Two forces vec F_1 = 5hat i + 10hat j - 20hat | vedclass

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

Question

(B) The angle $	heta$ between two vectors is found using the dot product formula: $\cos 	heta = \frac{\vec F_1 \cdot \vec F_2}{|\vec F_1| |\vec F_2|

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(B) The angle $	heta$ between two vectors is found using the dot product formula: $\cos 	heta = \frac{\vec F_1 \cdot \vec F_2}{|\vec F_1| |\vec F_2|}$.First,calculate the dot product: $\vec F_1 \cdot \vec F_2 = (5 	imes 10) + (10 	imes -5) + (-20 	imes -15) = 50 - 50 + 300 = 300$.Next,calculate the magnitudes: $|\vec F_1| = \sqrt{5^2 + 10^2 + (-20)^2} = \sqrt{25 + 100 + 400} = \sqrt{525} \approx 22.91$.$|\vec F_2| = \sqrt{10^2 + (-5)^2 + (-15)^2} = \sqrt{100 + 25 + 225} = \sqrt{350} \approx 18.71$.Now,substitute these into the formula: $\cos 	heta = \frac{300}{\sqrt{525} 	imes \sqrt{350}} = \frac{300}{\sqrt{183750}} \approx \frac{300}{428.66} \approx 0.70$. Since $\cos 45^\circ \approx 0.707$,the angle $	heta$ is approximately $45^\circ$.

Two forces $\vec F_1 = 5\hat i + 10\hat j - 20\hat k$ and $\vec F_2 = 10\hat i - 5\hat j - 15\hat k$ act on a single point. The angle between $\vec F_1$ and $\vec F_2$ is nearly . . . . . . $^\circ$.

Similar Questions

$(\vec{a}-\vec{b}) \times(\vec{a}+\vec{b})$ is equal to :

Find the unit vector perpendicular to the two vectors $\vec{A} = \hat{i} - \hat{j} + \hat{k}$ and $\vec{B} = \hat{i} + \hat{j} + \hat{k}$.

The angle between two vectors given by $6\hat i + 6\hat j - 3\hat k$ and $7\hat i + 4\hat j + 4\hat k$ is

The resultant of two vectors having magnitudes $2$ and $3$ is $1$. What is their cross product?

Two vectors $a \hat{i} + b \hat{j} + \hat{k}$ and $2 \hat{i} - 3 \hat{j} + 4 \hat{k}$ are perpendicular to each other. Given $3a + 2b = 7$,the ratio of $a$ to $b$ is $\frac{x}{2}$. The value of $x$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module