Let the values of p$p$, for which the shortest distance between the lines (x + 1)/(3) = (y)/(4) = (z)/(5)$\frac{x + 1}{3} = \frac{y}{4} = \frac{z}{5}$ and r = (p i + 2 j + hatk) + λ (2hati + 3hatj + 4hatk)$\vec{r} = (p\hat{i} + 2\hat{j} + hat{k}) + \lambda (2hat{i} + 3hat{j} + 4hat{k})$ is 1sqrt6$\frac{1}{sqrt{6}}$, be a, b, (a < b$a < b$). Then the length of the latus rectum of the ellipse (x²)/(a²) + (y²)/(b²) = 1$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is:

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Core Logic

Given point A(5, 4, 2)$A(5, 4, 2)$ and line (L)$(L)$:

Any general point P$P$ on this line has coordinates: (-1 + 2λ, 3 + 3λ, 1 - λ)$(-1 + 2\lambda, 3 + 3\lambda, 1 - \lambda)$

Step 1: Finding the foot of the perpendicular

Vector AP = P - A = (-1 + 2λ - 5) i + (3 + 3λ - 4) j + (1 - λ - 2) k$\vec{AP} = P - A = (-1 + 2\lambda - 5)\hat{i} + (3 + 3\lambda - 4)\hat{j} + (1 - \lambda - 2)\hat{k}$

Since AP$AP$ is perpendicular to line (L)$(L)$, the dot product of AP$\vec{AP}$ with the direction vector of the line (2 i + 3 j - k)$(2\hat{i} + 3\hat{j} - \hat{k})$ must be zero:

Step 2: Coordinates of the foot

Substitute λ = 1$\lambda = 1$ into general point P$P$ to get (α, β, γ)$(\alpha, \beta, \gamma)$: α = -1 + 2(1) = 1$\alpha = -1 + 2(1) = 1$ β = 3 + 3(1) = 6$\beta = 3 + 3(1) = 6$ γ = 1 - 1 = 0$\gamma = 1 - 1 = 0$ Foot of perpendicular is (1, 6, 0)$(1, 6, 0)$.

Step 3: Calculate the projection

Let u = α i + β j + γ k = i + 6 j + 0 k$\vec{u} = \alpha\hat{i} + \beta\hat{j} + \gamma\hat{k} = \hat{i} + 6\hat{j} + 0\hat{k}$ Let w = 6 i + 2 j + 3 k$\vec{w} = 6\hat{i} + 2\hat{j} + 3\hat{k}$

Pattern Recognition

Foot of perpendicular problems algorithm: 1) Frame general vector P(λ)$P(\lambda)$. 2) Construct distance vector AP$\vec{AP}$. 3) Dot product with direction vector d = 0$\vec{d} = 0$. 4) Solve for λ$\lambda$.

Chapter Mix

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

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Core Logic

Step 1: Apply Perpendicularity Condition

Let point P = (-2, r, 1)$P = (-2, r, 1)$. The vector PR = λ - 1, λ + 5 - r, λ + 1$\overrightarrow{PR} = \langle \lambda - 1, \lambda + 5 - r, \lambda + 1 \rangle$. Since PR · 1, 1, 1 = 0$\overrightarrow{PR} \cdot \langle 1, 1, 1 \rangle = 0$:

Step 2: Calculate Distance

Substitute λ$\lambda$ back into point R$R$:

We are given PR = √((14)/(3)) PR² = (14)/(3)$PR = \sqrt{\frac{14}{3}} \implies PR^2 = \frac{14}{3}$. PR² = ( (r - 14)/(3) + 2 )² + ( (r + 10)/(3) - r )² + ( (r + 1)/(3) - 1 )² = (14)/(3)$PR^2 = \left( \frac{r - 14}{3} + 2 \right)^2 + \left( \frac{r + 10}{3} - r \right)^2 + \left( \frac{r + 1}{3} - 1 \right)^2 = \frac{14}{3}$

Dividing by 6:

Step 3: Solve for r

The sum of all possible values of r$r$ is 3 + 7 = 10$3 + 7 = 10$.

Pattern Recognition

When a line makes equal angles with coordinate axes, its direction cosines are (1/√(3), 1/√(3), 1/√(3))$(1/\sqrt{3}, 1/\sqrt{3}, 1/\sqrt{3})$, making its simpler direction ratios (1, 1, 1)$(1, 1, 1)$. Use projection vector methods or direct dot product to find perpendicular foot.

