Question

If the line $\frac{x+1}{2} = \frac{y+1}{3} = \frac{z+1}{4}$ meets the plane $x + 2y + 3z = 14$ at $P$, then the distance between $P$ and the origin is:

• A. $\sqrt{14}$
• B. $\sqrt{15}$
• C. $\sqrt{13}$
• D. $\sqrt{12}$
• E. $\sqrt{17}$

Hint:

Whenever a problem asks for a point on a line, immediately switch to the parametric form. It converts a 3D geometry problem into a simple 1D linear equation.

Answer 1

The Correct Option is A

Concept: To find the intersection point $P$, express the general coordinates of a point on the line using a parameter $\lambda$, substitute these into the plane equation to find $\lambda$, and then find the distance from the origin $(0, 0, 0)$.

Step 1: Express the general point on the line.
Let $\frac{x+1}{2} = \frac{y+1}{3} = \frac{z+1}{4} = \lambda$. Then $x = 2\lambda - 1$, $y = 3\lambda - 1$, $z = 4\lambda - 1$.

Step 2: Substitute into the plane equation.
Plane: $x + 2y + 3z = 14$
\[ (2\lambda - 1) + 2(3\lambda - 1) + 3(4\lambda - 1) = 14 \] \[ 2\lambda - 1 + 6\lambda - 2 + 12\lambda - 3 = 14 \] \[ 20\lambda - 6 = 14 \quad \Rightarrow \quad 20\lambda = 20 \quad \Rightarrow \quad \lambda = 1 \]

Step 3: Find point $P$ and its distance from the origin.
Substituting $\lambda = 1$: $x = 2(1) - 1 = 1$, $y = 3(1) - 1 = 2$, $z = 4(1) - 1 = 3$. Point $P = (1, 2, 3)$. Distance from origin $O(0, 0, 0)$: \[ OP = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14} \]