The equation of a plane passing through the line of intersection of the planes x + 2y + 3z = 2 and x

Q: Wait, why do we use the 'Family of Planes' method instead of finding the line of intersection first?

That is a brilliant question! While you *could* find the line of intersection (by setting one variable to zero or using cross products), it is computationally tedious and prone to errors. The 'Family of Planes' method is the elegant shortcut. It encapsulates all possible planes passing through that line into a single parameter $\lambda$. It is the difference between building a house brick-by-brick versus using a pre-fabricated frame.

Q: Why do we square both sides in the final steps? Can't we just solve the absolute value equation directly?

You absolutely can! However, squaring is a powerful algebraic maneuver here. It simultaneously eliminates the absolute value bars and the square root in the denominator. Just remember: when you square both sides, you are essentially solving for $\lambda^2$. In this specific problem, the quadratic terms cancelled out, leaving us with a linear equation, which made life much easier!

Q: What happens if the distance given was zero?

If the distance were zero, it would mean the plane passes through the point $(3, 1, -1)$. In that case, you wouldn't need the distance formula at all! You would simply substitute the point $(3, 1, -1)$ directly into the family equation $(x + 2y + 3z - 2) + \lambda(x - y + z - 3) = 0$ and solve for $\lambda$ immediately. Always check if the point lies on the line of intersection first!

Question

The equation of a plane passing through the line of intersection of the planes $x + 2y + 3z = 2$ and $x - y + z = 3$ and at a distance $\frac{2}{\sqrt{3}}$ from the point $(3, 1, -1)$ is

Options:

(A) $5x - 11y + z = 17$

(B) $\sqrt{2}x + y = 3\sqrt{2} - 1$

(C) $x + y + z = \sqrt{3}$

(D) $x - \sqrt{2}y = 1 - \sqrt{2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The problem involves finding a plane that passes through the line of intersection of two given planes:
$P_1: x + 2y + 3z - 2 = 0$
$P_2: x - y + z - 3 = 0$

The family of planes passing through the intersection of $P_1$ and $P_2$ is represented by the equation $P_1 + \lambda P_2 = 0$, where $\lambda$ is a parameter. This gives us:
$$(x + 2y + 3z - 2) + \lambda(x - y + z - 3) = 0$$

### Organizing the Equation

To simplify, we group the terms by their respective variables to reach the standard form $Ax + By + Cz + D = 0$:
$$(1 + \lambda)x + (2 - \lambda)y + (3 + \lambda)z - (2 + 3\lambda) = 0$$

### Applying the Distance Constraint

We are given that the perpendicular distance from the point $(3, 1, -1)$ to this plane is $\frac{2}{\sqrt{3}}$. The formula for the distance $d$ from a point $(x_0, y_0, z_0)$ to a plane $Ax + By + Cz + D = 0$ is:
$$d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$$

Substituting our point $(3, 1, -1)$ and the coefficients into the formula, the numerator becomes:
$$|(1 + \lambda)(3) + (2 - \lambda)(1) + (3 + \lambda)(-1) - (2 + 3\lambda)|$$

Expanding this expression:
$$3 + 3\lambda + 2 - \lambda - 3 - \lambda - 2 - 3\lambda = |-2\lambda|$$

### Solving for the Parameter

The denominator is $\sqrt{(1 + \lambda)^2 + (2 - \lambda)^2 + (3 + \lambda)^2}$. Expanding the squares, we get:
$$\sqrt{(1 + 2\lambda + \lambda^2) + (4 - 4\lambda + \lambda^2) + (9 + 6\lambda + \lambda^2)} = \sqrt{3\lambda^2 + 4\lambda + 14}$$

Setting the distance equal to $\frac{2}{\sqrt{3}}$:
$$\frac{|-2\lambda|}{\sqrt{3\lambda^2 + 4\lambda + 14}} = \frac{2}{\sqrt{3}}$$

