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Question

(B) Given that the particles are at an equal distance from the origin,their magnitudes are equal: $|\overrightarrow{A}| = |\overrightarrow{B}|$.$|\ove

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) Given that the particles are at an equal distance from the origin,their magnitudes are equal: $|\overrightarrow{A}| = |\overrightarrow{B}|$.$|\overrightarrow{A}|^2 = |\overrightarrow{B}|^2 \implies 2^2 + (3n)^2 + 2^2 = 2^2 + (-2)^2 + (4p)^2$.$4 + 9n^2 + 4 = 4 + 4 + 16p^2 \implies 8 + 9n^2 = 8 + 16p^2 \implies 9n^2 = 16p^2$.Taking the square root,$3n = \pm 4p$,so $p = \pm \frac{3n}{4}$.Since the vectors are at a right angle,their dot product is zero: $\overrightarrow{A} \cdot \overrightarrow{B} = 0$.$(2)(2) + (3n)(-2) + (2)(4p) = 0 \implies 4 - 6n + 8p = 0$.Case $1$: Substitute $p = \frac{3n}{4}$ into the dot product equation: $4 - 6n + 8(\frac{3n}{4}) = 0 \implies 4 - 6n + 6n = 0 \implies 4 = 0$ (Impossible).Case $2$: Substitute $p = -\frac{3n}{4}$ into the dot product equation: $4 - 6n + 8(-\frac{3n}{4}) = 0 \implies 4 - 6n - 6n = 0 \implies 12n = 4 \implies n = \frac{1}{3}$.Therefore,$n^{-1} = \frac{1}{n} = 3$.

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

Similar Questions

$A$ vector $\vec{A}$ having magnitude $6$ units is added to vector $\vec{B}$,which is along the $x$-axis. The resultant of $\vec{A}$ and $\vec{B}$ is along the $y$-axis. If the magnitude of the resultant of $\vec{A}$ and $\vec{B}$ is three times that of $\vec{B}$,then the magnitude of $\vec{B}$ is:

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

Column $I$ Column $II$

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

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

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

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

Given $A = 3 \hat{i} + 4 \hat{j}$ and $B = 6 \hat{i} + 8 \hat{j}$,which of the following statements is correct?

The three vectors $\vec{A}=3 \hat{i}-2 \hat{j}+\hat{k}$,$\vec{B}=\hat{i}-3 \hat{j}+5 \hat{k}$ and $\vec{C}=2 \hat{i}-\hat{j}+4 \hat{k}$ will form

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