A vector vec{A} having magnitude 6 units is a | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $\vec{B} = B\hat{i}$ and $\vec{A} = A_x\hat{i} + A_y\hat{j}$. Given $|\vec{A}| = 6$,so $A_x^2 + A_y^2 = 36$. Since the resultant $\vec{R} = \v

Vedclass · Accepted Answer

$\sqrt{3.6}$

Vedclass · Answer

(C) Let $\vec{B} = B\hat{i}$ and $\vec{A} = A_x\hat{i} + A_y\hat{j}$. Given $|\vec{A}| = 6$,so $A_x^2 + A_y^2 = 36$. Since the resultant $\vec{R} = \vec{A} + \vec{B} = (A_x + B)\hat{i} + A_y\hat{j}$ is along the $y$-axis,the $x$-component must be zero: $A_x + B = 0$,so $A_x = -B$. Substituting $A_x = -B$ into the magnitude equation: $(-B)^2 + A_y^2 = 36$,which gives $A_y^2 = 36 - B^2$. The resultant vector is $\vec{R} = A_y\hat{j}$,so its magnitude is $|\vec{R}| = |A_y| = \sqrt{36 - B^2}$. Given $|\vec{R}| = 3B$,we have $\sqrt{36 - B^2} = 3B$. Squaring both sides: $36 - B^2 = 9B^2$,which simplifies to $10B^2 = 36$. Thus,$B^2 = 3.6$,so $B = \sqrt{3.6}$ units.

$A$ 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:

Similar Questions

Five equal forces of $10 \, N$ each are applied at one point and all are lying in one plane. If the angles between them are equal,the resultant force will be ........... $N$.

If $\overrightarrow{ F }=2 \hat{ i }+\hat{ j }-\hat{ k }$ and $\overrightarrow{ r }=3 \hat{ i }+2 \hat{ j }-2 \hat{ k }$,then the scalar and vector products of $\overrightarrow{ F }$ and $\overrightarrow{ r }$ have the magnitudes respectively as

The resultant of two vectors $A$ and $B$ is perpendicular to the vector $A$ and its magnitude is equal to half the magnitude of vector $B$. The angle between $A$ and $B$ is ....... $^o$

If $\vec{A}=\hat{i}+\hat{j}+3 \hat{k}$,$\vec{B}=-\hat{i}+\hat{j}+4 \hat{k}$ and $\vec{C}=2 \hat{i}-2 \hat{j}-8 \hat{k}$,then the angle between the vectors $\vec{P}=\vec{A}+\vec{B}+\vec{C}$ and $\vec{Q}=(\vec{A} \times \vec{B})$ is (in degree) (in $^{\circ}$)

Which of the following is not true? Given $\overrightarrow A = 3\hat i + 4\hat j$ and $\overrightarrow B = 6\hat i + 8\hat j$,where $A$ and $B$ are the magnitudes of $\overrightarrow A$ and $\overrightarrow B$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module