Squaring both sides to eliminate the absolute value and square root:
$$\frac{4\lambda^2}{3\lambda^2 + 4\lambda + 14} = \frac{4}{3} \Rightarrow \frac{\lambda^2}{3\lambda^2 + 4\lambda + 14} = \frac{1}{3}$$

Cross-multiplying yields $3\lambda^2 = 3\lambda^2 + 4\lambda + 14$. The quadratic terms cancel, leaving:
$$4\lambda + 14 = 0 \Rightarrow \lambda = -\frac{7}{2}$$

### Final Calculation

Substitute $\lambda = -\frac{7}{2}$ back into the family equation:
$$(x + 2y + 3z - 2) - \frac{7}{2}(x - y + z - 3) = 0$$

Multiplying by $2$ to clear the fraction:
$$2(x + 2y + 3z - 2) - 7(x - y + z - 3) = 0$$
$$2x + 4y + 6z - 4 - 7x + 7y - 7z + 21 = 0$$

Combining like terms results in $-5x + 11y - z + 17 = 0$. Multiplying by $-1$, we obtain the final equation of the plane:
**$5x - 11y + z = 17$**

Animated Solution for Mathematics - Three Dimensional Geometry: The equation of a plane passing through the line of intersection of the planes $x + 2 y + 3 z = 2$ and $x - y + z = 3$ and at a distance $\frac{2}{3}$ from the point $(3, 1, - 1)$ is

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Visualized Solution

$V i s u a l i z in g t h e I n t er sec t i o n$

$T h e F ami l y o f P l an es E q u a t i o n$

$S u b s t i t u t in g t h e P l an es$

$G r o u p in g T er m s$

$T h e D i s t an ce C o n s t r ain t$

$D i s t an ce F or m u l a$

$A ppl y in g t h e D i s t an ce F or m u l a$

$S im pl i f y in g t h e N u m er a t or$

$S im pl i f y in g t h e D e n o mina t or$

$S q u a r in g B o t h S i d es$

$S o l v in g f or λ$

$S u b s t i t u t in g λ B a c k$

$F ina l E q u a t i o n o f t h e P l an e$

The Sigma Insight: Intersection of a Line and a Plane

Analyzing the Setup

Organizing the Equation

Applying the Distance Constraint

Solving for the Parameter

Final Calculation

Similar Questions

The distance of the point $(1, 0, 2)$ from the point of intersection of the line $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12}$ and the plane $x - y + z = 16$ , is

The distance of the point $(1, 1, 9)$ from the point of intersection of the line $\frac{x - 3}{1} = \frac{y - 4}{2} = \frac{z - 5}{2}$ and the plane $x + y + z = 17$ is:

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The distance of the point $(1, - 2, 3)$ from the plane $x - y + z = 5$ measured parallel to the line $\frac{x}{2} = \frac{y}{3} = \frac{z}{- 6}$ is:

The image of the line $\frac{x - 1}{3} = \frac{y - 3}{1} = \frac{z - 4}{- 5}$ in the plane $2 x - y + z + 3 = 0$ is the line:

On which of the following lines lies the point of intersection of the line, $\frac{x - 4}{2} = \frac{y - 5}{2} = \frac{z - 3}{1}$ and the plane, $x + y + z = 2$ ?

If the line of intersection of the planes $a x + b y = 3$ and $a x + b y + cz = 0, a > 0$ makes an angle $3 0^{\circ}$ with the plane $y - z + 2 = 0$ , then the direction cosines of the line are :

The distance of the point $P (3, 4, 4)$ from the point of intersection of the line joining the points $Q (3, - 4, - 5)$ and $R (2, - 3, 1)$ and the plane $2 x + y + z = 7$ , is equal to ___

If the lines $\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z + 3}{1}$ and $\frac{x - a}{2} = \frac{y + 2}{3} = \frac{z - 3}{1}$ intersects at the point $P$ , then the distance of the point $P$ from the plane $z = a$ is :

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Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Family of Planes Equation

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

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Applying the Distance Formula

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

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

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Equation of the Plane