143544
Three forces \(F_{1}, F_{2}\) and \(F_{3}\) together keep a body in equilibrium. If \(F_{1}=3 \mathrm{~N}\) along the positive \(x\) axis, \(F_{2}=4 \mathrm{~N}\) along the positive \(y\)-axis, then the third force \(F_{3}\) is
1 \(5 \mathrm{~N}\) making an angle \(\theta=\tan ^{-1}\left(\frac{3}{4}\right)\) with negative \(y\)-axis
2 \(5 \mathrm{~N}\) making an angle \(\theta=\tan ^{-1}\left(\frac{4}{3}\right)\) with negative \(y\)-axis
3 \(7 \mathrm{~N}\) making an angle \(\theta=\tan ^{-1}\left(\frac{3}{4}\right)\) with negative \(y\)-axis
4 \(7 \mathrm{~N}\) making an angle \(\theta=\tan ^{-1}\left(\frac{4}{3}\right)\) with negative \(y\)-axis
Explanation:
C \(F_{1}, F_{2}, F_{3}\) keep a body in equilibrium then resultant of force, \(\Sigma \mathrm{F}=0\) \(\mathrm{F}_{1}+\mathrm{F}_{2}+\mathrm{F}_{3}=0\) \(3+4+\mathrm{F}_{3}=0\) \(\mathrm{~F}_{3}=-7 \mathrm{~N}\) Magnitude of \(\mathrm{F}_{3}=7 \mathrm{~N}\) \(\theta\) is angle made with- \(\mathrm{Y}\) axis \(\tan \theta=\frac{3}{4}\) \(\theta=\tan ^{-1} 3 / 4\) \(\mathrm{F}_{3}\) make angle with negative y-axis.
J and K CET- 2010
Motion in Plane
143545
Magnitudes of four pairs of displacement vectors are given. Which pair of displacement vectors, under vector addition, fails to give a resultant vector of magnitude \(3 \mathrm{~cm}\) ?
1 \(2 \mathrm{~cm}, 7 \mathrm{~cm}\)
2 \(1 \mathrm{~cm}, 4 \mathrm{~cm}\)
3 \(2 \mathrm{~cm}, 3 \mathrm{~cm}\)
4 \(2 \mathrm{~cm}, 4 \mathrm{~cm}\)
Explanation:
A The magnitude \(\mathrm{R}\) of the resultant of two vectors \(A\) and \(B\) depends upon the magnitudes of \(A\) and \(B\) and the angle \(\theta\) between them and is given by \(\mathrm{R}^{2}=\mathrm{A}^{2}+\mathrm{B}^{2}+2 \mathrm{AB} \cos \theta\) When \(\theta=0, R\) is maximum and given by \(\mathrm{R}_{\text {max }}^{2}=\mathrm{A}^{2}+\mathrm{B}^{2}+2 \mathrm{AB}=(\mathrm{A}+\mathrm{B})^{2}\) \(\mathrm{R}_{\max }=\mathrm{A}+\mathrm{B}\) When \(\theta=180^{\circ}, \mathrm{R}\) is minimum and given by \(\mathrm{R}_{\min }^{2}=\mathrm{A}^{2}+\mathrm{B}^{2}-2 \mathrm{AB}=(\mathrm{A}-\mathrm{B})^{2}\) \(\mathrm{R}_{\min }=\mathrm{A}-\mathrm{B}\) Thus, the magnitude of resultant will lie between A - B and \(\mathrm{A}+\mathrm{B}\). Now, \(\text { Checking option (a) }\) \(|\mathrm{A}-\mathrm{B}|=|2-7|=5\) \(|\mathrm{~A}+\mathrm{B}|=|2+7|=9\) So, \(\quad 5 \leq \mathrm{R} \leq 9\) and \(\mathrm{R}=4\) Hence, the option (a) is the correct answer.
