Let the values of λ$\lambda$ for which the shortest distance between the lines (x - 1)/(2) = (y - 2)/(3) = (z - 3)/(4) (x - λ)/(3) = (y - 4)/(4) = (z - 5)/(5)$$\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}\quad\text{and}\quad\frac{x - \lambda}{3} = \frac{y - 4}{4} = \frac{z - 5}{5}$$ is 1√(6)$\frac{1}{\sqrt{6}}$ be λ₁$\lambda_{1}$ and λ₂$\lambda_{2}$. Then the radius of the circle passing through the points (0, 0)$(0, 0)$, (λ₁, λ₂)$(\lambda_{1}, \lambda_{2})$ and (λ₂, λ₁)$(\lambda_{2}, \lambda_{1})$ is

Related Formula

Distance along a line from intersection point M$M$ using parametric coordinates.

Core Logic

Find intersection point M$M$ of L₁$L_1$ and L₂$L_2$:

Point of intersection M = (3, 5, 9)$M = (3, 5, 9)$.

Any parametric point P$P$ on L₁$L_1$ is (2K-1, 3K-1, 6K-3)$(2K-1, 3K-1, 6K-3)$.

Step 1: Solve for Points P and Q

Distance PM = 7$PM = 7$:

• For K=1$K=1$: P = (1, 2, 3)$P = (1, 2, 3)$, sum = 6$= 6$
• For K=3$K=3$: Q = (5, 8, 15)$Q = (5, 8, 15)$, sum = 28$= 28$

Step 2: Total Sum

Total sum = 6 + 28 = 34$= 6 + 28 = 34$.

Pattern Recognition

Intersection point gives central reference; unit direction vector gives required offset points along the line.

Chapter Mix

Class 12 Maths: Three Dimensional Geometry

Related Formula

Core Logic

The line is standardly written as (x)/(1) = (y - 1)/(2) = (z - 2)/(3) = λ$\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3} = \lambda$. Point B$B$ (4, 9, α)$(4, 9, \alpha)$ lies on this line, so we can solve for α$\alpha$ by substituting it in:

Thus, B$B$ is (4, 9, 14)$(4, 9, 14)$.

Step 1: Find the Perpendicular Foot (D)

Let D$D$ be the foot of the perpendicular from A(1, 6, 3)$A(1, 6, 3)$ to the line BC$BC$. The coordinates of a general point on the line are D = (λ, 2λ + 1, 3λ + 2)$D = (\lambda, 2\lambda + 1, 3\lambda + 2)$. The direction vector of AD$AD$ is AD = (λ - 1) i + (2λ - 5) j + (3λ - 1) k$\vec{AD} = (\lambda - 1)\hat{i} + (2\lambda - 5)\hat{j} + (3\lambda - 1)\hat{k}$. Since AD$AD$ is perpendicular to the line whose direction ratios are 1, 2, 3$\langle 1, 2, 3 \rangle$:

Step 2: Calculate Height (AD)

Substitute λ = 1$\lambda = 1$ into D$D$: D = (1, 3, 5)$D = (1, 3, 5)$. The perpendicular distance AD$AD$ is:

Step 3: Calculate Area of Triangle

We are given that the base BC$BC$ is 10$10$ units.

Pattern Recognition

When a triangle area is requested given a line equation containing the base, dropping a generic lambda-perpendicular establishes height (AD$AD$), decoupling the problem from finding the exact coordinate of C$C$.

Chapter Mix

Class 12 Maths: Three Dimensional Geometry

Related Formula

Core Logic

From the first linear equation, isolate n$n$: n = 4 + m$n = 4\ell + m$ Substitute this into the second quadratic relation:

Factorizing the quadratic equation:

This gives two cases, generating the direction ratios of the two lines.

Step 1: Determine Direction Ratios

Case 1: m = -8$m = -8\ell$ Then n = 4 - 8 = -4$n = 4\ell - 8\ell = -4\ell$. Direction Ratios (D.R.s) for line L₁$L_1$: ( , -8 , -4 ) ≡ (1, -8, -4)$(\ell, -8\ell, -4\ell) \equiv (1, -8, -4)$.

Case 2: m = -(5)/(2)$m = -\frac{5}{2}\ell$ Then n = 4 - (5)/(2) = (3)/(2)$n = 4\ell - \frac{5}{2}\ell = \frac{3}{2}\ell$. Direction Ratios (D.R.s) for line L₂$L_2$: ( , -(5)/(2) , (3)/(2) ) ≡ (1, -(5)/(2), (3)/(2)) ≡ (2, -5, 3)$(\ell, -\frac{5}{2}\ell, \frac{3}{2}\ell) \equiv \left(1, -\frac{5}{2}, \frac{3}{2}\right) \equiv (2, -5, 3)$.

Step 2: Calculate Angle Between Lines

Use the angle formula with D.R.s d₁ = (1, -8, -4)$d_1 = (1, -8, -4)$ and d₂ = (2, -5, 3)$d_2 = (2, -5, 3)$:

Pattern Recognition

When direction cosines are entangled in one linear and one quadratic equation, isolate the single-degree variable from the linear plane equation and plug it into the cone equation to split it into two generating lines.

