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Question

(C) Given $\overrightarrow{A} + \overrightarrow{B} = \overrightarrow{C}$,which implies $\overrightarrow{B} = \overrightarrow{C} - \overrightarrow{A}$.

Vedclass · Accepted Answer

$\frac{3\pi}{4} 	ext{ radian}$

Vedclass · Answer

(C) Given $\overrightarrow{A} + \overrightarrow{B} = \overrightarrow{C}$,which implies $\overrightarrow{B} = \overrightarrow{C} - \overrightarrow{A}$.Since $\overrightarrow{C} \perp \overrightarrow{A}$,the vectors $\overrightarrow{C}$ and $-\overrightarrow{A}$ form a right-angled triangle with $\overrightarrow{B}$ as the hypotenuse.Thus,$B^2 = C^2 + A^2$.Given $|\overrightarrow{A}| = |\overrightarrow{C}|$,we substitute $C = A$ into the equation:$B^2 = A^2 + A^2 = 2A^2$,so $B = \sqrt{2}A$.Using the law of cosines for the vector addition $\overrightarrow{C} = \overrightarrow{A} + \overrightarrow{B}$:$C^2 = A^2 + B^2 + 2AB \cos 	heta$,where $	heta$ is the angle between $\overrightarrow{A}$ and $\overrightarrow{B}$.Substituting $C^2 = A^2$ and $B^2 = 2A^2$:$A^2 = A^2 + 2A^2 + 2A(\sqrt{2}A) \cos 	heta$.$A^2 = 3A^2 + 2\sqrt{2}A^2 \cos 	heta$.$-2A^2 = 2\sqrt{2}A^2 \cos 	heta$.$\cos 	heta = -\frac{1}{\sqrt{2}}$.Therefore,$	heta = \frac{3\pi}{4} 	ext{ radians}$.

Column $I$	Column $II$
$(A)$ $\vec{A} \cdot \vec{B} = \|\vec{A} \times \vec{B}\|$	$(p)$ $\theta = 45^{\circ}$ or $135^{\circ}$
$(B)$ $\vec{A} \cdot \vec{B} = B^2$	$(q)$ $\theta = 0^{\circ}$
$(C)$ $\|\vec{A} + \vec{B}\| = \|\vec{A} - \vec{B}\|$	$(r)$ $\vec{A} = \vec{B}$
$(D)$ $\|\vec{A} \times \vec{B}\| = AB$	$(s)$ $\theta = 90^{\circ}$

Column $I$	Column $II$
$(A) \|A+B\|$	$(p) \frac{\sqrt{3}}{2} x^2$
$(B) \|A-B\|$	$(q) x$
$(C) A \cdot B$	$(r) \sqrt{3} x$
$(D) \|A \times B\|$	$(s) \frac{x^2}{2}$

Given that $\overrightarrow{A} + \overrightarrow{B} = \overrightarrow{C}$ and that $\overrightarrow{C}$ is $\perp$ to $\overrightarrow{A}$. Further,if $|\overrightarrow{A}| = |\overrightarrow{C}|$,then what is the angle between $\overrightarrow{A}$ and $\overrightarrow{B}$?

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