Find the angle between two vectors vec A = 2h | vedclass

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

Question

(B) The angle $	heta$ between two vectors $\vec A$ and $\vec B$ is given by the formula $\cos 	heta = \frac{\vec A \cdot \vec B}{|\vec A| |\vec B|}$

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(B) The angle $	heta$ between two vectors $\vec A$ and $\vec B$ is given by the formula $\cos 	heta = \frac{\vec A \cdot \vec B}{|\vec A| |\vec B|}$.First,calculate the dot product: $\vec A \cdot \vec B = (2)(1) + (1)(0) + (-1)(-1) = 2 + 0 + 1 = 3$.Next,calculate the magnitudes: $|\vec A| = \sqrt{2^2 + 1^2 + (-1)^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$ and $|\vec B| = \sqrt{1^2 + 0^2 + (-1)^2} = \sqrt{1 + 0 + 1} = \sqrt{2}$.Now,substitute these values into the formula: $\cos 	heta = \frac{3}{\sqrt{6} \cdot \sqrt{2}} = \frac{3}{\sqrt{12}} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$.Since $\cos 	heta = \frac{\sqrt{3}}{2}$,the angle $	heta = 30^o$.

Find the angle between two vectors $\vec A = 2\hat i + \hat j - \hat k$ and $\vec B = \hat i - \hat k$ in degrees.

Similar Questions

If $\overrightarrow A = 3\hat i + \hat j + 2\hat k$ and $\overrightarrow B = 2\hat i - 2\hat j + 4\hat k$,then find $|\overrightarrow A \times \overrightarrow B |$.

Define the scalar product and obtain the magnitude of a vector from it. Mention the direction of scalar product.

Why is $\vec{v} \times \vec{p} = 0$ for a particle moving in a straight line,and how does this relate to the angular momentum of a rotating particle?

If the vector $2\hat{i} + 3\hat{j} + 8\hat{k}$ is perpendicular to the vector $4\hat{j} - 4\hat{i} + \alpha\hat{k}$,then what is the value of $\alpha$?

If $\overrightarrow{F}=(4 \hat{i}-10 \hat{j}) \text{ N}$ and $\overrightarrow{r}=(-5 \hat{i}-3 \hat{j}) \text{ m}$,then $(\overrightarrow{r} \times \overrightarrow{F})$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module