Let the line passing through the points (-1, 2, 1)$(-1, 2, 1)$ and \parallel to the line (x - 1)/(2) = (y + 1)/(3) = (z)/(4)$\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4}$ intersect the line (x + 2)/(3) = (y - 3)/(2) = (z - 4)/(1)$\frac{x + 2}{3} = \frac{y - 3}{2} = \frac{z - 4}{1}$ at the point P$P$. Then the distance of P$P$ from the point Q(4, -5, 1)$Q(4, -5, 1)$ is :

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