Let a line passing through the point (4, 1, 0) intersect the line L1: (x-1)/2 = (y-2)/3 = (z-3)/4 at

Q: Wait, why do we use two different parameters $p$ and $q$?

Great question! The points $A$ and $B$ lie on two completely independent lines, $L_1$ and $L_2$. There is no reason to assume they are at the same 'distance' or 'position' along their respective lines. Using separate parameters $p$ and $q$ allows us to explore all possible locations for $A$ and $B$ independently before the collinearity condition forces them to align.

Q: Why can't we just use the equation of a line passing through two points?

You absolutely could, but you don't know the points $A$ and $B$ yet! The problem gives you the lines they lie on, not the points themselves. We use the parametric form to 'search' along the lines until we find the specific points $A$ and $B$ that satisfy the collinearity condition with $P$. It is a search-and-find strategy.

Q: Is there a geometric shortcut for the determinant?

While determinants often have geometric interpretations (like volume), in this specific problem, the algebraic approach is the most robust. Once you have the coordinates, the determinant is just a calculation. Don't look for a shortcut that might lead to a trap; trust the algebra!

Question

Let a line passing through the point $(4, 1, 0)$ intersect the line $L_1: \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ at the point $A(\alpha, \beta, \gamma)$ and the line $L_2: x - 6 = y = -z + 4$ at the point $B(a, b, c)$. Then $\begin{vmatrix} 1 & 0 & 1 \ \alpha & \beta & \gamma \ a & b & c \end{vmatrix}$ is equal to

Options:

(A) $8$

(B) $16$

(C) $12$

(D) $6$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, three-dimensional space. You have a fixed point $P(4, 1, 0)$ and two distinct lines, $L_1$ and $L_2$, stretching out into the void.

A new line is drawn through $P$, acting as a bridge that connects $L_1$ at point $A$ and $L_2$ at point $B$. Because $P, A,$ and $B$ all lie on this single line, they are **collinear**. This collinearity is the key that unlocks the entire problem.

### The Parametric Language

To solve this, we must use parametric coordinates to define the positions of $A$ and $L_1$. For $L_1$, we define the ratio as a parameter $p$:

$$\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4} = p$$

This allows us to write any point $A$ on $L_1$ as $A(2p+1, 3p+2, 4p+3)$.

Similarly, for $L_2$, we introduce a parameter $q$:

$$x-6 = y = \frac{z-4}{-1} = q$$

This gives us the general coordinates for any point $B$ on $L_2$ as $B(q+6, q, 4-q)$. We now possess the "DNA" of any point on these lines.

### The Collinearity Bridge

Since $P, A,$ and $B$ are collinear, the vector $\vec{PA}$ must be parallel to the vector $\vec{PB}$. In the world of vectors, being parallel means their direction ratios are proportional.

First, we calculate the vector $\vec{PA}$ by subtracting the coordinates of $P(4, 1, 0)$ from $A(2p+1, 3p+2, 4p+3)$:

$$\vec{PA} = (2p-3, 3p+1, 4p+3)$$

Next, we calculate $\vec{PB}$ by subtracting $P$ from $B(q+6, q, 4-q)$:

$$\vec{PB} = (q+2, q-1, 4-q)$$

### The Algebraic Odyssey

We now equate the ratios of the components of these vectors:

$$\frac{2p-3}{q+2} = \frac{3p+1}{q-1} = \frac{4p+3}{4-q}$$

By cross-multiplying the first two fractions, we obtain:

$$(2p-3)(q-1) = (3p+1)(q+2)$$

Expanding this yields $2pq - 2p - 3q + 3 = 3pq + 6p + q + 2$, which simplifies to the equation $pq + 8p + 4q - 1 = 0$. Solving this system with the remaining ratios leads us to the values **$p = -1$** and **$q = 3$**.

### The Final Calculation

With $p = -1$, our point $A$ becomes $A(-1, -1, -1)$. With $q = 3$, our point $B$ becomes $B(9, 3, 1)$.

We are now ready to compute the determinant $\Delta$:

$$\Delta = \begin{vmatrix} 1 & 0 & 1 \ -1 & -1 & -1 \ 9 & 3 & 1 \end{vmatrix}$$

Expanding along the first row, we calculate:

$$\Delta = 1((-1)(1) - (-1)(3)) - 0 + 1((-1)(3) - (-1)(9))$$

This simplifies to $1(-1 + 3) + 1(-3 + 9) = 1(2) + 1(6) = 8$. The final result of the calculation is **$8$**.

Select Answer:

Visualized Solution

Visualizing the Setup

The Intersecting Line

Parametric Coordinates of $A$

Parametric Coordinates of $B$

The Collinearity Condition

Direction Ratios of $P A$

Direction Ratios of $P B$

Setting up the Proportionality

Solving for Parameters $p$ and $q$

Finding Coordinates of $A$ and $B$

Setting up the Determinant

Final Calculation

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

The Parametric Language

The Collinearity Bridge

The Algebraic Odyssey

The Final Calculation

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