If the distance between a point P and the point (1, 1, 1) on the line is 13, then the coordinates of P are

Engineering

Mathematics

Equation of Straight Line in 3D

Distance from a Plane

Question

If the distance between a point P and the point (1, 1, 1) on the line $\frac{x - 1}{3} = \frac{y - 1}{4} = \frac{z - 1}{12}$ is 13, then the coordinates of P are

(3, 4, 12)

$(\frac{3}{13}, \frac{4}{13}, \frac{12}{13})$

(4, 5, 13)

(40, 53, 157)

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According to equation :

We have the line equation $= \frac{x - 1}{3} = \frac{y - 1}{4} = \frac{z - 1}{12}$ and point is 13

Now,

Line of equation

$\Rightarrow \frac{x - 1}{3} = \frac{y - 1}{4} = \frac{z - 1}{12} = 13$

Now find x,y & z coordinates

$\Rightarrow \frac{x - 1}{3} = 13$

∴ x = 39 + 1 = 40

$\Rightarrow \frac{y - 1}{4} = 13$

∴ y = 52 + 1 = 53

$\Rightarrow \frac{z - 1}{12} = 13$

∴ y = 156 + 1 = 157

Now, we get new coordinates of P : (40,53,157)

so that the correct option is D.