Let (5,(a)/(4))$(5,\frac{a}{4})$ be the circumcenter of a triangle with vertices A(a,-2)$A(a,-2)$, B(a,6)$B(a,6)$ and C((a)/(4),-2)$C(\frac{a}{4},-2)$. Let α$\alpha$ denote the circumradius, β$\beta$ denote the area and γ$\gamma$ denote the perimeter of the triangle. Then α+β+γ$\alpha+\beta+\gamma$ is

Related Formula

For any line making angles α, β, γ$\alpha, \beta, \gamma$ with the coordinate axes, the direction cosines satisfy:

Core Logic

Given that:

Substituting these values into the identity:

Step 1: Solving the Trigonometric Equation

Using the half-angle identity 2 ²((α)/(2)) = 1 + α$2\cos^2\left(\frac{\alpha}{2}\right) = 1 + \cos \alpha$:

This yields two possible cases:

• α = 0 α = (π)/(2)$\cos \alpha = 0 \implies \alpha = \frac{\pi}{2}$
• α = -1 α = π$\cos \alpha = -1 \implies \alpha = \pi$

Step 2: Finding Values of β$\beta$ and their Sum

Now we find corresponding values for β = (α)/(2)$\beta = \frac{\alpha}{2}$:

• If α = (π)/(2) β₁ = (π)/(4)$\alpha = \frac{\pi}{2} \implies \beta_1 = \frac{\pi}{4}$
• If α = π β₂ = (π)/(2)$\alpha = \pi \implies \beta_2 = \frac{\pi}{2}$

Sum of all possible values:

Pattern Recognition

Direction cosines are bounded between [-1, 1]$[-1, 1]$. Always utilize the identities relating double angles or half angles (2 ²θ = 1 + 2θ$2\cos^2\theta = 1 + \cos 2\theta$) to simplify quadratic forms involving different multiples of the coordinate angles.

Chapter Mix

Class 12 Mathematics: Three Dimensional Geometry Class 11 Mathematics: Trigonometric Functions

Related Formula

Distance 'along a line' means the direction vector of the line segment joining target point P$P$ to intersecting point Q$Q$ must be parallel to the given direction ratios:

Core Logic

Let point P = (7, 10, 11)$P = (7, 10, 11)$. Any general point Q$Q$ on the first line (x-4)/(1) = (y-4)/(0) = (z-2)/(3) = λ$\frac{x-4}{1} = \frac{y-4}{0} = \frac{z-2}{3} = \lambda$ is:

Direction ratios of PQ$PQ$:

Step 1: Equating direction vectors

Since distance is measured parallel to the line with direction vector (2, 3, 6)$(2, 3, 6)$, PQ$\vec{PQ}$ must be parallel to (2, 3, 6)$(2, 3, 6)$:

From the middle term:

Let's check consistency with the third term:

Step 2: Distance calculation

For λ = -1$\lambda = -1$, the intersection point Q$Q$ is: Q = (3, 4, -1)$Q = (3, 4, -1)$

Distance PQ$PQ$:

Pattern Recognition

When solving distance parallel to a line in 3D, always write down the parametric coordinates of the general point first. Match the ratio of direction cosines directly to avoid setting up complicated systems of coordinate planes.

Chapter Mix

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

Related Formula

Shortest distance between two skewed lines with vector equations r = a₁ + λ b₁$\vec{r} = \vec{a}_1 + \lambda \vec{b}_1$ and \vec{r} = \vec{a}_2 + \mu \vec{b}_2 is:

Core Logic

From the given lines: Line 1 passes through A(1, 2, 3)$A(1, 2, 3)$ with direction vector b₁ = 2 i + 3 j + 4 k$\vec{b}_1 = 2\hat{i} + 3\hat{j} + 4\hat{k}$. Line 2 passes through B(0, 0, 5)$B(0, 0, 5)$ with direction vector b₂ = i + α j + k$\vec{b}_2 = \hat{i} + \alpha\hat{j} + \hat{k}$.

The vector connecting the two fixed points is:

Step 1: Compute Cross Product of Direction Vectors

Step 2: Apply Shortest Distance Formula

Taking dot product in numerator:

Squaring both sides:

Step 3: Calculate the Sum of Roots

The sum of all possible values of α$\alpha$ is given by the relation:

Pattern Recognition

Whenever asked for the sum of all possible parameter values resulting from a vector condition, look to directly read the linear term coefficient via Vieta's relations rather than explicitly factoring or computing the roots individually.

Chapter Mix

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

Related Formula

General coordinates of any variable point on a 3D line given symmetric form equations:

Core Logic

Let line L intersect Line 1 at point A(2λ + 1, 3λ - 1, 4λ + 1)$A(2\lambda + 1, 3\lambda - 1, 4\lambda + 1)$ and Line 2 at point B(μ + 3, 2μ + 4, μ)$B(\mu + 3, 2\mu + 4, \mu)$. Since line L passes through C(1, 1, 1)$C(1, 1, 1)$, the direction ratios computed from vector segment AC$AC$ must be proportional to the direction ratios computed from vector segment BC$BC$.

Step 1: Determine Direction Ratio Parameters

Direction ratios of BC$BC$ segment:

Equating directional proportionality ratios:

Step 2: Solve for Parameter Intersection values

From the first and third fractional groups:

Substitute μ = -5$\mu = -5$ back into the second parameter group linkage to evaluate the target structural direction indicators, which gives the simplified direction ratio vector for BC$BC$ as:

Step 3: Construct Line Equation and Verify Choice

Equation of line L passing through C(1, 1, 1)$C(1, 1, 1)$ with direction vector (3, 7, 6)$(3, 7, 6)$:

Let's check option (7, 15, 13)$(7, 15, 13)$:

Since all values match perfectly, (7, 15, 13)$(7, 15, 13)$ lies on the line L.

Pattern Recognition

By comparing the first and third fractional terms containing λ$\lambda$ in the denominator, you can solve for μ$\mu$ independently without tracking complex cross-multiplied quadratic λμ$\lambda\mu$ variations.

Chapter Mix

Class 12 Mathematics: Three Dimensional Geometry

Related Formula

Core Logic

Identify points A(1, 2, 3)$A(1, 2, 3)$ and B(λ, 4, 5)$B(\lambda, 4, 5)$ on the lines with directions p = 2 i + 3 j + 4 k$\vec{p} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ and q = 3 i + 4 j + 5 k$\vec{q} = 3\hat{i} + 4\hat{j} + 5\hat{k}$ respectively. Use the shortest distance formula to determine λ₁$\lambda_1$ and λ₂$\lambda_2$.

Step 1: Calculate Cross Product and Direction Vector

Step 2: Solve for Lambda

Step 3: Radius of the Passing Circle

The circle passes through (0,0)$(0,0)$, (4,2)$(4,2)$ and (2,4)$(2,4)$. Using the circumradius formula R = (abc)/(4Δ)$R = \frac{abc}{4\Delta}$:

Pattern Recognition

Shortest distance values create symmetric configurations. When finding a circle passing through (0,0)$(0,0)$, (x,y)$(x,y)$, and (y,x)$(y,x)$, the symmetry about y=x$y=x$ simplifies radius calculations immediately.

Chapter Mix

Class 12 Mathematics: Three Dimensional Geometry Class 11 Mathematics: Circles