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

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(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:

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