Chapter Mix

Class 12 Maths: Three Dimensional Geometry

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Core Logic

Write general points on L₁$L_1$ and L₂$L_2$ representing points C$C$ and D$D$. The vector CD$\overrightarrow{CD}$ must be parallel to the given direction vector of L₃$L_3$. This yields proportional equations to solve for the line parameters.

Step 1: Write Line Equations

Line L₁$L_{1}$: (x-2)/(-3) = (y-6)/(2) = (z-7)/(4) = λ₁$\frac{x-2}{-3} = \frac{y-6}{2} = \frac{z-7}{4} = \lambda_{1}$ Point C$C$ on L₁$L_{1}$: (-3λ₁+2, 2λ₁+6, 4λ₁+7)$(-3\lambda_{1}+2, 2\lambda_{1}+6, 4\lambda_{1}+7)$ Line L₂$L_{2}$: (x-4)/(2) = (y-3)/(1) = (z-5)/(3) = λ₂$\frac{x-4}{2} = \frac{y-3}{1} = \frac{z-5}{3} = \lambda_{2}$ Point D$D$ on L₂$L_{2}$: (2λ₂+4, λ₂+3, 3λ₂+5)$(2\lambda_{2}+4, \lambda_{2}+3, 3\lambda_{2}+5)$

Step 2: Proportionality of Vector CD

The vector CD = 2λ₂+3λ₁+2, λ₂-2λ₁-3, 3λ₂-4λ₁-2$\overrightarrow{CD} = \langle 2\lambda_{2}+3\lambda_{1}+2, \lambda_{2}-2\lambda_{1}-3, 3\lambda_{2}-4\lambda_{1}-2 \rangle$. Since L₃$L_3$ is parallel to -3 i + 5 j + 16 k$-3\hat{i} + 5\hat{j} + 16\hat{k}$, the components are proportional:

Step 3: Solve for lambda values

From the first two expressions:

From the last two expressions:

Substitute λ₂ = 12λ₁ + 38$\lambda_{2} = 12\lambda_{1} + 38$ into the first equation:

Coordinates of C$C$: (11, 0, -5)$(11, 0, -5)$ Coordinates of D$D$: (8, 5, 11)$(8, 5, 11)$

Step 4: Calculate Magnitude squared

Pattern Recognition

For intersecting lines via a transversal of known direction, represent intersection points generally using independent parameters λ$\lambda$ and μ$\mu$. The difference vector MUST be proportional to the given direction ratio.

Chapter Mix

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

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Core Logic

Let any point on the first line (x - 1)/(2) = (y + 1)/(-3) = (z)/(1) = λ$\frac{x - 1}{2} = \frac{y + 1}{-3} = \frac{z}{1} = \lambda$ be given by P(2λ + 1, -3λ - 1, λ)$P(2\lambda + 1, -3\lambda - 1, \lambda)$.

The distance of P$P$ from the point A(1, -1, 0)$A(1, -1, 0)$ is 4√(14)$4\sqrt{14}$.

Step 1: Finding Point P

For λ = 4$\lambda = 4$, point is P₁(9, -13, 4)$P_1(9, -13, 4)$. Distance from origin: √(81 + 169 + 16) = √(266)$\sqrt{81 + 169 + 16} = \sqrt{266}$ For λ = -4$\lambda = -4$, point is P₂(-7, 11, -4)$P_2(-7, 11, -4)$. Distance from origin: √(49 + 121 + 16) = √(186)$\sqrt{49 + 121 + 16} = \sqrt{186}$

Since P$P$ is nearer to the origin, we choose λ = -4$\lambda = -4$. Therefore, P(α, β, γ) = (-7, 11, -4)$P(\alpha, \beta, \gamma) = (-7, 11, -4)$.

Step 2: Shortest Distance Calculation

We need the shortest distance between Line 1: (x + 7)/(1) = (y - 11)/(2) = (z + 4)/(3)$\frac{x + 7}{1} = \frac{y - 11}{2} = \frac{z + 4}{3}$ and Line 2: (x + 5)/(2) = (y - 10)/(1) = (z - 3)/(1)$\frac{x + 5}{2} = \frac{y - 10}{1} = \frac{z - 3}{1}$.

