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

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Core Logic

Let R$R$ be the foot of the perpendicular drawn from point P(1,0,3)$P(1,0,3)$ to the line joining A(4,7,1)$A(4,7,1)$ and B(3,5,3)$B(3,5,3)$. Since Q(α,β,γ)$Q(\alpha,\beta,\gamma)$ is the reflection (image) of P$P$ across the line, point R$R$ serves as the midpoint of the line segment PQ$PQ$.

Step 1: Find the Equation of the Line AB

The direction ratios of the line AB are:

Using point B(3,5,3)$B(3,5,3)$, the symmetric equation of the line AB is:

Any general point R$R$ on this line can be written in terms of parameter λ$\lambda$:

Step 2: Find the Foot of the Perpendicular R

The direction ratios of the line segment PR are:

Since PR is perpendicular to the line AB, the dot product of their direction vectors must equal zero:

Step 3: Coordinates of Foot of Perpendicular

Substitute λ = -(4)/(3)$\lambda = -\frac{4}{3}$ into the general coordinates of R$R$:

Step 4: Solve for the Image Coordinates and Sum

Since R$R$ is the midpoint of PQ$PQ$, where P = (1, 0, 3)$P = (1, 0, 3)$ and Q = (α, β, γ)$Q = (\alpha, \beta, \gamma)$:

• (α + 1)/(2) = (5)/(3) α = (10)/(3) - 1 = (7)/(3)$\frac{\alpha + 1}{2} = \frac{5}{3} \implies \alpha = \frac{10}{3} - 1 = \frac{7}{3}$
• (β + 0)/(2) = (7)/(3) β = (14)/(3)$\frac{\beta + 0}{2} = \frac{7}{3} \implies \beta = \frac{14}{3}$
• (γ + 3)/(2) = (17)/(3) γ = (34)/(3) - 3 = (25)/(3)$\frac{\gamma + 3}{2} = \frac{17}{3} \implies \gamma = \frac{34}{3} - 3 = \frac{25}{3}$

Now, let us calculate the sum:

Pattern Recognition

Standard Midpoint reflection: The image coordinates are given directly by ximage = 2 xfoot - xₚₒᵢₙₜ$x_{\text{image}} = 2 x_{\text{foot}} - x_{\text{point}}$, yimage = 2 yfoot - yₚₒᵢₙₜ$y_{\text{image}} = 2 y_{\text{foot}} - y_{\text{point}}$, and zimage = 2 zfoot - zₚₒᵢₙₜ$z_{\text{image}} = 2 z_{\text{foot}} - z_{\text{point}}$. Finding the parameter λ$\lambda$ by using the perpendicular vector dot-product rule is the fastest and most robust method.

Chapter Mix

Class 12 Mathematics: Three Dimensional Geometry

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Core Logic

Shortest distance between two skew lines is the projection of the vector joining any two points of the lines onto their common normal.

Step 1: Find the vector joining the two points

Let the points be P₁(7, 6, 2)$P_1(7, 6, 2)$ on L₁$L_1$ and P₂(5, 3, 4)$P_2(5, 3, 4)$ on L₂$L_2$:

Step 2: Find the common normal vector

The direction is given by the cross product of the direction vectors:

Step 3: Calculate the distance

Calculate the dot product of the vectors:

Calculate the magnitude of the cross product:

Now, compute the shortest distance:

Pattern Recognition

Matrix determinant check: In skew lines problems, the numerator can also be computed as the determinant of the 3x3 matrix composed of ( r₂- r₁)$(\vec{r}_2-\vec{r}_1)$, a$\vec{a}$, and b$\vec{b}$.

Chapter Mix

Class 12 Mathematics: Three Dimensional Geometry

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Core Logic

First, we compute the total area of the triangle ABC from the given coordinate points. We then apply the geometric maximum area property for inscribed parallelograms.

Step 1: Calculate the area of triangle ABC

The coordinate points are A(4, -2)$A(4, -2)$, B(1, 1)$B(1, 1)$, and C(9, -3)$C(9, -3)$:

Step 2: Apply the maximum area theorem

The maximum area of a parallelogram inscribed in a triangle of area Δ$\Delta$ is always exactly half the area of the triangle:

Pattern Recognition

Inscribed shapes maximization: The maximum area of any inscribed parallelogram AFDE$AFDE$ on the sides of a triangle ABC$ABC$ occurs when the vertices D, E, F$D, E, F$ are exactly the midpoints of the respective sides of the triangle.

Chapter Mix

Class 11 Mathematics: Straight Lines

Related Formula

Area of a triangle given base length b$b$ and perpendicular height h$h$:

Core Logic

Since Q$Q$ and R$R$ lie on the line, the length QR=5$QR=5$ forms the base of the triangle. The perpendicular distance from point P$P$ to the line represents the height h$h$.

Step 1: Define Perpendicular Foot coordinates

Let M$M$ be the foot of the perpendicular from P(0,2,3)$P(0,2,3)$ to the line. Parametric form of any point on the line:

Direction ratios of line segment PM$PM$:

Step 2: Solve for Parameter using Perpendicularity

Since PM$PM$ is perpendicular to the given line (DRs: 5, 2, 3$5, 2, 3$):

Step 3: Compute Perpendicular Distance

For λ = 1$\lambda = 1$, the coordinates of M$M$ are (2, 3, -1)$(2, 3, -1)$. Calculate length PM$PM$:

Step 4: Formulate the Area Equation

Pattern Recognition

Finding the foot of a perpendicular via parametric variables is a guaranteed shortcut for 3D area problems instead of cross-product calculations, keeping computation times minimal.

Chapter Mix

Class 12 Mathematics: Three Dimensional Geometry

Related Formula

De Gua's Theorem (3D extension of Pythagorean theorem) for right corner tetrahedron states:

Core Logic

Since the three edges meeting at vertex A$A$ are mutually perpendicular, we can align them directly with orthogonal coordinate axes to compute the slanted face area.

Step 1: Apply Square Area Summation

Let the target area be Δ$\Delta$.

Pattern Recognition

This is exactly analogous to finding the length of a vector given three perpendicular component lengths. Squaring, adding, and taking the square root yields the value immediately.

Chapter Mix

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