What is the angle between the resultant of ov | vedclass

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

Question

(C) Let $\overrightarrow{A}$ and $\overrightarrow{B}$ be two vectors in a plane. The sum $\overrightarrow{A} + \overrightarrow{B}$ is a vector that li

Vedclass · Accepted Answer

$\frac{\pi}{2} 	ext{ rad}$

Vedclass · Answer

(C) Let $\overrightarrow{A}$ and $\overrightarrow{B}$ be two vectors in a plane. The sum $\overrightarrow{A} + \overrightarrow{B}$ is a vector that lies in the same plane as $\overrightarrow{A}$ and $\overrightarrow{B}$.By the definition of the cross product,$\overrightarrow{A} 	imes \overrightarrow{B}$ is a vector that is perpendicular to the plane containing both $\overrightarrow{A}$ and $\overrightarrow{B}$.Since the vector $(\overrightarrow{A} + \overrightarrow{B})$ lies in the plane and the vector $(\overrightarrow{A} 	imes \overrightarrow{B})$ is perpendicular to the plane,the angle between them is $90^{\circ}$ or $\frac{\pi}{2} 	ext{ rad}$.Mathematically,the dot product of these two vectors is:$(\overrightarrow{A} + \overrightarrow{B}) \cdot (\overrightarrow{A} 	imes \overrightarrow{B}) = \overrightarrow{A} \cdot (\overrightarrow{A} 	imes \overrightarrow{B}) + \overrightarrow{B} \cdot (\overrightarrow{A} 	imes \overrightarrow{B})$Using the property of the scalar triple product,$\overrightarrow{A} \cdot (\overrightarrow{A} 	imes \overrightarrow{B}) = 0$ and $\overrightarrow{B} \cdot (\overrightarrow{A} 	imes \overrightarrow{B}) = 0$.Therefore,$(\overrightarrow{A} + \overrightarrow{B}) \cdot (\overrightarrow{A} 	imes \overrightarrow{B}) = 0$,which confirms that the vectors are orthogonal,and the angle between them is $90^{\circ}$.

What is the angle between the resultant of $\overrightarrow{A} + \overrightarrow{B}$ and $\overrightarrow{A} \times \overrightarrow{B}$?

Similar Questions

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: $(\tan \frac{\pi}{4} = 1, \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}})$

If $|\hat{a} \cdot \hat{b}| = \frac{1}{2}$,then $|\hat{a} - \hat{b}|$ may be $:-$

Why is the product of two vectors not commutative?

Two vectors $\vec{A}$ and $\vec{B}$ are at right angles to each other,when

Consider three vectors $A = \hat{i} + \hat{j} - 2\hat{k}$,$B = \hat{i} - \hat{j} + \hat{k}$,and $C = 2\hat{i} - 3\hat{j} + 4\hat{k}$. $A$ vector $X$ of the form $X = \alpha A + \beta B$ (where $\alpha$ and $\beta$ are scalars) is perpendicular to $C$. The ratio of $\alpha$ to $\beta$ is: (in $: 1$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module