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: \[ OR = \sqrt{4^2 + (-2)^2 + 6^2} = \sqrt{56} = 2\sqrt{14} \]

Answer 2

Step 1: Understand the problem
Point \(R(4, y, z)\) lies on the line joining points \(P(2, -3, 4)\) and \(Q(8, 0, 10)\). We need to find the distance of \(R\) from the origin.

Step 2: Parametric form of line PQ
The line joining \(P\) and \(Q\) can be written as:
\[ (x, y, z) = (2, -3, 4) + t(8 - 2, 0 + 3, 10 - 4) = (2, -3, 4) + t(6, 3, 6) \]
So,
\[ x = 2 + 6t, \quad y = -3 + 3t, \quad z = 4 + 6t \]

Step 3: Find parameter \(t\) for \(x = 4\)
Given \(x = 4\), substitute:
\[ 4 = 2 + 6t \implies 6t = 2 \implies t = \frac{1}{3} \]

Step 4: Find corresponding \(y\) and \(z\)
\[ y = -3 + 3\left(\frac{1}{3}\right) = -3 + 1 = -2 \] \[ z = 4 + 6\left(\frac{1}{3}\right) = 4 + 2 = 6 \]

Step 5: Coordinates of \(R\)
\[ R = (4, -2, 6) \]

Step 6: Calculate distance of \(R\) from origin
\[ \text{Distance} = \sqrt{4^2 + (-2)^2 + 6^2} = \sqrt{16 + 4 + 36} = \sqrt{56} = 2\sqrt{14} \]

Final answer:
\[ 2\sqrt{14} \]