Here, a₁ = -7 i + 11 j - 4 k$\vec{a_1} = -7\hat{i} + 11\hat{j} - 4\hat{k}$ and b₁ = i + 2 j + 3 k$\vec{b_1} = \hat{i} + 2\hat{j} + 3\hat{k}$. a₂ = -5 i + 10 j + 3 k$\vec{a_2} = -5\hat{i} + 10\hat{j} + 3\hat{k}$ and b₂ = 2 i + j + k$\vec{b_2} = 2\hat{i} + \hat{j} + \hat{k}$.

Shortest distance d$d$ is given by the determinant form:

Evaluate the determinant:

Now find | b₁ × b₂| = | matrix i & j & k 1 & 2 & 3 2 & 1 & 1 matrix | = i(2-3) - j(1-6) + k(1-4) = - i + 5 j - 3 k$|\vec{b_1} \times \vec{b_2}| = \left| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 2 & 1 & 1 \end{matrix} \right| = \hat{i}(2-3) - \hat{j}(1-6) + \hat{k}(1-4) = -\hat{i} + 5\hat{j} - 3\hat{k}$

Magnitude is √((-1)² + 5² + (-3)²) = √(1 + 25 + 9) = √(35)$\sqrt{(-1)^2 + 5^2 + (-3)^2} = \sqrt{1 + 25 + 9} = \sqrt{35}$.

Pattern Recognition

To find points on a line at a given distance from a fixed point on the line itself, the algebraic distance parameter λ$\lambda$ translates to d² = λ²(a²+b²+c²)$d^2 = \lambda^2(a^2+b^2+c^2)$, meaning 14λ²$14\lambda^2$ equates instantly to the squared given distance. Bypasses the complex distance formula.

Chapter Mix

Class 12 Maths: Three Dimensional Geometry

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Core Logic

Given line L$L$: (x - 6)/(3) = (y - 7)/(2) = (z - 7)/(-2)$\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{-2}$. Notice the standard form requires (z-7)/(-2)$(z-7)/(-2)$.

Point P = (1, 2, a)$P = (1, 2, a)$ and its image is Q = (5, b, c)$Q = (5, b, c)$. The midpoint M$M$ of PQ$PQ$ must lie exactly on the line L$L$.

Step 1: Using the Midpoint on the Line

Substitute M$M$ into the line equation:

From -1 = (b - 12)/(4)$-1 = \frac{b - 12}{4}$, we get -4 = b - 12 b = 8$-4 = b - 12 \implies b = 8$.

From -1 = (c + a - 14)/(-4)$-1 = \frac{c + a - 14}{-4}$, we get 4 = c + a - 14 c + a = 18$4 = c + a - 14 \implies c + a = 18$.

Step 2: Using the Orthogonality Condition

The vector PQ$\overrightarrow{PQ}$ must be perpendicular to the line's direction vector v = 3 i + 2 j - 2 k$\vec{v} = 3\hat{i} + 2\hat{j} - 2\hat{k}$.

PQ = (5 - 1) i + (b - 2) j + (c - a) k = 4 i + 6 j + (c - a) k$$\overrightarrow{PQ} = (5 - 1)\hat{i} + (b - 2)\hat{j} + (c - a)\hat{k} = 4\hat{i} + 6\hat{j} + (c - a)\hat{k}$$ (since b = 8$b = 8$)

Now set the dot product to zero:

Step 3: Solving for variables

We have a system of linear equations:

• c + a = 18$c + a = 18$
• c - a = 12$c - a = 12$

Adding both: 2c = 30 c = 15$2c = 30 \implies c = 15$. Substituting c$c$: 15 + a = 18 a = 3$15 + a = 18 \implies a = 3$.

Therefore, a = 3, b = 8, c = 15$a = 3, b = 8, c = 15$.

Calculate a² + b² + c²$a^2 + b^2 + c^2$:

Pattern Recognition

Any 'image of a point' problem revolves around two strict constraints: 1) The line bisects the segment joining the point and its image (Midpoint lies on the line), and 2) The segment is orthogonal to the line (Dot product = 0). Directly imposing these generates decoupled simple linear equations.

Chapter Mix

Class 12 Maths: Three Dimensional Geometry