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

The line passing through a$\vec{a}$ and \parallel to direction vector v = l i + m j + n k$\vec{v} = l\hat{\mathbf{i}} + m\hat{\mathbf{j}} + n\hat{\mathbf{k}}$ is written in symmetric form as:

Core Logic

Formulate the equation of the line passing through (-1, 2, 1)$(-1, 2, 1)$ with direction vector components (2, 3, 4)$(2, 3, 4)$:

Any generic point on this line can be written as:

The second given line equation is:

Any generic point on line L₂$L_2$ is:

Step 1: Compute the Point of Intersection

At the point of intersection P$P$, equate the coordinates from both lines:

Solving equations (3) and (4) simultaneously gives:

Verify these values using equation (5):

Thus, substituting λ = 1$\lambda = 1$ gives the coordinates of point P$P$:

Step 2: Calculate Euclidean Distance PQ

Find the distance between P(1, 5, 5)$P(1, 5, 5)$ and Q(4, -5, 1)$Q(4, -5, 1)$ using the 3D distance formula:

Pattern Recognition

When finding the intersection point of two 3D lines, always use the third coordinate equation to verify the parameter values obtained from the first two equations to ensure consistency.

Chapter Mix

Class 12 Mathematics: Three Dimensional Geometry

Related Formula

Equation of a line passing through P(x₀, y₀, z₀)$P(x_0, y_0, z_0)$ along direction vector v = a i+b j+c k$\vec{v} = a\hat{i}+b\hat{j}+c\hat{k}$:

Core Logic

Let the given point be P((15)/(7), (32)/(7), 7)$P\left(\frac{15}{7}, \frac{32}{7}, 7\right)$. We need to find the distance measured along the line passing through P$P$ parallel to the vector v = i+4 j+7 k$\vec{v} = \hat{i}+4\hat{j}+7\hat{k}$.

Equation of this line PQ$PQ$ is:

Any general point Q$Q$ on this line can be written as:

Step 1: Find Intersection Point Q with Given Line

Point Q$Q$ must lie on the given target line L: (x+1)/(3) = (y+3)/(5) = (z+5)/(7)$L: \frac{x+1}{3} = \frac{y+3}{5} = \frac{z+5}{7}$.

Substitute coordinates of Q$Q$ into the first and third fractions:

Multiplying by 21:

Step 2: Calculate Coordinates of Q and Distance squared

Substituting λ = -1$\lambda = -1$ into coordinates of Q$Q$:

Now, compute the distance squared (PQ)²$(PQ)^2$:

Pattern Recognition

Instead of finding coordinates of Q$Q$, we could also use the vector form directly: PQ = λ( i + 4 j + 7 k)$\vec{PQ} = \lambda(\hat{i} + 4\hat{j} + 7\hat{k})$. Distance squared is PQ² = λ²(1² + 4² + 7²) = (-1)²(1 + 16 + 49) = 66$PQ^2 = \lambda^2(1^2 + 4^2 + 7^2) = (-1)^2(1 + 16 + 49) = 66$. This saves a lot of fractional coordinate subtraction arithmetic!

Chapter Mix

Class 12 Mathematics: Three Dimensional Geometry

Related Formula

Core Logic

Express general points on L₁$L_1$ and L₂$L_2$ vector components:

Step 1: Equate components to find point of intersection

Substituting (3) into (1):

Point of intersection r₀ = 2(1) i + (1+7) j + (1+3) k = 2 i + 8 j + 4 k$\vec{r}_0 = 2(1)\hat{i} + (1+7)\hat{j} + (1+3)\hat{k} = 2\hat{i} + 8\hat{j} + 4\hat{k}$.

Step 2: Formulate line L3

Direction vector v = a + b = 3 i + 9 j + 4 k$\vec{v} = \vec{a} + \vec{b} = 3\hat{i} + 9\hat{j} + 4\hat{k}$.

Step 3: Match options

For α = 2$\alpha = 2$:

This matches coordinates (8, 26, 12)$(8, 26, 12)$.

Pattern Recognition

Instead of checking all lines simultaneously, match options via quick ratio checks: (x - 2)/(3) = (y - 8)/(9) = (z - 4)/(4)$\frac{x - 2}{3} = \frac{y - 8}{9} = \frac{z - 4}{4}$. This eliminates false choices instantly.

Chapter Mix

Class 12 Mathematics: Three Dimensional Geometry Class 12 Mathematics: Vector Algebra

Related Formula

Core Logic

Compute direction ratios for L₃$L_3$ using cross product of direction vectors of L₁$L_1$ and L₂$L_2$:

Step 1: Establish intersection constraints

Let A$A$ be a general point on L₃$L_3$ and B$B$ be a general point on L₁$L_1$:

Since L₃$L_3$ intersects L₁$L_1$, at the point of intersection A ≡ B$A \equiv B$:

Step 2: Solve final algebraic target equation

Substitute expressions into the target mod expression:

Taking absolute value yields |-25| = 25$|-25| = 25$.

Pattern Recognition

The cancellation of internal variables (λ$\lambda$ and k$k$) shows that the absolute expression describes a invariant geometric plane coordinate constraint, allowing simple baseline parameter values (like k=0, λ=0$k=0, \lambda=0$) to evaluate the question instantly.

Chapter Mix

Class 12 Mathematics: Three Dimensional Geometry

Related Formula

The shortest distance between two skew lines r = a₁ + s b₁$\vec{r} = \vec{a}_1 + s\vec{b}_1$ and r = a₂ + t b₂$\vec{r} = \vec{a}_2 + t\vec{b}_2$ is given by:

Core Logic

Identify the vectors from the equations of the lines:

• Line 1 passes through a₁ = λ i + 2 j + k$\vec{a}_1 = \lambda\hat{i} + 2\hat{j} + \hat{k}$ with direction b₁ = -2 i + j + k$\vec{b}_1 = -2\hat{i} + \hat{j} + \hat{k}$.
• Line 2 passes through a₂ = √(3) i + j + 2 k$\vec{a}_2 = \sqrt{3}\hat{i} + \hat{j} + 2\hat{k}$ with direction b₂ = i - 2 j + k$\vec{b}_2 = \hat{i} - 2\hat{j} + \hat{k}$.

Calculate the coordinate difference vector:

Step 1: Compute the Cross Product of the Direction Vectors

Find the cross product matrix:

Compute its magnitude:

Step 2: Solve the Absolute Distance Equation for lambda

Substitute these values into the shortest distance formula:

This gives:

Unfolding the absolute value gives two cases:

• Case A: √(3) - λ = √(3) λ = 0$\sqrt{3} - \lambda = \sqrt{3} \implies \lambda = 0$
• Case B: √(3) - λ = -√(3) λ = 2√(3)$\sqrt{3} - \lambda = -\sqrt{3} \implies \lambda = 2\sqrt{3}$

Step 3: Calculate the Final Sum

The sum of all possible values of λ$\lambda$ is:

Pattern Recognition

Sees: Shortest distance setup between vector paths containing free variables. Shortcut: In equations like |c - λ| = d$|c - \lambda| = d$, the sum of the roots is simply equal to 2c$2c$, because the roots are symmetrically balanced around the center point c$c$. Thus, Sum = 2 × √(3) = 2√(3)$\text{Sum} = 2 \times \sqrt{3} = 2\sqrt{3}$ holds instantly without calculating individual values.

Chapter Mix

Class 12 Mathematics: Three Dimensional Geometry Class 12 Mathematics: Vector Algebra