J and K CET- 2009
Motion in Plane
143546
A body is under the action of two mutually perpendicular forces of \(3 \mathrm{~N}\) and \(4 \mathrm{~N}\). The resultant force acting on the body is
1 \(7 \mathrm{~N}\)
2 \(1 \mathrm{~N}\)
3 \(5 \mathrm{~N}\)
4 zero
Explanation:
C The two forces be \(\mathrm{A}=3 \mathrm{~N} \text { and } \mathrm{B}=4 \mathrm{~N}\) \(\mathrm{A}\) is mutually perpendicular to \(\mathrm{B}\). \(\therefore \theta=90^{\circ}\) \(\mathrm{R}=\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}+2 \mathrm{AB} \cos \theta}\) \(\mathrm{R}=\sqrt{4^{2}+(3)^{2}+2 \mathrm{AB} \cos 90^{\circ}}\) \(\mathrm{R}=\sqrt{16+9+0}\) \(\mathrm{R}=5 \mathrm{~N}\)
J and K CET- 2008
Motion in Plane
143549
Two vectors are given by \(\vec{A}=3 \hat{i}+\hat{j}+3 \hat{k}\) and \(\vec{B}=3 \hat{i}+5 \hat{j}-2 \hat{k}\). Find the third vector \(\vec{C}\) if \(\overrightarrow{\mathbf{A}}+3 \overrightarrow{\mathrm{B}}-\overrightarrow{\mathbf{C}}=\mathbf{0}\)
143544
Three forces \(F_{1}, F_{2}\) and \(F_{3}\) together keep a body in equilibrium. If \(F_{1}=3 \mathrm{~N}\) along the positive \(x\) axis, \(F_{2}=4 \mathrm{~N}\) along the positive \(y\)-axis, then the third force \(F_{3}\) is
1 \(5 \mathrm{~N}\) making an angle \(\theta=\tan ^{-1}\left(\frac{3}{4}\right)\) with negative \(y\)-axis
2 \(5 \mathrm{~N}\) making an angle \(\theta=\tan ^{-1}\left(\frac{4}{3}\right)\) with negative \(y\)-axis
3 \(7 \mathrm{~N}\) making an angle \(\theta=\tan ^{-1}\left(\frac{3}{4}\right)\) with negative \(y\)-axis
4 \(7 \mathrm{~N}\) making an angle \(\theta=\tan ^{-1}\left(\frac{4}{3}\right)\) with negative \(y\)-axis
Explanation:
C \(F_{1}, F_{2}, F_{3}\) keep a body in equilibrium then resultant of force, \(\Sigma \mathrm{F}=0\) \(\mathrm{F}_{1}+\mathrm{F}_{2}+\mathrm{F}_{3}=0\) \(3+4+\mathrm{F}_{3}=0\) \(\mathrm{~F}_{3}=-7 \mathrm{~N}\) Magnitude of \(\mathrm{F}_{3}=7 \mathrm{~N}\) \(\theta\) is angle made with- \(\mathrm{Y}\) axis \(\tan \theta=\frac{3}{4}\) \(\theta=\tan ^{-1} 3 / 4\) \(\mathrm{F}_{3}\) make angle with negative y-axis.
J and K CET- 2010
Motion in Plane
143545
Magnitudes of four pairs of displacement vectors are given. Which pair of displacement vectors, under vector addition, fails to give a resultant vector of magnitude \(3 \mathrm{~cm}\) ?
1 \(2 \mathrm{~cm}, 7 \mathrm{~cm}\)
2 \(1 \mathrm{~cm}, 4 \mathrm{~cm}\)
3 \(2 \mathrm{~cm}, 3 \mathrm{~cm}\)
4 \(2 \mathrm{~cm}, 4 \mathrm{~cm}\)
Explanation:
A The magnitude \(\mathrm{R}\) of the resultant of two vectors \(A\) and \(B\) depends upon the magnitudes of \(A\) and \(B\) and the angle \(\theta\) between them and is given by \(\mathrm{R}^{2}=\mathrm{A}^{2}+\mathrm{B}^{2}+2 \mathrm{AB} \cos \theta\) When \(\theta=0, R\) is maximum and given by \(\mathrm{R}_{\text {max }}^{2}=\mathrm{A}^{2}+\mathrm{B}^{2}+2 \mathrm{AB}=(\mathrm{A}+\mathrm{B})^{2}\) \(\mathrm{R}_{\max }=\mathrm{A}+\mathrm{B}\) When \(\theta=180^{\circ}, \mathrm{R}\) is minimum and given by \(\mathrm{R}_{\min }^{2}=\mathrm{A}^{2}+\mathrm{B}^{2}-2 \mathrm{AB}=(\mathrm{A}-\mathrm{B})^{2}\) \(\mathrm{R}_{\min }=\mathrm{A}-\mathrm{B}\) Thus, the magnitude of resultant will lie between A - B and \(\mathrm{A}+\mathrm{B}\). Now, \(\text { Checking option (a) }\) \(|\mathrm{A}-\mathrm{B}|=|2-7|=5\) \(|\mathrm{~A}+\mathrm{B}|=|2+7|=9\) So, \(\quad 5 \leq \mathrm{R} \leq 9\) and \(\mathrm{R}=4\) Hence, the option (a) is the correct answer.
