Let a line with direction ratios a, -4a, -7 be perpendicular to the lines with direction ratios 3, -

Q: Wait, why do we use the dot product for perpendicular lines?

Great question! In 3D geometry, the dot product of two vectors is defined as $\vec{A} \cdot \vec{B} = |A||B| \cos(\theta)$. If the lines are perpendicular, the angle $\theta$ is $90^\circ$, and $\cos(90^\circ) = 0$. Thus, the dot product must be zero. It is the most elegant way to enforce orthogonality.

Q: What if $a$ or $b$ had multiple values? Would the answer change?

In this specific problem, the equations lead to a unique value for $b^2$ and $a^2$. If you had found multiple possible values for $a$ or $b$, you would need to check if they all satisfy the original conditions. Always verify your solutions by plugging them back into the original equations!

Q: Why use the parametric form $k$ for the intersection?

The parametric form is your best friend in 3D geometry. It allows you to represent any point on the line using a single variable $k$. When you need to find where a line hits a plane, you just need to find the specific value of $k$ that satisfies the plane's equation. It turns a 3D problem into a simple 1D algebraic one.

Question

Let a line with direction ratios $a, -4a, -7$ be perpendicular to the lines with direction ratios $3, -1, 2b$ and $b, a, -2$. If the point of intersection of the line $\frac{x+1}{a^2+b^2} = \frac{y-2}{a^2-b^2} = \frac{z}{1}$ and the plane $x - y + z = 0$ is $(\alpha, \beta, \gamma)$, then $\alpha + \beta + \gamma$ is equal to ______.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing in the vast, silent expanse of 3D space. You are looking at a line, a mysterious entity whose direction is defined by the variables $a$ and $b$.

This line is constrained by the rigid laws of geometry, forced to be perpendicular to two other lines. Our goal is to uncover the identity of this line and find its intersection with a plane.

### The Language of Orthogonality

We start with the condition of perpendicularity. In 3D space, two lines with direction ratios $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$ are perpendicular if and only if their dot product is zero:

$$l_1l_2 + m_1m_2 + n_1n_2 = 0$$

For our first pair of lines, with direction ratios $(a, -4a, -7)$ and $(3, -1, 2b)$, the condition becomes:

$$(a)(3) + (-4a)(-1) + (-7)(2b) = 0$$

Simplifying this, we get $3a + 4a - 14b = 0$, which reduces to $7a = 14b$, or simply $a = 2b$. This is a beautiful, simple relationship that anchors our understanding.

### The Algebraic Bridge

Now, we apply the same logic to the second pair of lines with direction ratios $(a, -4a, -7)$ and $(b, a, -2)$. The dot product gives us:

$$(a)(b) + (-4a)(a) + (-7)(-2) = 0$$

This expands to $ab - 4a^2 + 14 = 0$. Here is where the magic happens. We substitute our relationship $a = 2b$ into this equation:

$$(2b)(b) - 4(2b)^2 + 14 = 0$$

This simplifies to $2b^2 - 16b^2 + 14 = 0$, leading to $-14b^2 = -14$, or $b^2 = 1$. Consequently, $a^2 = 4b^2 = 4$. We have successfully decoded the variables.

### The Piercing Point

With $a^2 = 4$ and $b^2 = 1$, the equation of our line becomes:

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

This simplifies to:

$$\frac{x+1}{5} = \frac{y-2}{3} = \frac{z}{1}$$

To find where this line pierces the plane $x - y + z = 0$, we introduce the parameter $k$. By setting the line equal to $k$, we get the general point $(5k-1, 3k+2, k)$.

This point must satisfy the plane equation. Substituting these coordinates into $x - y + z = 0$, we get:

$$(5k-1) - (3k+2) + k = 0$$

Solving for $k$, we find $3k - 3 = 0$, so $k = 1$.

### The Final Triumph

With $k = 1$, the intersection point $(\alpha, \beta, \gamma)$ is found by substituting $k$ back into our general point:

$$\alpha = 5(1) - 1 = 4, \quad \beta = 3(1) + 2 = 5, \quad \gamma = 1$$

The sum $\alpha + \beta + \gamma = 4 + 5 + 1 = 10$.

We have navigated the complexities of 3D space and arrived at the answer with precision and clarity. The final result is **10**.

Enter Numerical Value:

Visualized Solution

The 3D Geometry Setup

Condition for Perpendicular Lines

Applying Condition to First Line

Simplifying the First Equation

Applying Condition to Second Line

Simplifying the Second Equation

Solving for $b^{2}$ and $a^{2}$

Constructing the Line Equation

Parametric Form of the Line

Intersection with the Plane

Solving for the Parameter $k$

Finding the Intersection Point

Final Calculation

The Sigma Insight: Intersection of a Line and a Plane

Analyzing the Setup

The Language of Orthogonality

The Algebraic Bridge

The Piercing Point

The Final Triumph

Similar Questions

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