The angle between the two vectors vec A = 3ha | vedclass

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

Question

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

Vedclass · Accepted Answer

$90$

Vedclass · Answer

(A) The angle $	heta$ between two vectors $\vec A$ and $\vec B$ is given by the formula: $\cos 	heta = \frac{\vec A \cdot \vec B}{|A||B|}$.First,calculate the dot product $\vec A \cdot \vec B = (3)(3) + (4)(4) + (5)(-5) = 9 + 16 - 25 = 0$.Next,calculate the magnitudes $|A| = \sqrt{3^2 + 4^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50}$ and $|B| = \sqrt{3^2 + 4^2 + (-5)^2} = \sqrt{9 + 16 + 25} = \sqrt{50}$.Substituting these values into the formula: $\cos 	heta = \frac{0}{\sqrt{50} \cdot \sqrt{50}} = \frac{0}{50} = 0$.Since $\cos 	heta = 0$,the angle $	heta = 90^\circ$.

The angle between the two vectors $\vec A = 3\hat i + 4\hat j + 5\hat k$ and $\vec B = 3\hat i + 4\hat j - 5\hat k$ will be....... $^o$

Similar Questions

The value of $\hat{i} \times (\hat{i} \times \vec{a}) + \hat{j} \times (\hat{j} \times \vec{a}) + \hat{k} \times (\hat{k} \times \vec{a})$ is

If $|\vec A \times \vec B| = \sqrt 3 \vec A \cdot \vec B,$ then the value of $|\vec A + \vec B|$ is

Let $\vec{P} = P \sin \theta \hat{i} - P \cos \theta \hat{j}$ be any vector. Another vector $\vec{Q}$ which is perpendicular to $\vec{P}$ is

Find the angle between force $F = (3 \hat{i} + 4 \hat{j} - 5 \hat{k})$ units and displacement $d = (5 \hat{i} + 4 \hat{j} + 3 \hat{k})$ units. Also,find the projection of $F$ on $d$.

The angle between force $\vec{F}=3 \hat{i}+4 \hat{j}-5 \hat{k}$ and displacement $\vec{d}=5 \hat{i}+4 \hat{j}+3 \hat{k}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module