The coordinates of P equiv(1, 2, 3) and O equ | vedclass

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(A) Given points are $O \equiv(0, 0, 0)$ and $P \equiv(1, 2, 3)$.First,calculate the length of the vector $\overline{OP}$:$|\overline{OP}| = \sqrt{(1-

Vedclass · Accepted Answer

$\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}$

Vedclass · Answer

(A) Given points are $O \equiv(0, 0, 0)$ and $P \equiv(1, 2, 3)$.First,calculate the length of the vector $\overline{OP}$:$|\overline{OP}| = \sqrt{(1-0)^2 + (2-0)^2 + (3-0)^2} = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$.The direction cosines $(l, m, n)$ of a vector $\overline{OP} = (x, y, z)$ are given by $\frac{x}{|\overline{OP}|}, \frac{y}{|\overline{OP}|}, \frac{z}{|\overline{OP}|}$.Thus,the direction cosines are $\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}$.

The coordinates of $P \equiv(1, 2, 3)$ and $O \equiv(0, 0, 0)$ are given. Find the direction cosines of $\overline{OP}$.

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