Find the angle between vec{A} = 3hat{i} - hat | vedclass

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

Question

(B) The unit vector along the $Z$-axis is $\hat{k} = 0\hat{i} + 0\hat{j} + 1\hat{k}$.Let $	heta$ be the angle between $\vec{A}$ and the $Z$-axis.The

Vedclass · Accepted Answer

$	an^{-1}\left(\frac{\sqrt{10}}{4}ight)$

Vedclass · Answer

(B) The unit vector along the $Z$-axis is $\hat{k} = 0\hat{i} + 0\hat{j} + 1\hat{k}$.Let $	heta$ be the angle between $\vec{A}$ and the $Z$-axis.The dot product formula is $\vec{A} \cdot \hat{k} = |\vec{A}| |\hat{k}| \cos 	heta$.Calculating the dot product: $\vec{A} \cdot \hat{k} = (3)(0) + (-1)(0) + (4)(1) = 4$.The magnitude of $\vec{A}$ is $|\vec{A}| = \sqrt{3^2 + (-1)^2 + 4^2} = \sqrt{9 + 1 + 16} = \sqrt{26}$.The magnitude of $\hat{k}$ is $1$.Thus,$\cos 	heta = \frac{4}{\sqrt{26} 	imes 1} = \frac{4}{\sqrt{26}}$.Using the identity $\sin^2 	heta + \cos^2 	heta = 1$,we find $\sin 	heta = \sqrt{1 - \left(\frac{4}{\sqrt{26}}ight)^2} = \sqrt{1 - \frac{16}{26}} = \sqrt{\frac{10}{26}} = \frac{\sqrt{10}}{\sqrt{26}}$.Therefore,$	an 	heta = \frac{\sin 	heta}{\cos 	heta} = \frac{\sqrt{10}/\sqrt{26}}{4/\sqrt{26}} = \frac{\sqrt{10}}{4}$.Hence,$	heta = 	an^{-1}\left(\frac{\sqrt{10}}{4}ight)$.

Find the angle between $\vec{A} = 3\hat{i} - \hat{j} + 4\hat{k}$ and the $Z$-axis.

Similar Questions

The magnitude of the vector $|2\hat{i} - \hat{j} - 5\hat{k}|$ is .....

Two vectors of same magnitude act at a point. Twice the product of the magnitudes of two vectors is equal to the square of the magnitude of their resultant. The angle between the two vectors is (in $^{\circ}$)

$A$ vector makes $45^{\circ}$ and $60^{\circ}$ angles with the $x$ and $y$ axes respectively. The unit vector along the given vector is:

When $\vec A \cdot \vec B = - |A||B|$,then

The minimum number of switches in the simplified form of the following switching circuit is

Vedclass Test Series

Exam Paper Generator

Online Exam Module