Vector P = 6 hat{i} + 4 sqrt{2} hat{j} + 4 sq | vedclass

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

Question

(C) The angle $\gamma$ that a vector $P = P_x \hat{i} + P_y \hat{j} + P_z \hat{k}$ makes with the $z$-axis is given by $\cos \gamma = \frac{P_z}{|P|}$

Vedclass · Accepted Answer

$\cos^{-1}\left(\frac{2 \sqrt{2}}{5}\right)$

Vedclass · Answer

(C) The angle $\gamma$ that a vector $P = P_x \hat{i} + P_y \hat{j} + P_z \hat{k}$ makes with the $z$-axis is given by $\cos \gamma = \frac{P_z}{|P|}$.Given $P = 6 \hat{i} + 4 \sqrt{2} \hat{j} + 4 \sqrt{2} \hat{k}$.The magnitude of the vector $P$ is $|P| = \sqrt{(6)^2 + (4 \sqrt{2})^2 + (4 \sqrt{2})^2}$.$|P| = \sqrt{36 + (16 	imes 2) + (16 	imes 2)} = \sqrt{36 + 32 + 32} = \sqrt{100} = 10$.The $z$-component is $P_z = 4 \sqrt{2}$.Therefore,$\cos \gamma = \frac{4 \sqrt{2}}{10} = \frac{2 \sqrt{2}}{5}$.Thus,$\gamma = \cos^{-1}\left(\frac{2 \sqrt{2}}{5}ight)$.

Vector $P = 6 \hat{i} + 4 \sqrt{2} \hat{j} + 4 \sqrt{2} \hat{k}$ makes an angle with the $z$-axis equal to:

Similar Questions

The magnitude of vectors $\overrightarrow{ OA }, \overrightarrow{ OB }$ and $\overrightarrow{ OC }$ in the given figure are equal. The direction of $\overrightarrow{ OA }+\overrightarrow{ OB }-\overrightarrow{ OC }$ with the $x$-axis will be

What is the maximum number of rectangular components into which a vector can be split in space?

The $y$-component of vector $\vec{A}$ is $+3.0 \ m$. If $\vec{A}$ makes an angle of $30^{\circ}$ counterclockwise from the positive $y$-axis, the magnitude of $\vec{A}$ is (assume $\vec{A}$ is in the $x-y$ plane).

What is the component of $3\hat{i} + 4\hat{j}$ along $\hat{i} + \hat{j}$?

Two forces of magnitude $P$ and $Q$ acting at a point have a resultant $R$. The resolved part of $R$ in the direction of $P$ is of magnitude $Q$. The angle between the forces is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module