For any two vectors vec{A} and vec{B},if vec{ | vedclass

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

Question

(A) Given the condition $\vec{A} \cdot \vec{B} = |\vec{A} 	imes \vec{B}|$.Using the definitions of dot and cross products,we have $AB \cos 	heta = A

Vedclass · Accepted Answer

$\sqrt{A^{2} + B^{2} + \sqrt{2} AB}$

Vedclass · Answer

(A) Given the condition $\vec{A} \cdot \vec{B} = |\vec{A} 	imes \vec{B}|$.Using the definitions of dot and cross products,we have $AB \cos 	heta = AB \sin 	heta$.Dividing both sides by $AB$ (assuming $A, B 
eq 0$),we get $\cos 	heta = \sin 	heta$,which implies $	an 	heta = 1$.Thus,$	heta = 45^{\circ}$ or $\frac{\pi}{4}$ radians.The magnitude of the resultant vector $\vec{R} = \vec{A} + \vec{B}$ is given by $|\vec{R}| = \sqrt{A^{2} + B^{2} + 2AB \cos 	heta}$.Substituting $	heta = 45^{\circ}$,we get $|\vec{R}| = \sqrt{A^{2} + B^{2} + 2AB \cos 45^{\circ}}$.Since $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$,the expression becomes $|\vec{R}| = \sqrt{A^{2} + B^{2} + 2AB 	imes \frac{1}{\sqrt{2}}}$.Simplifying,we get $|\vec{R}| = \sqrt{A^{2} + B^{2} + \sqrt{2} AB}$.

Column-$I$	Column-$II$
$(a) \|\vec{A} \times \vec{B}\| = 0$	$(i) \theta = 30^{\circ}$
$(b) \|\vec{A} \times \vec{B}\| = 8$	$(ii) \theta = 45^{\circ}$
$(c) \|\vec{A} \times \vec{B}\| = 4$	$(iii) \theta = 90^{\circ}$
$(d) \|\vec{A} \times \vec{B}\| = 4\sqrt{2}$	$(iv) \theta = 0^{\circ}$

For any two vectors $\vec{A}$ and $\vec{B}$,if $\vec{A} \cdot \vec{B} = |\vec{A} \times \vec{B}|$,the magnitude of $(\vec{A} + \vec{B})$ is: $(\tan \frac{\pi}{4} = 1, \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}})$

Similar Questions

The diagonals of a parallelogram are represented by vectors $\vec{A} = 5\hat{i} - 4\hat{j} + 3\hat{k}$ and $\vec{B} = 3\hat{i} - 2\hat{j} - \hat{k}$. What is the area of the parallelogram (in $\sqrt{3}$)?

The area of the parallelogram having diagonals $\vec{d_1} = 3\hat{i} + \hat{j} - 2\hat{k}$ and $\vec{d_2} = \hat{i} - 3\hat{j} + 4\hat{k}$ is:

If $\vec{A}, \vec{B}$ and $\vec{C}$ are vectors having unit magnitude. If $\vec{A} + \vec{B} + \vec{C} = \vec{0}$,then $\vec{A} \cdot \vec{B} + \vec{B} \cdot \vec{C} + \vec{C} \cdot \vec{A}$ will be:

$A$ vector $\overrightarrow{F}_1$ is along the positive $X$-axis. If its vector product with another vector $\overrightarrow{F}_2$ is zero,then $\overrightarrow{F}_2$ could be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module