J and K CET- 2009
Motion in Plane
143546
A body is under the action of two mutually perpendicular forces of \(3 \mathrm{~N}\) and \(4 \mathrm{~N}\). The resultant force acting on the body is
1 \(7 \mathrm{~N}\)
2 \(1 \mathrm{~N}\)
3 \(5 \mathrm{~N}\)
4 zero
Explanation:
C The two forces be \(\mathrm{A}=3 \mathrm{~N} \text { and } \mathrm{B}=4 \mathrm{~N}\) \(\mathrm{A}\) is mutually perpendicular to \(\mathrm{B}\). \(\therefore \theta=90^{\circ}\) \(\mathrm{R}=\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}+2 \mathrm{AB} \cos \theta}\) \(\mathrm{R}=\sqrt{4^{2}+(3)^{2}+2 \mathrm{AB} \cos 90^{\circ}}\) \(\mathrm{R}=\sqrt{16+9+0}\) \(\mathrm{R}=5 \mathrm{~N}\)
J and K CET- 2008
Motion in Plane
143549
Two vectors are given by \(\vec{A}=3 \hat{i}+\hat{j}+3 \hat{k}\) and \(\vec{B}=3 \hat{i}+5 \hat{j}-2 \hat{k}\). Find the third vector \(\vec{C}\) if \(\overrightarrow{\mathbf{A}}+3 \overrightarrow{\mathrm{B}}-\overrightarrow{\mathbf{C}}=\mathbf{0}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Motion in Plane
143544
Three forces \(F_{1}, F_{2}\) and \(F_{3}\) together keep a body in equilibrium. If \(F_{1}=3 \mathrm{~N}\) along the positive \(x\) axis, \(F_{2}=4 \mathrm{~N}\) along the positive \(y\)-axis, then the third force \(F_{3}\) is
1 \(5 \mathrm{~N}\) making an angle \(\theta=\tan ^{-1}\left(\frac{3}{4}\right)\) with negative \(y\)-axis
2 \(5 \mathrm{~N}\) making an angle \(\theta=\tan ^{-1}\left(\frac{4}{3}\right)\) with negative \(y\)-axis
3 \(7 \mathrm{~N}\) making an angle \(\theta=\tan ^{-1}\left(\frac{3}{4}\right)\) with negative \(y\)-axis
4 \(7 \mathrm{~N}\) making an angle \(\theta=\tan ^{-1}\left(\frac{4}{3}\right)\) with negative \(y\)-axis
Explanation:
C \(F_{1}, F_{2}, F_{3}\) keep a body in equilibrium then resultant of force, \(\Sigma \mathrm{F}=0\) \(\mathrm{F}_{1}+\mathrm{F}_{2}+\mathrm{F}_{3}=0\) \(3+4+\mathrm{F}_{3}=0\) \(\mathrm{~F}_{3}=-7 \mathrm{~N}\) Magnitude of \(\mathrm{F}_{3}=7 \mathrm{~N}\) \(\theta\) is angle made with- \(\mathrm{Y}\) axis \(\tan \theta=\frac{3}{4}\) \(\theta=\tan ^{-1} 3 / 4\) \(\mathrm{F}_{3}\) make angle with negative y-axis.
J and K CET- 2010
Motion in Plane
143545
Magnitudes of four pairs of displacement vectors are given. Which pair of displacement vectors, under vector addition, fails to give a resultant vector of magnitude \(3 \mathrm{~cm}\) ?
