CBSE 2025, Set 3

Question

Find the distance of the point $(-1,-5,-10)$ from the point of intersection of the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-4}{5}=\frac{y-1}{2}=z$.

Edvaya · Accepted Answer

The correct answer is Subjective. **Step 1:** Find the general points on both lines.
 Let Line 1 be $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4} = \lambda$.
 A general point on Line 1 is $P(2\lambda+1, 3\lambda+2, 4\lambda+3)$.

Let Line 2 be $\frac{x-4}{5} = \frac{y-1}{2} = \frac{z-0}{1} = \mu$.
 A general point on Line 2 is $Q(5\mu+4, 2\mu+1, \mu)$.

**Step 2:** Find the point of intersection.
 At the point of intersection, the coordinates of $P$ and $Q$ must be identical for some $\lambda$ and $\mu$:
 $2\lambda + 1 = 5\mu + 4 \implies 2\lambda - 5\mu = 3$ --- (Equation 1)
 $3\lambda + 2 = 2\mu + 1 \implies 3\lambda - 2\mu = -1$ --- (Equation 2)
 $4\lambda + 3 = \mu \implies \mu = 4\lambda + 3$ --- (Equation 3)

Substitute Equation 3 into Equation 1:
 $2\lambda - 5(4\lambda + 3) = 3$
 $2\lambda - 20\lambda - 15 = 3$
 $-18\lambda = 18 \implies \lambda = -1$

Using $\lambda = -1$, the point of intersection $I$ is:
 $I = (2(-1)+1, 3(-1)+2, 4(-1)+3) = (-1, -1, -1)$.

**Step 3:** Calculate the distance.
 We need to find the distance between the given point $A(-1, -5, -10)$ and the intersection point $I(-1, -1, -1)$.
 Using the 3D distance formula:
 $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$$
 $$d = \sqrt{(-1 - (-1))^2 + (-1 - (-5))^2 + (-1 - (-10))^2}$$
 $$d = \sqrt{0^2 + 4^2 + 9^2}$$
 $$d = \sqrt{0 + 16 + 81} = \sqrt{97} 	ext{ units}.$$

The required distance is $\sqrt{97}$ units.

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Mathematics > Coordinate Geometry > Three Dimensional Geometry

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