Question

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

• A. \(\pm \frac{10}{13}\)
• B. \(\pm \frac{28}{13}\)
• C. \(\pm \frac{19}{13}\)
• D. \(\pm \frac{57}{13}\)

Hint:

For a point on a line through origin, use parametric form \(Q = tP\). Solve for t using distance, then compute sum of coordinates.

Answer 1

The Correct Option is D

Step 1: Find the length OP.
\[|OP| = \sqrt{3^2+12^2+4^2} = \sqrt{9+144+16} = \sqrt{169} = 13\]

Step 2: Get the unit vector along OP.
\[\hat{u} = \left(\frac{3}{13}, \frac{12}{13}, \frac{4}{13}\right)\]

Step 3: Scale to length 3.
Since Q lies on line OP with \(OQ=3\), Q equals \(3\hat{u}\) measured from O, in either direction along the line:
\[Q = \pm 3\left(\frac{3}{13}, \frac{12}{13}, \frac{4}{13}\right) = \pm\left(\frac{9}{13}, \frac{36}{13}, \frac{12}{13}\right)\]

Step 4: Sum the coordinates.
\[\frac{9}{13}+\frac{36}{13}+\frac{12}{13} = \frac{57}{13}\]
Including both directions along the line:
\[\boxed{\pm\frac{57}{13}}\]