1 \(2 \mathrm{~cm}, 7 \mathrm{~cm}\)
2 \(1 \mathrm{~cm}, 4 \mathrm{~cm}\)
3 \(2 \mathrm{~cm}, 3 \mathrm{~cm}\)
4 \(2 \mathrm{~cm}, 4 \mathrm{~cm}\)
Explanation:
A The magnitude \(\mathrm{R}\) of the resultant of two vectors \(A\) and \(B\) depends upon the magnitudes of \(A\) and \(B\) and the angle \(\theta\) between them and is given by \(\mathrm{R}^{2}=\mathrm{A}^{2}+\mathrm{B}^{2}+2 \mathrm{AB} \cos \theta\) When \(\theta=0, R\) is maximum and given by \(\mathrm{R}_{\text {max }}^{2}=\mathrm{A}^{2}+\mathrm{B}^{2}+2 \mathrm{AB}=(\mathrm{A}+\mathrm{B})^{2}\) \(\mathrm{R}_{\max }=\mathrm{A}+\mathrm{B}\) When \(\theta=180^{\circ}, \mathrm{R}\) is minimum and given by \(\mathrm{R}_{\min }^{2}=\mathrm{A}^{2}+\mathrm{B}^{2}-2 \mathrm{AB}=(\mathrm{A}-\mathrm{B})^{2}\) \(\mathrm{R}_{\min }=\mathrm{A}-\mathrm{B}\) Thus, the magnitude of resultant will lie between A - B and \(\mathrm{A}+\mathrm{B}\). Now, \(\text { Checking option (a) }\) \(|\mathrm{A}-\mathrm{B}|=|2-7|=5\) \(|\mathrm{~A}+\mathrm{B}|=|2+7|=9\) So, \(\quad 5 \leq \mathrm{R} \leq 9\) and \(\mathrm{R}=4\) Hence, the option (a) is the correct answer.
J and K CET- 2009
Motion in Plane
143546
A body is under the action of two mutually perpendicular forces of \(3 \mathrm{~N}\) and \(4 \mathrm{~N}\). The resultant force acting on the body is
1 \(7 \mathrm{~N}\)
2 \(1 \mathrm{~N}\)
3 \(5 \mathrm{~N}\)
4 zero
Explanation:
C The two forces be \(\mathrm{A}=3 \mathrm{~N} \text { and } \mathrm{B}=4 \mathrm{~N}\) \(\mathrm{A}\) is mutually perpendicular to \(\mathrm{B}\). \(\therefore \theta=90^{\circ}\) \(\mathrm{R}=\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}+2 \mathrm{AB} \cos \theta}\) \(\mathrm{R}=\sqrt{4^{2}+(3)^{2}+2 \mathrm{AB} \cos 90^{\circ}}\) \(\mathrm{R}=\sqrt{16+9+0}\) \(\mathrm{R}=5 \mathrm{~N}\)
J and K CET- 2008
Motion in Plane
143549
Two vectors are given by \(\vec{A}=3 \hat{i}+\hat{j}+3 \hat{k}\) and \(\vec{B}=3 \hat{i}+5 \hat{j}-2 \hat{k}\). Find the third vector \(\vec{C}\) if \(\overrightarrow{\mathbf{A}}+3 \overrightarrow{\mathrm{B}}-\overrightarrow{\mathbf{C}}=\mathbf{0}\)
143544
Three forces \(F_{1}, F_{2}\) and \(F_{3}\) together keep a body in equilibrium. If \(F_{1}=3 \mathrm{~N}\) along the positive \(x\) axis, \(F_{2}=4 \mathrm{~N}\) along the positive \(y\)-axis, then the third force \(F_{3}\) is
1 \(5 \mathrm{~N}\) making an angle \(\theta=\tan ^{-1}\left(\frac{3}{4}\right)\) with negative \(y\)-axis
2 \(5 \mathrm{~N}\) making an angle \(\theta=\tan ^{-1}\left(\frac{4}{3}\right)\) with negative \(y\)-axis
3 \(7 \mathrm{~N}\) making an angle \(\theta=\tan ^{-1}\left(\frac{3}{4}\right)\) with negative \(y\)-axis
4 \(7 \mathrm{~N}\) making an angle \(\theta=\tan ^{-1}\left(\frac{4}{3}\right)\) with negative \(y\)-axis
Explanation:
C \(F_{1}, F_{2}, F_{3}\) keep a body in equilibrium then resultant of force, \(\Sigma \mathrm{F}=0\) \(\mathrm{F}_{1}+\mathrm{F}_{2}+\mathrm{F}_{3}=0\) \(3+4+\mathrm{F}_{3}=0\) \(\mathrm{~F}_{3}=-7 \mathrm{~N}\) Magnitude of \(\mathrm{F}_{3}=7 \mathrm{~N}\) \(\theta\) is angle made with- \(\mathrm{Y}\) axis \(\tan \theta=\frac{3}{4}\) \(\theta=\tan ^{-1} 3 / 4\) \(\mathrm{F}_{3}\) make angle with negative y-axis.
