Two vectors A and B have equal magnitude x. T | vedclass

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Question

(A) Given: $|A| = |B| = x$ and $	heta = 60^{\circ}$.$(A) |A+B| = \sqrt{x^2 + x^2 + 2x^2 \cos(60^{\circ})} = \sqrt{2x^2 + 2x^2(0.5)} = \sqrt{3x^2} = \

Vedclass · Accepted Answer

$(A \rightarrow r, B \rightarrow q, C \rightarrow s, D \rightarrow p)$

Vedclass · Answer

(A) Given: $|A| = |B| = x$ and $	heta = 60^{\circ}$.$(A) |A+B| = \sqrt{x^2 + x^2 + 2x^2 \cos(60^{\circ})} = \sqrt{2x^2 + 2x^2(0.5)} = \sqrt{3x^2} = \sqrt{3}x$. Thus,$(A ightarrow r)$.$(B) |A-B| = \sqrt{x^2 + x^2 - 2x^2 \cos(60^{\circ})} = \sqrt{2x^2 - 2x^2(0.5)} = \sqrt{x^2} = x$. Thus,$(B ightarrow q)$.$(C) A \cdot B = |A||B| \cos(60^{\circ}) = x \cdot x \cdot 0.5 = \frac{x^2}{2}$. Thus,$(C ightarrow s)$.$(D) |A 	imes B| = |A||B| \sin(60^{\circ}) = x \cdot x \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}x^2$. Thus,$(D ightarrow p)$.Therefore,the correct matching is $(A ightarrow r, B ightarrow q, C ightarrow s, D ightarrow p)$.

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

Column-$I$	Column-$II$
$(1)$ Resultant of two mutually perpendicular vectors	$(a)$ Along the bisector of the angle between them
$(2)$ Direction of $\overrightarrow A \times \overrightarrow B$	$(b)$ Coplanar
	$(c)$ Perpendicular to the plane containing $\overrightarrow A$ and $\overrightarrow B$

Two vectors $A$ and $B$ have equal magnitude $x$. The angle between them is $60^{\circ}$. Match the following two columns:

Column $I$ Column $II$

$(A) |A+B|$ $(p) \frac{\sqrt{3}}{2} x^2$

$(B) |A-B|$ $(q) x$

$(C) A \cdot B$ $(r) \sqrt{3} x$

$(D) |A \times B|$ $(s) \frac{x^2}{2}$

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Two vectors $A$ and $B$ have equal magnitude $x$. The angle between them is $60^{\circ}$. Match the following two columns: Column $I$ Column $II$ $(A) |A+B|$ $(p) \frac{\sqrt{3}}{2} x^2$ $(B) |A-B|$ $(q) x$ $(C) A \cdot B$ $(r) \sqrt{3} x$ $(D) |A \times B|$ $(s) \frac{x^2}{2}$