}=30+0-20\) \(\mathrm{~W}=10 \text { units }\)
143609
The angle subtended by the vector \(A=4 \hat{i}+\) \(3 \hat{\mathbf{j}}+12 \hat{\mathrm{k}}\) with \(\mathrm{x}\)-axis is
1 \(\sin ^{-1}\left(\frac{3}{13}\right)\)
2 \(\sin ^{-1}\left(\frac{4}{13}\right)\)
3 \(\cos ^{-1}\left(\frac{4}{13}\right)\)
4 \(\cos ^{-1}\left(\frac{3}{13}\right)\)
Explanation:
C Given that, \(\overrightarrow{\mathrm{A}}=4 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+12 \hat{\mathrm{k}}\) Angle with \(\mathrm{x}\)-axis \(\cos \theta =\frac{\mathrm{A}_{\mathrm{x}}}{|\overrightarrow{\mathrm{A}}|}\) \(\cos \theta =\frac{4}{\sqrt{(4)^{2}+(3)^{2}+(12)^{2}}}\) \(\cos \theta =\frac{4}{13}\) \(\theta =\cos ^{-1}\left(\frac{4}{13}\right)\)
WB JEE 2008
Motion in Plane
143611
If \(a_{1}\) and \(a_{2}\) are two non-collinear unit vectors and if \(\left|a_{1}+a_{2}\right|=\sqrt{3}\), Then the value of \(\left(a_{1}-a_{2}\right) \cdot\left(2 a_{1}+a_{2}\right)\) is
1 2
2 \(\sqrt{3}\)
3 \(\frac{1}{2}\)
4 \(\vec{a}_{1}\), and \(\vec{a}_{2} 1\)
Explanation:
C Given that, \(\left|\mathrm{a}_{1}+\mathrm{a}_{2}\right|=\sqrt{3}\) \(\overrightarrow{\mathrm{a}}_{1}\) and \(\overrightarrow{\mathrm{a}}_{2}\) are non-collinear vector \(\therefore\left|\overrightarrow{\mathrm{a}}_{1}\right|=\left|\overrightarrow{\mathrm{a}}_{2}\right|=1\) Let the angle between \(a_{1}\) and \(a_{2}\) is \(\theta\) \(\mathrm{a}_{1}{ }^{2}+\mathrm{a}_{2}{ }^{2}+2 \mathrm{a}_{1} \mathrm{a}_{2} \cos \theta=(\sqrt{3})^{2}\) \(1+1+2 \cos \theta=3\) \(2 \cos \theta=3-2\) \(2 \cos \theta=1\) \(\cos \theta=\frac{1}{2}\) Now, \(=\left(a_{1}-a_{2}\right) \cdot\left(2 a_{1}+a_{2}\right)\) \(=2 a_{1}^{2}-2\left(a_{1} \cdot a_{2}\right)+\left(a_{1} \cdot a_{2}\right)-a_{2}^{2}\) \(=2 a_{1}^{2}-\left(a_{1} \cdot a_{2}\right)-a_{2}^{2}\) Putting the value of \(a_{1}=1\) and \(a_{2}=1\) and \(\cos \theta=1 / 2\) \(=2 a_{1}^{2}-a_{2}^{2}-a_{1} a_{2} \cos \theta\) \(=2-1-\frac{1}{2}\) \(\left(a_{1}-a_{2}\right) \cdot\left(2 a_{1}+a_{2}\right)=1-\frac{1}{2}\) \(\left(a_{1}-a_{2}\right) \cdot\left(2 a_{1}+a_{2}\right)=\frac{1}{2}\)
C We know that, \(\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}=|\overrightarrow{\mathrm{A}}||\overrightarrow{\mathrm{B}}| \cos \theta\) \(\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}=|\overrightarrow{\mathrm{A}} \| \overrightarrow{\mathrm{B}}| \cos 90^{\circ}\) \(\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}=0 \quad\left(\because \cos 90^{\circ}=0\right)\) \(\therefore \quad \overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}=0\) Hence, two vector are perpendicular if their dot product is equal to zero.