J and K CET- 2010
Motion in Plane
143545
Magnitudes of four pairs of displacement vectors are given. Which pair of displacement vectors, under vector addition, fails to give a resultant vector of magnitude \(3 \mathrm{~cm}\) ?
1 \(2 \mathrm{~cm}, 7 \mathrm{~cm}\)
2 \(1 \mathrm{~cm}, 4 \mathrm{~cm}\)
3 \(2 \mathrm{~cm}, 3 \mathrm{~cm}\)
4 \(2 \mathrm{~cm}, 4 \mathrm{~cm}\)
Explanation:
A The magnitude \(\mathrm{R}\) of the resultant of two vectors \(A\) and \(B\) depends upon the magnitudes of \(A\) and \(B\) and the angle \(\theta\) between them and is given by \(\mathrm{R}^{2}=\mathrm{A}^{2}+\mathrm{B}^{2}+2 \mathrm{AB} \cos \theta\) When \(\theta=0, R\) is maximum and given by \(\mathrm{R}_{\text {max }}^{2}=\mathrm{A}^{2}+\mathrm{B}^{2}+2 \mathrm{AB}=(\mathrm{A}+\mathrm{B})^{2}\) \(\mathrm{R}_{\max }=\mathrm{A}+\mathrm{B}\) When \(\theta=180^{\circ}, \mathrm{R}\) is minimum and given by \(\mathrm{R}_{\min }^{2}=\mathrm{A}^{2}+\mathrm{B}^{2}-2 \mathrm{AB}=(\mathrm{A}-\mathrm{B})^{2}\) \(\mathrm{R}_{\min }=\mathrm{A}-\mathrm{B}\) Thus, the magnitude of resultant will lie between A - B and \(\mathrm{A}+\mathrm{B}\). Now, \(\text { Checking option (a) }\) \(|\mathrm{A}-\mathrm{B}|=|2-7|=5\) \(|\mathrm{~A}+\mathrm{B}|=|2+7|=9\) So, \(\quad 5 \leq \mathrm{R} \leq 9\) and \(\mathrm{R}=4\) Hence, the option (a) is the correct answer.
J and K CET- 2009
Motion in Plane
143546
A body is under the action of two mutually perpendicular forces of \(3 \mathrm{~N}\) and \(4 \mathrm{~N}\). The resultant force acting on the body is
1 \(7 \mathrm{~N}\)
2 \(1 \mathrm{~N}\)
3 \(5 \mathrm{~N}\)
4 zero
Explanation:
C The two forces be \(\mathrm{A}=3 \mathrm{~N} \text { and } \mathrm{B}=4 \mathrm{~N}\) \(\mathrm{A}\) is mutually perpendicular to \(\mathrm{B}\). \(\therefore \theta=90^{\circ}\) \(\mathrm{R}=\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}+2 \mathrm{AB} \cos \theta}\) \(\mathrm{R}=\sqrt{4^{2}+(3)^{2}+2 \mathrm{AB} \cos 90^{\circ}}\) \(\mathrm{R}=\sqrt{16+9+0}\) \(\mathrm{R}=5 \mathrm{~N}\)
J and K CET- 2008
Motion in Plane
143549
Two vectors are given by \(\vec{A}=3 \hat{i}+\hat{j}+3 \hat{k}\) and \(\vec{B}=3 \hat{i}+5 \hat{j}-2 \hat{k}\). Find the third vector \(\vec{C}\) if \(\overrightarrow{\mathbf{A}}+3 \overrightarrow{\mathrm{B}}-\overrightarrow{\mathbf{C}}=\mathbf{0}\)
143544
Three forces \(F_{1}, F_{2}\) and \(F_{3}\) together keep a body in equilibrium. If \(F_{1}=3 \mathrm{~N}\) along the positive \(x\) axis, \(F_{2}=4 \mathrm{~N}\) along the positive \(y\)-axis, then the third force \(F_{3}\) is
1 \(5 \mathrm{~N}\) making an angle \(\theta=\tan ^{-1}\left(\frac{3}{4}\right)\) with negative \(y\)-axis
2 \(5 \mathrm{~N}\) making an angle \(\theta=\tan ^{-1}\left(\frac{4}{3}\right)\) with negative \(y\)-axis
3 \(7 \mathrm{~N}\) making an angle \(\theta=\tan ^{-1}\left(\frac{3}{4}\right)\) with negative \(y\)-axis
4 \(7 \mathrm{~N}\) making an angle \(\theta=\tan ^{-1}\left(\frac{4}{3}\right)\) with negative \(y\)-axis
Explanation:
C \(F_{1}, F_{2}, F_{3}\) keep a body in equilibrium then resultant of force, \(\Sigma \mathrm{F}=0\) \(\mathrm{F}_{1}+\mathrm{F}_{2}+\mathrm{F}_{3}=0\) \(3+4+\mathrm{F}_{3}=0\) \(\mathrm{~F}_{3}=-7 \mathrm{~N}\) Magnitude of \(\mathrm{F}_{3}=7 \mathrm{~N}\) \(\theta\) is angle made with- \(\mathrm{Y}\) axis \(\tan \theta=\frac{3}{4}\) \(\theta=\tan ^{-1} 3 / 4\) \(\mathrm{F}_{3}\) make angle with negative y-axis.
