Let the angle between two nonzero vectors ove | vedclass

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Question

(C) Given that the angle between $\overrightarrow{A}$ and $\overrightarrow{B}$ is $	heta = 120^{\circ}$.The magnitude of the resultant $\overrightarr

Vedclass · Accepted Answer

$\overrightarrow{C}$ must be less than $|\overrightarrow{A} - \overrightarrow{B}|$

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(C) Given that the angle between $\overrightarrow{A}$ and $\overrightarrow{B}$ is $	heta = 120^{\circ}$.The magnitude of the resultant $\overrightarrow{C} = \overrightarrow{A} + \overrightarrow{B}$ is given by:$|\overrightarrow{C}| = \sqrt{A^{2} + B^{2} + 2AB \cos 120^{\circ}} = \sqrt{A^{2} + B^{2} - AB}$.The magnitude of the difference vector $|\overrightarrow{A} - \overrightarrow{B}|$ is given by:$|\overrightarrow{A} - \overrightarrow{B}| = \sqrt{A^{2} + B^{2} - 2AB \cos 120^{\circ}} = \sqrt{A^{2} + B^{2} - 2AB (-1/2)} = \sqrt{A^{2} + B^{2} + AB}$.Comparing the two expressions,since $A$ and $B$ are nonzero magnitudes,$\sqrt{A^{2} + B^{2} - AB} Therefore,$|\overrightarrow{C}| < |\overrightarrow{A} - \overrightarrow{B}|$.

Column-$I$	Column-$II$
$(1)$ Resultant of two vectors is maximum	$(a)$ $180^o$
$(2)$ Resultant of two vectors is minimum	$(b)$ $90^o$
	$(c)$ $0^o$

Let the angle between two nonzero vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ be $120^{\circ}$ and their resultant be $\overrightarrow{C}$. Which of the following is true?

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If $\vec{P}+\vec{Q}=\vec{P}-\vec{Q}$,then

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 $(b)$ $90^o$

$(c)$ $0^o$

The vectors $\vec{A}$ and $\vec{B}$ are such that $|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}|$. The angle between the two vectors is: (in $^{\circ}$)

If $\overrightarrow A = 2\hat i + \hat j$,$\overrightarrow B = 3\hat j - \hat k$,and $\overrightarrow C = 6\hat i - 2\hat k$,find the value of $\overrightarrow A - 2\overrightarrow B + 3\overrightarrow C$.

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