If the equation of a plane P, passing through the intersection of the planes x + 4y - z + 7 = 0 and

Q: Wait, why do we use the family of planes formula $P_1 + \lambda P_2 = 0$ instead of finding the line of intersection first?

That is a great question! While you *could* find the line of intersection, it is often a tedious process involving finding two points on the line or the direction vector. The family of planes approach is a powerful shortcut in JEE geometry because it directly gives you the equation of any plane passing through that line without needing to explicitly define the line itself. It saves time and reduces the chance of calculation errors.

Q: Why are the coefficients proportional? Can't we just equate them directly?

Remember that the equation of a plane $Ax + By + Cz + D = 0$ is not unique. If you multiply the entire equation by a constant, it still represents the same plane. Therefore, when we compare our derived equation with the given one, we must account for this scaling factor by setting the ratios of the coefficients equal to each other, rather than assuming the coefficients themselves are identical.

Q: Why do we use the absolute value in the distance formula?

Distance is a physical quantity, and by definition, it must be non-negative. The expression $Ax_1 + By_1 + Cz_1 + D$ can result in a negative value depending on which side of the plane the point lies. The absolute value ensures that we get the magnitude of the perpendicular distance, which is always positive.

Question

If the equation of a plane $P$, passing through the intersection of the planes $x + 4y - z + 7 = 0$ and $3x + y + 5z = 8$ is $ax + by + 6z = 15$ for some $a, b \in R$, then the distance of the point $(3, 2, -1)$ from the plane $P$ is

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are holding a book, and you open it to the middle. The two pages are like two planes in 3D space, and the spine where they meet is their line of intersection.

We are given two planes:
$$P_1: x + 4y - z + 7 = 0$$
$$P_2: 3x + y + 5z - 8 = 0$$

Our goal is to find a third plane $P$ that passes through this spine. In the world of JEE Advanced, we use the **Family of Planes** method. Any plane passing through the intersection of $P_1$ and $P_2$ can be expressed as:
$$P_1 + \lambda P_2 = 0$$

The parameter $\lambda$ acts like a dial that rotates our new plane around the spine until it hits the exact orientation we need.

### The Algebraic Dance

Substituting the expressions for $P_1$ and $P_2$, we get:
$$(x + 4y - z + 7) + \lambda(3x + y + 5z - 8) = 0$$

Grouping the $x$, $y$, and $z$ terms together, we obtain:
$$x(1 + 3\lambda) + y(4 + \lambda) + z(-1 + 5\lambda) + (7 - 8\lambda) = 0$$

The problem provides the target plane: $ax + by + 6z - 15 = 0$. Since our equation and this target equation represent the same plane, their coefficients must be proportional:
$$\frac{1+3\lambda}{a} = \frac{4+\lambda}{b} = \frac{-1+5\lambda}{6} = \frac{7-8\lambda}{-15}$$

### The Golden Ticket

We isolate the last two ratios to solve for $\lambda$:
$$\frac{-1+5\lambda}{6} = \frac{7-8\lambda}{-15}$$

Cross-multiplying gives:
$$-15(-1 + 5\lambda) = 6(7 - 8\lambda)$$
$$15 - 75\lambda = 42 - 48\lambda$$

Rearranging the terms:
$$-27\lambda = 27 \Rightarrow \lambda = -1$$

Plugging $\lambda = -1$ back into our grouped equation:
$$x(1 - 3) + y(4 - 1) + z(-1 - 5) + (7 + 8) = 0$$
$$-2x + 3y - 6z + 15 = 0$$

Multiplying by $-1$ for standard form, we get the plane equation:
$$2x - 3y + 6z - 15 = 0$$

### Final Calculation

We now calculate the distance from the point $(3, 2, -1)$ to the plane using the formula:
$$d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$$

Substituting the values:
$$d = \frac{|2(3) - 3(2) + 6(-1) - 15|}{\sqrt{2^2 + (-3)^2 + 6^2}}$$

The numerator simplifies as follows:
$$|6 - 6 - 6 - 15| = |-21| = 21$$

The denominator is:
$$\sqrt{4 + 9 + 36} = \sqrt{49} = 7$$

Thus, the final distance is:
$$d = \frac{21}{7} = 3$$

The final answer is **3**.

Enter Numerical Value:

Visualized Solution

Visualizing the Intersecting Planes

The Family of Planes

Substituting the Plane Equations

Grouping the Variables

Comparing with the Target Plane

Establishing Proportionality

Isolating $λ$

Calculating $λ$

Finding the Final Plane Equation

The Distance Formula

Substituting Point and Plane

Final Calculation

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Algebraic Dance

The Golden Ticket

Final Calculation

Similar Questions

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Enter Numerical Value:

Visualized Solution

Visualizing the Intersecting Planes

The Family of Planes

Substituting the Plane Equations

Grouping the Variables

Comparing with the Target Plane

Establishing Proportionality

Isolating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ

Finding the Final Plane Equation

The Distance Formula

Substituting Point and Plane

Final Calculation

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Algebraic Dance

The Golden Ticket

Final Calculation

Similar Questions

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