J and K CET- 2010
Motion in Plane
143545
Magnitudes of four pairs of displacement vectors are given. Which pair of displacement vectors, under vector addition, fails to give a resultant vector of magnitude \(3 \mathrm{~cm}\) ?
1 \(2 \mathrm{~cm}, 7 \mathrm{~cm}\)
2 \(1 \mathrm{~cm}, 4 \mathrm{~cm}\)
3 \(2 \mathrm{~cm}, 3 \mathrm{~cm}\)
4 \(2 \mathrm{~cm}, 4 \mathrm{~cm}\)
Explanation:
A The magnitude \(\mathrm{R}\) of the resultant of two vectors \(A\) and \(B\) depends upon the magnitudes of \(A\) and \(B\) and the angle \(\theta\) between them and is given by \(\mathrm{R}^{2}=\mathrm{A}^{2}+\mathrm{B}^{2}+2 \mathrm{AB} \cos \theta\) When \(\theta=0, R\) is maximum and given by \(\mathrm{R}_{\text {max }}^{2}=\mathrm{A}^{2}+\mathrm{B}^{2}+2 \mathrm{AB}=(\mathrm{A}+\mathrm{B})^{2}\) \(\mathrm{R}_{\max }=\mathrm{A}+\mathrm{B}\) When \(\theta=180^{\circ}, \mathrm{R}\) is minimum and given by \(\mathrm{R}_{\min }^{2}=\mathrm{A}^{2}+\mathrm{B}^{2}-2 \mathrm{AB}=(\mathrm{A}-\mathrm{B})^{2}\) \(\mathrm{R}_{\min }=\mathrm{A}-\mathrm{B}\) Thus, the magnitude of resultant will lie between A - B and \(\mathrm{A}+\mathrm{B}\). Now, \(\text { Checking option (a) }\) \(|\mathrm{A}-\mathrm{B}|=|2-7|=5\) \(|\mathrm{~A}+\mathrm{B}|=|2+7|=9\) So, \(\quad 5 \leq \mathrm{R} \leq 9\) and \(\mathrm{R}=4\) Hence, the option (a) is the correct answer.
J and K CET- 2009
Motion in Plane
143546
A body is under the action of two mutually perpendicular forces of \(3 \mathrm{~N}\) and \(4 \mathrm{~N}\). The resultant force acting on the body is
1 \(7 \mathrm{~N}\)
2 \(1 \mathrm{~N}\)
3 \(5 \mathrm{~N}\)
4 zero
Explanation:
C The two forces be \(\mathrm{A}=3 \mathrm{~N} \text { and } \mathrm{B}=4 \mathrm{~N}\) \(\mathrm{A}\) is mutually perpendicular to \(\mathrm{B}\). \(\therefore \theta=90^{\circ}\) \(\mathrm{R}=\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}+2 \mathrm{AB} \cos \theta}\) \(\mathrm{R}=\sqrt{4^{2}+(3)^{2}+2 \mathrm{AB} \cos 90^{\circ}}\) \(\mathrm{R}=\sqrt{16+9+0}\) \(\mathrm{R}=5 \mathrm{~N}\)
J and K CET- 2008
Motion in Plane
143549
Two vectors are given by \(\vec{A}=3 \hat{i}+\hat{j}+3 \hat{k}\) and \(\vec{B}=3 \hat{i}+5 \hat{j}-2 \hat{k}\). Find the third vector \(\vec{C}\) if \(\overrightarrow{\mathbf{A}}+3 \overrightarrow{\mathrm{B}}-\overrightarrow{\mathbf{C}}=\mathbf{0}\)
