If P=(3,12,4) and Q is a point on the line OP | vedclass

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(D) Given $P=(3,12,4)$ and $O=(0,0,0)$.The distance $OP = \sqrt{3^2 + 12^2 + 4^2} = \sqrt{9 + 144 + 16} = \sqrt{169} = 13$.Since $Q$ lies on $OP$ and

Vedclass · Accepted Answer

$\pm \frac{57}{13}$

Vedclass · Answer

(D) Given $P=(3,12,4)$ and $O=(0,0,0)$.The distance $OP = \sqrt{3^2 + 12^2 + 4^2} = \sqrt{9 + 144 + 16} = \sqrt{169} = 13$.Since $Q$ lies on $OP$ and $OQ=3$,the point $Q$ divides the segment $OP$ in the ratio $3 : (13-3) = 3 : 10$ internally or $3 : -10$ externally.The coordinates of $Q$ are given by the section formula: $Q = \left( \frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n}, \frac{m z_2 + n z_1}{m+n} ight)$.For internal division $(3:10)$: $Q = \left( \frac{3 	imes 3}{13}, \frac{3 	imes 12}{13}, \frac{3 	imes 4}{13} ight) = \left( \frac{9}{13}, \frac{36}{13}, \frac{12}{13} ight)$.Sum of coordinates $= \frac{9+36+12}{13} = \frac{57}{13}$.For external division $(3:-10)$: $Q = \left( \frac{3 	imes 3}{-7}, \frac{3 	imes 12}{-7}, \frac{3 	imes 4}{-7} ight) = \left( -\frac{9}{7}, -\frac{36}{7}, -\frac{12}{7} ight)$.However,checking the options,the intended answer is $\frac{57}{13}$.

If $P=(3,12,4)$ and $Q$ is a point on the line $OP$ such that $OQ=3$,then the sum of all the coordinates of $Q$ is

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