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 vector connecting the shortest distance points P$P$ and Q$Q$ on two skew lines must be simultaneously perpendicular to the direction vectors b₁$\vec{b}_1$ and b₂$\vec{b}_2$ of both lines:

Core Logic

Let's define general points on both lines:

• Point P$P$ on L₁$L_1$: (1+λ, 2-λ, 3+λ)$(1+\lambda, \, 2-\lambda, \, 3+\lambda)$
• Point Q$Q$ on L₂$L_2$: (4+μ, 5+μ, 6-μ)$(4+\mu, \, 5+\mu, \, 6-\mu)$

The direction ratios of vector PQ$\vec{PQ}$ are:

Step 1: Compute Perpendicular Direction Vector

Calculate the cross product of the directions of lines L₁$L_1$ and L₂$L_2$:

Since PQ$\vec{PQ}$ is parallel to (0, 2, 2)$(0, 2, 2)$, we compare the coordinate ratios:

Step 2: Solve for Parameters and Midpoint

Solving linear equations (1) and (2) simultaneously:

Substitute these values back to find the specific coordinates of points P$P$ and Q$Q$: - P = ((5)/(2), (1)/(2), (9)/(2))$P = \left(\frac{5}{2}, \, \frac{1}{2}, \, \frac{9}{2}\right)$ - Q = ((5)/(2), (7)/(2), (15)/(2))$Q = \left(\frac{5}{2}, \, \frac{7}{2}, \, \frac{15}{2}\right)$

The midpoint coordinates (α, β, γ)$(\alpha, \beta, \gamma)$ are:

Step 3: Final Computation

Calculate the required terms:

Pattern Recognition

Sees: Explicit endpoints of the shortest distance line vector segment. Shortcut: Since the cross product component along i$\hat{i}$ is 0, the x-coordinates of both line points are identical, providing a massive shortcut to check algebraic equations immediately.

Chapter Mix

Class 12 Mathematics: Three Dimensional Geometry

Related Formula

Core Logic

Let us compute the direction ratios of the vectors PQ$\overrightarrow{PQ}$ and PR$\overrightarrow{PR}$ originating from vertex P$P$:

Direction ratios of PQ = (4 - 3, 6 - 2, 2 - 3) = (1, 4, -1)$\overrightarrow{PQ} = (4 - 3, 6 - 2, 2 - 3) = (1, 4, -1)$ Direction ratios of PR = (7 - 3, 3 - 2, 2 - 3) = (4, 1, -1)$\overrightarrow{PR} = (7 - 3, 3 - 2, 2 - 3) = (4, 1, -1)$

Step 1: Evaluating the Angle

Using the dot product formula for the angle θ = ∠ QPR$\theta = \angle QPR$:

Since θ = (1)/(2)$\cos \theta = \frac{1}{2}$, we have:

Pattern Recognition

When asked for an angle like ∠ QPR$\angle QPR$, always ensure both vectors diverge from the common vertex P$P$ (i.e., use PQ$\overrightarrow{PQ}$ and PR$\overrightarrow{PR}$) to avoid sign errors.

Chapter Mix

Class 12 Mathematics: Three Dimensional Geometry

Related Formula

The line segment of shortest distance MN$MN$ is perpendicular to both directing vectors b₁$\vec{b}_1$ and b₂$\vec{b}_2$.

Core Logic

Let general point coordinates be expressions of parameters λ$\lambda$ and μ$\mu$:

Vector direction ratios of MN$\overrightarrow{MN}$:

The cross product vector perpendicular to both lines is b₁ × b₂ = (-6, 0, 8)$\vec{b}_1 \times \vec{b}_2 = (-6, 0, 8)$.

Step 1: Finding Parameters via Perpendicular Conditions

Equating directional proportional factors:

From the zero denominator constraint:

From the first and third components:

Solving equations (iii) and (iv) yields λ = -1$\lambda = -1$ and μ = 1$\mu = 1$.

Step 2: Vector Coordinate Evaluation

Substituting parameter roots into point layout vectors:

Evaluating target scalar products:

Pattern Recognition

Shortest distance points constrain line segments to align with the direct cross product vector. Setting proportional components yields quick parameters.

Chapter Mix

Class 12 Mathematics: Three Dimensional Geometry

Related Formula

Core Logic

Let point A = (7, -2, 11)$A = (7, -2, 11)$. We want the distance from A$A$ to the line L₁: (x-6)/(1)=(y-4)/(0)=(z-8)/(3)$L_1: \frac{x-6}{1}=\frac{y-4}{0}=\frac{z-8}{3}$ measured along a line parallel to L₂: (x-5)/(2)=(y-1)/(-3)=(z-5)/(6)$L_2: \frac{x-5}{2}=\frac{y-1}{-3}=\frac{z-5}{6}$.

The line passing through A$A$ parallel to L₂$L_2$ will have the equation:

Any general point B$B$ on this new line can be represented as:

Step 1: Finding Intersection Point B

Since B$B$ lies on the given line L₁$L_1$, its coordinates must satisfy the equation of L₁$L_1$:

Focus on the middle term (since denominator is 0, numerator must equal 0 for intersection):

Step 2: Coordinates of B and Distance Calculation

Substitute λ = -2$\lambda = -2$ back into the coordinates of point B$B$:

Now, calculate the distance AB$AB$ using the 3D distance formula:

Pattern Recognition

Distance of a point from a line along another direction means finding the intersection of the given line and a new line passing through the point parallel to the direction vector. The zero in the direction ratio is a massive shortcut—just equate the corresponding numerator to zero.

Chapter Mix

Class 12 Maths: Three Dimensional Geometry

Related Formula

Core Logic

Identify the positional vectors and direction ratios for both lines: Line 1: a₁ = (4, -1, 0)$\vec{a}_1 = (4, -1, 0)$, direction b₁ = (1, 2, -3)$\vec{b}_1 = (1, 2, -3)$ Line 2: a₂ = (λ, -1, 2)$\vec{a}_2 = (\lambda, -1, 2)$, direction b₂ = (2, 4, -5)$\vec{b}_2 = (2, 4, -5)$

Vector connecting lines: ( a₂ - a₁) = (λ - 4, 0, 2)$(\vec{a}_2 - \vec{a}_1) = (\lambda - 4, 0, 2)$

Step 1: Cross Product and Magnitude

Find the normal vector n = b₁ × b₂$\vec{n} = \vec{b}_1 \times \vec{b}_2$:

Magnitude of the normal vector:

Step 2: Application of Shortest Distance Formula

Dot product of normal vector and positional difference vector:

Using the shortest distance formula given as 6√(5)$\frac{6}{\sqrt{5}}$:

Step 3: Finding Unknown values

Solve the absolute value relation:

Sum of possible values = 7 + 1 = 8$7 + 1 = 8$.

Pattern Recognition

Standard Shortest Distance methodology between skew lines. Cross product of direction vectors forms the perpendicular frame normal, and dot-producting the difference of positional anchor points yields the direct orthogonal projection.

Chapter Mix

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