The angle between the two vectors overrightar | vedclass

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

Question

(C) The angle $	heta$ between two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ is given by the dot product formula: $\overrightarrow{A} \cdo

Vedclass · Accepted Answer

$90$

Vedclass · Answer

(C) The angle $	heta$ between two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ is given by the dot product formula: $\overrightarrow{A} \cdot \overrightarrow{B} = |\overrightarrow{A}| |\overrightarrow{B}| \cos 	heta$.First,calculate the dot product: $\overrightarrow{A} \cdot \overrightarrow{B} = (5\hat{i} + 5\hat{j}) \cdot (5\hat{i} - 5\hat{j}) = (5 	imes 5) + (5 	imes -5) = 25 - 25 = 0$.Since the dot product is $0$,we have $0 = |\overrightarrow{A}| |\overrightarrow{B}| \cos 	heta$.This implies $\cos 	heta = 0$,which means $	heta = 90^\circ$.

The angle between the two vectors $\overrightarrow{A} = 5\hat{i} + 5\hat{j}$ and $\overrightarrow{B} = 5\hat{i} - 5\hat{j}$ will be ....... $^\circ$.

Similar Questions

The edges of a parallelepiped are represented by the vectors $\hat{i} + 2\hat{j}$,$4\hat{j}$,and $\hat{j} + 3\hat{k}$. Find its volume.

Find the scalar and vector products of two vectors $\vec{a} = (3 \hat{i} - 4 \hat{j} + 5 \hat{k})$ and $\vec{b} = (-2 \hat{i} + \hat{j} - 3 \hat{k})$.

Three vectors $\vec{A}, \vec{B}$ and $\vec{C}$ are such that $\vec{A} \cdot \vec{B} = 0$ and $\vec{A} \cdot \vec{C} = 0$. Then $\vec{A}$ is parallel to:

The area of the parallelogram determined by vectors $A = 2\hat{i} + \hat{j} - 3\hat{k}$ and $B = 12\hat{j} - 2\hat{k}$ is approximately:

Which of the following is the unit vector perpendicular to $\vec{A}$ and $\vec{B}$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module