Given below in Column $-I$ are the relations between vectors $\vec a$,$\vec b$,and $\vec c$,and in Column $-II$ are the orientations of $\vec a$,$\vec b$,and $\vec c$ in the $XY-$ plane. Match the relation in Column $-I$ to the correct orientations in Column $-II$.
Column $-I$ Column $-II$ $(a) \vec a + \vec b = \vec c$ $(i)$ Vector $\vec a$ is along $+Y$,$\vec c$ is along $+X$,and $\vec b$ connects the origin to the tip of $\vec c$ $(b) \vec a - \vec c = \vec b$ $(ii)$ Vector $\vec a$ is along $+X$,$\vec b$ is along $+Y$,and $\vec c$ connects the origin to the tip of $\vec b$ $(c) \vec b - \vec a = \vec c$ $(iii)$ Vector $\vec c$ is along $+X$,$\vec a$ is along $+Y$,and $\vec b$ connects the tip of $\vec c$ to the tip of $\vec a$ $(d) \vec a + \vec b + \vec c = 0$ $(iv)$ Vector $\vec a$ is along $-X$,$\vec b$ is along $-Y$,and $\vec c$ connects the origin to the tip of $\vec b$
| Column $-I$ | Column $-II$ |
| $(a) \vec a + \vec b = \vec c$ | $(i)$ Vector $\vec a$ is along $+Y$,$\vec c$ is along $+X$,and $\vec b$ connects the origin to the tip of $\vec c$ |
| $(b) \vec a - \vec c = \vec b$ | $(ii)$ Vector $\vec a$ is along $+X$,$\vec b$ is along $+Y$,and $\vec c$ connects the origin to the tip of $\vec b$ |
| $(c) \vec b - \vec a = \vec c$ | $(iii)$ Vector $\vec c$ is along $+X$,$\vec a$ is along $+Y$,and $\vec b$ connects the tip of $\vec c$ to the tip of $\vec a$ |
| $(d) \vec a + \vec b + \vec c = 0$ | $(iv)$ Vector $\vec a$ is along $-X$,$\vec b$ is along $-Y$,and $\vec c$ connects the origin to the tip of $\vec b$ |
(A-II, B-III, C-I, D-IV) Using the triangle law of vector addition,where the resultant vector is the one that closes the triangle in the opposite direction of the other two vectors connected head-to-tail.
$(a)$ For $\vec a + \vec b = \vec c$,the vectors $\vec a$ and $\vec b$ must be connected head-to-tail,and $\vec c$ is the resultant. In diagram $(ii)$,$\vec a$ is along $+X$,$\vec b$ is along $+Y$,and $\vec c$ is the resultant. Thus,$(a) \rightarrow (ii)$.
$(b)$ For $\vec a - \vec c = \vec b$,we can write $\vec a = \vec b + \vec c$. In diagram $(iii)$,$\vec c$ is along $+X$,$\vec a$ is along $+Y$,and $\vec b$ connects the tip of $\vec c$ to the tip of $\vec a$. Thus,$\vec c + \vec b = \vec a$,which implies $\vec a - \vec c = \vec b$. Thus,$(b) \rightarrow (iii)$.
$(c)$ For $\vec b - \vec a = \vec c$,we can write $\vec b = \vec a + \vec c$. In diagram $(i)$,$\vec a$ is along $+Y$,$\vec c$ is along $+X$,and $\vec b$ is the resultant. Thus,$(c) \rightarrow (i)$.
$(d)$ For $\vec a + \vec b + \vec c = 0$,the vectors must form a closed loop. In diagram $(iv)$,$\vec a$ is along $-X$,$\vec b$ is along $-Y$,and $\vec c$ is the vector closing the loop. Thus,$(d) \rightarrow (iv)$.
$(a)$ For $\vec a + \vec b = \vec c$,the vectors $\vec a$ and $\vec b$ must be connected head-to-tail,and $\vec c$ is the resultant. In diagram $(ii)$,$\vec a$ is along $+X$,$\vec b$ is along $+Y$,and $\vec c$ is the resultant. Thus,$(a) \rightarrow (ii)$.
$(b)$ For $\vec a - \vec c = \vec b$,we can write $\vec a = \vec b + \vec c$. In diagram $(iii)$,$\vec c$ is along $+X$,$\vec a$ is along $+Y$,and $\vec b$ connects the tip of $\vec c$ to the tip of $\vec a$. Thus,$\vec c + \vec b = \vec a$,which implies $\vec a - \vec c = \vec b$. Thus,$(b) \rightarrow (iii)$.
$(c)$ For $\vec b - \vec a = \vec c$,we can write $\vec b = \vec a + \vec c$. In diagram $(i)$,$\vec a$ is along $+Y$,$\vec c$ is along $+X$,and $\vec b$ is the resultant. Thus,$(c) \rightarrow (i)$.
$(d)$ For $\vec a + \vec b + \vec c = 0$,the vectors must form a closed loop. In diagram $(iv)$,$\vec a$ is along $-X$,$\vec b$ is along $-Y$,and $\vec c$ is the vector closing the loop. Thus,$(d) \rightarrow (iv)$.
Explore More
Similar Questions
Match Column-$I$ with Column-$II$.
Column-$I$ Column-$II$ $(1)$ Resultant of two vectors is maximum $(a)$ $180^o$ $(2)$ Resultant of two vectors is minimum $(b)$ $90^o$ $(c)$ $0^o$
| Column-$I$ | Column-$II$ |
|---|---|
| $(1)$ Resultant of two vectors is maximum | $(a)$ $180^o$ |
| $(2)$ Resultant of two vectors is minimum | $(b)$ $90^o$ |
| $(c)$ $0^o$ |
Easy
View SolutionLet $\overrightarrow C = \overrightarrow A + \overrightarrow B$. Which of the following statements are correct?
$(A)$ It is possible to have $|\overrightarrow C| < |\overrightarrow A|$ and $|\overrightarrow C| < |\overrightarrow B|$.
$(B)$ $|\overrightarrow C|$ is always greater than $|\overrightarrow A|$.
$(C)$ $|\overrightarrow C|$ may be equal to $|\overrightarrow A| + |\overrightarrow B|$.
$(D)$ $|\overrightarrow C|$ is never equal to $|\overrightarrow A| + |\overrightarrow B|$.
$(A)$ It is possible to have $|\overrightarrow C| < |\overrightarrow A|$ and $|\overrightarrow C| < |\overrightarrow B|$.
$(B)$ $|\overrightarrow C|$ is always greater than $|\overrightarrow A|$.
$(C)$ $|\overrightarrow C|$ may be equal to $|\overrightarrow A| + |\overrightarrow B|$.
$(D)$ $|\overrightarrow C|$ is never equal to $|\overrightarrow A| + |\overrightarrow B|$.
Medium
View SolutionAmong three vectors,two vectors are equal in magnitude,and the magnitude of the third vector is $\sqrt{2}$ times the magnitude of the other two vectors. If $\overrightarrow{A} + \overrightarrow{B} + \overrightarrow{C} = 0$,find the angles between the vectors.
Difficult
View SolutionThe sum of the magnitudes of two forces is $25 \,N$. The resultant of these forces is perpendicular to the smaller force and has a magnitude of $10 \,N$. The two forces are:
DifficultTS EAMCET 2015
View SolutionTwo vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ are at an angle of $60^{\circ}$ with each other. Their resultant makes an angle of $45^{\circ}$ with $\overrightarrow{a}$. If $|\overrightarrow{b}| = 2 \text{ units}$,then $|\overrightarrow{a}|$ is:
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo