Finding Determinant of Line Intersection Points

This solution details the steps to calculate the value of a determinant involving the intersection points of a line with two other given lines in 3D space.

1. Define Points and Lines

• Point P: $(4, 1, 0)$
• Line 1 (L₁): $(x-1)/2 = (y-2)/3 = (z-3)/4$
• Line 2 (L₂): $x-6 = y = -z+4$
• Intersection of the line through P with L₁ is point A$(α, β, γ)$.
• Intersection of the line through P with L₂ is point B$(a, b, c)$.
• The determinant to evaluate is $D = det([[1, 0, 1], [α, β, γ], [a, b, c]])$.

2. Parametrize Lines and Find Coordinates of A and B

Parametric equations help find the coordinates of the intersection points.

• For L₁, let $(x-1)/2 = (y-2)/3 = (z-3)/4 = t₁$. This gives the coordinates of A as: $A(α, β, γ) = (1 + 2t₁, 2 + 3t₁, 3 + 4t₁)$.
• For L₂, let $x-6 = y = -z+4 = t₂$. This gives $x = 6 + t₂$, $y = t₂$, $z = 4 - t₂$. This gives the coordinates of B as: $B(a, b, c) = (6 + t₂, t₂, 4 - t₂)$.

3. Use the Collinearity Condition

Since a single line passes through P(4, 1, 0) and intersects L₁ at A and L₂ at B, the points P, A, and B must lie on the same line (i.e., they are collinear).

This collinearity implies that the vector $PA$ is parallel to the vector $PB$.

• Vector $PA$ = $(α-4, β-1, γ-0)$ $= (1 + 2t₁ - 4, 2 + 3t₁ - 1, 3 + 4t₁ - 0)$ $= (2t₁ - 3, 3t₁ + 1, 4t₁ + 3)$.
• Vector $PB$ = $(a-4, b-1, c-0)$ $= (6 + t₂ - 4, t₂ - 1, 4 - t₂ - 0)$ $= (t₂ + 2, t₂ - 1, 4 - t₂)$.

For vectors $PA$ and $PB$ to be parallel, their corresponding components must be proportional:

$ \frac{2t_1 - 3}{t_2 + 2} = \frac{3t_1 + 1}{t_2 - 1} = \frac{4t_1 + 3}{4 - t_2} $

4. Solve for Parameters t₁ and t₂

We solve the system of equations derived from the proportionality:

1. Equating the first two fractions: $(2t₁ - 3)(t₂ - 1) = (3t₁ + 1)(t₂ + 2)$ $2t₁t₂ - 2t₁ - 3t₂ + 3 = 3t₁t₂ + 6t₁ + t₂ + 2$ $t₁t₂ + 8t₁ + 4t₂ - 1 = 0$ (Equation 1)
2. Equating the second and third fractions: $(3t₁ + 1)(4 - t₂) = (4t₁ + 3)(t₂ - 1)$ $12t₁ - 3t₁t₂ + 4 - t₂ = 4t₁t₂ - 4t₁ + 3t₂ - 3$ $7t₁t₂ - 16t₁ + 4t₂ - 7 = 0$ (Equation 2)

Solving these two equations simultaneously yields $t₂ = -3t₁$.

Substituting $t₂ = -3t₁$ into Equation 1 leads to the quadratic equation:

$3t₁² + 4t₁ + 1 = 0$

Factoring gives $(3t₁ + 1)(t₁ + 1) = 0$.

The possible values for $t₁$ are $-1$ and $-1/3$.

Testing these values shows that $t₁ = -1/3$ leads to non-parallel vectors PA and PB. Therefore, the only valid solution is $t₁ = -1$.

Using $t₁ = -1$, we find $t₂ = -3(-1) = 3$.

5. Calculate Coordinates of A and B

Using the valid parameter values:

• For A, with $t₁ = -1$: $α = 1 + 2(-1) = -1$ $β = 2 + 3(-1) = -1$ $γ = 3 + 4(-1) = -1$ So, $A = (-1, -1, -1)$.
• For B, with $t₂ = 3$: $a = 6 + 3 = 9$ $b = 3$ $c = 4 - 3 = 1$ So, $B = (9, 3, 1)$.

6. Evaluate the Determinant

Substitute the coordinates of A and B into the determinant:

$ D = \begin{vmatrix} 1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{vmatrix} = \begin{vmatrix} 1 & 0 & 1 \\ -1 & -1 & -1 \\ 9 & 3 & 1 \end{vmatrix} $

Expand the determinant along the first row:

$ D = 1 \cdot \begin{vmatrix} -1 & -1 \\ 3 & 1 \end{vmatrix} - 0 \cdot \begin{vmatrix} -1 & -1 \\ 9 & 1 \end{vmatrix} + 1 \cdot \begin{vmatrix} -1 & -1 \\ 9 & 3 \end{vmatrix} $

$ D = 1 \cdot ((-1)(1) - (-1)(3)) + 1 \cdot ((-1)(3) - (-1)(9)) $

$ D = 1 \cdot (-1 + 3) + 1 \cdot (-3 + 9) $

$ D = 1 \cdot (2) + 1 \cdot (6) $

$ D = 2 + 6 = 8 $