UP CPMT-2007
Motion in Plane
143615
If force \(\mathbf{F}=\mathbf{5} \hat{\mathbf{i}}+\mathbf{3} \hat{\mathbf{j}}+4 \hat{\mathbf{k}}\) makes a displacement of \(s=6 \hat{i}-5 \hat{k}\) work done by the force is
1 10 units
2 \(122 \sqrt{5}\) units
3 \(5 \sqrt{122}\) units
4 20 units
Explanation:
A Given that, Force \((F)=5 \hat{i}+3 \hat{j}+4 \hat{k}\) Displacement \((\mathrm{s})=6 \hat{\mathrm{i}}-5 \hat{\mathrm{k}}\) The work done is given by the following relation. \(\mathrm{W}=\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{d}}\) \(\mathrm{W}=(5 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \cdot(6 \hat{\mathrm{i}}-5 \hat{\mathrm{k}})\) \(\mathrm{W}=(5 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \cdot(6 \hat{\mathrm{i}}+0 \hat{\mathrm{j}}-5 \hat{\mathrm{k}})\) \(\mathrm{W}=30+0-20\) \(\mathrm{~W}=10 \text { units }\)
143609
The angle subtended by the vector \(A=4 \hat{i}+\) \(3 \hat{\mathbf{j}}+12 \hat{\mathrm{k}}\) with \(\mathrm{x}\)-axis is
1 \(\sin ^{-1}\left(\frac{3}{13}\right)\)
2 \(\sin ^{-1}\left(\frac{4}{13}\right)\)
3 \(\cos ^{-1}\left(\frac{4}{13}\right)\)
4 \(\cos ^{-1}\left(\frac{3}{13}\right)\)
Explanation:
C Given that, \(\overrightarrow{\mathrm{A}}=4 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+12 \hat{\mathrm{k}}\) Angle with \(\mathrm{x}\)-axis \(\cos \theta =\frac{\mathrm{A}_{\mathrm{x}}}{|\overrightarrow{\mathrm{A}}|}\) \(\cos \theta =\frac{4}{\sqrt{(4)^{2}+(3)^{2}+(12)^{2}}}\) \(\cos \theta =\frac{4}{13}\) \(\theta =\cos ^{-1}\left(\frac{4}{13}\right)\)
WB JEE 2008
Motion in Plane
143611
If \(a_{1}\) and \(a_{2}\) are two non-collinear unit vectors and if \(\left|a_{1}+a_{2}\right|=\sqrt{3}\), Then the value of \(\left(a_{1}-a_{2}\right) \cdot\left(2 a_{1}+a_{2}\right)\) is
1 2
2 \(\sqrt{3}\)
3 \(\frac{1}{2}\)
4 \(\vec{a}_{1}\), and \(\vec{a}_{2} 1\)
Explanation:
C Given that, \(\left|\mathrm{a}_{1}+\mathrm{a}_{2}\right|=\sqrt{3}\) \(\overrightarrow{\mathrm{a}}_{1}\) and \(\overrightarrow{\mathrm{a}}_{2}\) are non-collinear vector \(\therefore\left|\overrightarrow{\mathrm{a}}_{1}\right|=\left|\overrightarrow{\mathrm{a}}_{2}\right|=1\) Let the angle between \(a_{1}\) and \(a_{2}\) is \(\theta\) \(\mathrm{a}_{1}{ }^{2}+\mathrm{a}_{2}{ }^{2}+2 \mathrm{a}_{1} \mathrm{a}_{2} \cos \theta=(\sqrt{3})^{2}\) \(1+1+2 \cos \theta=3\) \(2 \cos \theta=3-2\) \(2 \cos \theta=1\) \(\cos \theta=\frac{1}{2}\) Now, \(=\left(a_{1}-a_{2}\right) \cdot\left(2 a_{1}+a_{2}\right)\) \(=2 a_{1}^{2}-2\left(a_{1} \cdot a_{2}\right)+\left(a_{1} \cdot a_{2}\right)-a_{2}^{2}\) \(=2 a_{1}^{2}-\left(a_{1} \cdot a_{2}\right)-a_{2}^{2}\) Putting the value of \(a_{1}=1\) and \(a_{2}=1\) and \(\cos \theta=1 / 2\) \(=2 a_{1}^{2}-a_{2}^{2}-a_{1} a_{2} \cos \theta\) \(=2-1-\frac{1}{2}\) \(\left(a_{1}-a_{2}\right) \cdot\left(2 a_{1}+a_{2}\right)=1-\frac{1}{2}\) \(\left(a_{1}-a_{2}\right) \cdot\left(2 a_{1}+a_{2}\right)=\frac{1}{2}\)
C We know that, \(\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}=|\overrightarrow{\mathrm{A}}||\overrightarrow{\mathrm{B}}| \cos \theta\) \(\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}=|\overrightarrow{\mathrm{A}} \| \overrightarrow{\mathrm{B}}| \cos 90^{\circ}\) \(\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}=0 \quad\left(\because \cos 90^{\circ}=0\right)\) \(\therefore \quad \overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}=0\) Hence, two vector are perpendicular if their dot product is equal to zero.
UP CPMT-2007
Motion in Plane
143615
If force \(\mathbf{F}=\mathbf{5} \hat{\mathbf{i}}+\mathbf{3} \hat{\mathbf{j}}+4 \hat{\mathbf{k}}\) makes a displacement of \(s=6 \hat{i}-5 \hat{k}\) work done by the force is
1 10 units
2 \(122 \sqrt{5}\) units
3 \(5 \sqrt{122}\) units
4 20 units
Explanation:
A Given that, Force \((F)=5 \hat{i}+3 \hat{j}+4 \hat{k}\) Displacement \((\mathrm{s})=6 \hat{\mathrm{i}}-5 \hat{\mathrm{k}}\) The work done is given by the following relation. \(\mathrm{W}=\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{d}}\) \(\mathrm{W}=(5 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \cdot(6 \hat{\mathrm{i}}-5 \hat{\mathrm{k}})\) \(\mathrm{W}=(5 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \cdot(6 \hat{\mathrm{i}}+0 \hat{\mathrm{j}}-5 \hat{\mathrm{k}})\) \(\mathrm{W}=30+0-20\) \(\mathrm{~W}=10 \text { units }\)