Chapter Mix

Class 12 Maths: Three Dimensional Geometry

Related Formula

For a point and its image across a line, the midpoint of the segment connecting them lies exactly on the line. The segment itself is perpendicular to the line.

Core Logic

Thus, Q$Q$ is given by:

Now, the image of Q$Q$ across L₂$L_2$: (x - 2b)/(2) = (y - a)/(1) = (z + 2b)/(-5) = μ$\frac{x - 2b}{2} = \frac{y - a}{1} = \frac{z + 2b}{-5} = \mu$ is P(a, 2, a)$P(a, 2, a)$. The midpoint M₂$M_2$ of PQ$PQ$ lies on L₂$L_2$. But M₂$M_2$ is the exact same set of coordinates in space as M₁$M_1$ since they are both the midpoint of the segment PQ$PQ$!

Step 1: Equating Midpoints

General point on L₂$L_2$ is (2μ + 2b, μ + a, -5μ - 2b)$(2\mu + 2b, \mu + a, -5\mu - 2b)$. Since the midpoint must be the same: (a + (4λ - a))/(2) = 2μ + 2b 2λ = 2μ + 2b λ - μ = b$\frac{a + (4\lambda - a)}{2} = 2\mu + 2b \implies 2\lambda = 2\mu + 2b \implies \lambda - \mu = b$ (2 + (2λ - 2a - 2))/(2) = μ + a λ - a = μ + a λ - μ = 2a$\frac{2 + (2\lambda - 2a - 2)}{2} = \mu + a \implies \lambda - a = \mu + a \implies \lambda - \mu = 2a$ (a + (2λ - a))/(2) = -5μ - 2b λ = -5μ - 2b λ + 5μ = -2b$\frac{a + (2\lambda - a)}{2} = -5\mu - 2b \implies \lambda = -5\mu - 2b \implies \lambda + 5\mu = -2b$

Step 2: Solving System

From the first two equations, b = 2a$b = 2a$. Substitute b = 2a$b = 2a$ into the third: λ + 5μ = -4a$\lambda + 5\mu = -4a$. Also λ - μ = 2a$\lambda - \mu = 2a$. Subtracting gives 6μ = -6a μ = -a$6\mu = -6a \implies \mu = -a$. Thus λ = a$\lambda = a$.

This vector must be perpendicular to L₁$L_1$ (direction vector 2 i + j + k$2\hat{i} + \hat{j} + \hat{k}$): (a)(2) + (-2)(1) + (0)(1) = 0 2a - 2 = 0 a = 1$(a)(2) + (-2)(1) + (0)(1) = 0 \implies 2a - 2 = 0 \implies a = 1$.

Step 3: Final Values

Since a = 1$a = 1$, and b = 2a$b = 2a$, we have b = 2$b = 2$. Therefore, a + b = 1 + 2 = 3$a + b = 1 + 2 = 3$.

Pattern Recognition

If P$P$ is the image of Q$Q$ on L₁$L_1$, and Q$Q$ is the image of P$P$ on L₂$L_2$, then L₁$L_1$ and L₂$L_2$ must intersect segment PQ$PQ$ at the exact same midpoint. Using the midpoint consistency across both lines creates a rapid linear system.

Chapter Mix

Class 12 Maths: Three Dimensional Geometry

Related Formula

Core Logic

Find the point of intersection R$R$ by equating general points on L₁$L_1$ and L₂$L_2$: General point on L₁$L_1$: (2λ + 1, 3λ + 2, 4λ + 3)$(2\lambda + 1, 3\lambda + 2, 4\lambda + 3)$ General point on L₂$L_2$: (5μ + 4, 2μ + 1, μ)$(5\mu + 4, 2\mu + 1, \mu)$

Equating coordinates:

Solving yields λ = -1$\lambda = -1$ and μ = -1$\mu = -1$. Point of intersection R$R$ is (-1, -1, -1)$(-1, -1, -1)$.

Step 1: Finding Point P

Let point P$P$ be (2λ + 1, 3λ + 2, 4λ + 3)$(2\lambda + 1, 3\lambda + 2, 4\lambda + 3)$. PR² = 29$PR^2 = 29$

If λ = -2$\lambda = -2$, P(-3, -4, -5)$P(-3, -4, -5)$ (not in first octant, rejected). If λ = 0$\lambda = 0$, P(1, 2, 3)$P(1, 2, 3)$ (in first octant, accepted).

Step 2: Finding Point Q

Let Q$Q$ be (5μ + 4, 2μ + 1, μ)$(5\mu + 4, 2\mu + 1, \mu)$.

Step 3: Calculating QR^2

Point Q = ( 5(-(1)/(3))+4, 2(-(1)/(3))+1, -(1)/(3) ) = ( (7)/(3), (1)/(3), -(1)/(3) )$Q = \left( 5\left(-\frac{1}{3}\right)+4, 2\left(-\frac{1}{3}\right)+1, -\frac{1}{3} \right) = \left( \frac{7}{3}, \frac{1}{3}, -\frac{1}{3} \right)$. R(-1, -1, -1)$R(-1, -1, -1)$.

Pattern Recognition

When given fixed line parameters and distances, expressing all points in terms of their single parameter (λ$\lambda$ or μ$\mu$) converts 3D distance problems into simple single-variable quadratic equations.

Chapter Mix

Class 12 Maths: Three Dimensional Geometry