Let the equation of the plane passing through the line x - 2y - z - 5 = 0 = x + y + 3z - 5 and paral

Q: Why do we use the parameter $\lambda$ instead of just solving for x, y, and z directly?

Great question! The intersection of two planes is a line, not a single point. There are infinitely many planes passing through that line. By using $\lambda$, we create a 'family' of planes, allowing us to pick the one specific plane that satisfies our extra condition (parallelism).

Q: Why is the normal vector of the plane perpendicular to the direction vector of the line?

Think of the plane as a flat sheet of paper and the line as a pencil lying on the table. If the pencil is parallel to the paper, the normal (the vector sticking straight out of the paper) must be perpendicular to the pencil. It's a fundamental geometric requirement for parallelism.

Q: How do I know which cross product to take to find the line's direction?

A line formed by the intersection of two planes lies on both planes. Therefore, its direction vector must be perpendicular to the normal vectors of both planes. The cross product of two vectors always yields a third vector perpendicular to both, which is exactly what we need!

Question

Let the equation of the plane passing through the line $x - 2y - z - 5 = 0 = x + y + 3z - 5$ and parallel to the line $x + y + 2z - 7 = 0 = 2x + 3y + z - 2$ be $ax + by + cz = 65$. Then the distance of the point $(a, b, c)$ from the plane $2x + 2y - z + 16 = 0$ is ______.

Sigma Percentile · Accepted Answer

### Analyzing the Family of Planes

The family of planes passing through the intersection of two given planes, $P_1: x - 2y - z - 5 = 0$ and $P_2: x + y + 3z - 5 = 0$, is represented by the equation $P_1 + \lambda P_2 = 0$.

Substituting the given equations, we obtain:

$$(x - 2y - z - 5) + \lambda(x + y + 3z - 5) = 0$$

Rearranging this to group the coefficients of $x$, $y$, and $z$, we get:

$$(1 + \lambda)x + (-2 + \lambda)y + (-1 + 3\lambda)z - (5 + 5\lambda) = 0$$

This equation represents the "DNA" of our required plane, where $\lambda$ is a parameter that determines the specific orientation of the plane within the family.

### The Parallelism Constraint

Our plane must be parallel to the line formed by the intersection of $x + y + 2z - 7 = 0$ and $2x + 3y + z - 2 = 0$. To find the direction vector $\vec{v}$ of this line, we compute the cross product of the normals of these two planes: $\vec{n}_3 = \hat{i} + \hat{j} + 2\hat{k}$ and $\vec{n}_4 = 2\hat{i} + 3\hat{j} + \hat{k}$.

The direction vector $\vec{v}$ is given by:

$$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 2 \ 2 & 3 & 1 \end{vmatrix} = \hat{i}(1 - 6) - \hat{j}(1 - 4) + \hat{k}(3 - 2) = -5\hat{i} + 3\hat{j} + \hat{k}$$

Since our plane is parallel to this line, its normal vector $\vec{n} = (1 + \lambda)\hat{i} + (-2 + \lambda)\hat{j} + (-1 + 3\lambda)\hat{k}$ must be perpendicular to $\vec{v}$. Therefore, their dot product must be zero: $\vec{n} \cdot \vec{v} = 0$.

### Solving for the Parameter

Substituting the components into the dot product equation:

$$(1 + \lambda)(-5) + (-2 + \lambda)(3) + (-1 + 3\lambda)(1) = 0$$

Expanding the terms, we get:

$$-5 - 5\lambda - 6 + 3\lambda - 1 + 3\lambda = 0$$

Simplifying this expression leads to:

$$\lambda - 12 = 0 \Rightarrow \lambda = 12$$

### Final Calculation

Plugging $\lambda = 12$ back into our family equation:

$$(1 + 12)x + (-2 + 12)y + (-1 + 36)z - (5 + 60) = 0$$
$$13x + 10y + 35z = 65$$

Comparing this to $ax + by + cz = 65$, we identify the point $(a, b, c)$ as $(13, 10, 35)$. We now calculate the distance $d$ of this point from the plane $2x + 2y - z + 16 = 0$ using the formula:

$$d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$$

Substituting the values:

$$d = \frac{|2(13) + 2(10) - 35 + 16|}{\sqrt{2^2 + 2^2 + (-1)^2}} = \frac{|26 + 20 - 35 + 16|}{\sqrt{9}} = \frac{27}{3} = 9$$

The final distance is **9**.

Enter Numerical Value:

Visualized Solution

Family of Planes Concept

Equation of Family of Planes

Substituting the Planes

Finding the Normal Vector $n$

The Parallel Line

Direction Vector of the Line

Calculating Vector $v$

Parallelism Condition

Applying the Dot Product

Solving for $λ$

Final Plane Equation

Identifying Point $(a, b, c)$

Distance from Target Plane

Distance Formula Setup

Final Calculation

The Sigma Insight: Equation of a Plane

Analyzing the Family of Planes

The Parallelism Constraint

Solving for the Parameter

Final Calculation

Similar Questions

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Distance between two parallel planes $2 x + y + 2 z = 8$ and $4 x + 2 y + 4 z + 5 = 0$ is

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Enter Numerical Value:

Visualized Solution

Family of Planes Concept

Equation of Family of Planes

Substituting the Planes

Finding the Normal Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

The Parallel Line

Direction Vector of the Line

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

Parallelism Condition

Applying the Dot Product

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

Final Plane Equation

Identifying Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (a,b,c)

Distance from Target Plane

Distance Formula Setup

Final Calculation

The Sigma Insight: Equation of a Plane

Analyzing the Family of Planes

The Parallelism Constraint

Solving for the Parameter

Final Calculation

Similar Questions

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Solving for $λ$

Identifying Point $(a, b, c)$

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