Question

If a point R$(4,y,z)$ lies on the line joining $P(2,-3,4)$ and $Q(8,0,10)$, then distan of R from origin is

• A. $2\sqrt{14}$
• B. $6$
• C. $\sqrt{53}$
• D. $2\sqrt{21}$

Hint:

3D Point on Line and Distance.
Use parametric equations from direction vectors. Then, distance from origin is simply $\sqrtx^2 + y^2 + z^2$.

Answer 1

The Correct Option is A

Line PQ: Parametric form: \[ (x,y,z) = (2 + 6t, -3 + 3t, 4 + 6t) \] Given $x = 4 \Rightarrow 4 = 2 + 6t \Rightarrow t = \frac{1}{3}$ \[ y = -3 + 1 = -2, \quad z = 4 + 2 = 6 \Rightarrow R = (4,-2,6) \] Distance from origin: \[ OR = \sqrt{4^2 + (-2)^2 + 6^2} = \sqrt{56} = 2\sqrt{14} \]