If the distance between the plane, 23x - 10y - 2z + 48 = 0 and the plane containing the lines (x+1)/

Q: Wait, why don't we need to find the value of $\lambda$?

That is a brilliant observation! We only need the intersection point to calculate the distance. Since the lines intersect at a unique point regardless of the value of $\lambda$, we can find that point using only the $x$ and $y$ coordinates. The $\lambda$ term only affects the $z$ coordinate, which we can determine once we have the parameter $q$.

Q: What if the lines were skew? Would this method still work?

If the lines were skew, they would not intersect, and they would not lie on a single plane. The problem specifically states that the lines lie on a second plane, which implies they must be coplanar (intersecting or parallel). In this case, they intersect, so our method is perfectly valid.

Q: Is there a faster way to find the distance without finding the intersection point?

You could find the normal vector of the second plane by taking the cross product of the direction vectors of the two lines. However, that would require finding the equation of the plane first. Finding the intersection point is generally faster and less prone to algebraic errors in this specific case.

Question

If the distance between the plane, $23x - 10y - 2z + 48 = 0$ and the plane containing the lines $\frac{x+1}{2} = \frac{y-3}{4} = \frac{z+1}{3}$ and $\frac{x+3}{2} = \frac{y+2}{6} = \frac{z-1}{\lambda}, (\lambda \in R)$ is equal to $\frac{k}{\sqrt{633}}$, then $k$ is equal to _______

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, three-dimensional space. Before you, there are two perfectly parallel planes, stretching out into infinity like sheets of glass.

One plane is defined by the equation $23x - 10y - 2z + 48 = 0$. The other plane is a bit more mysterious; it contains two lines, $L_1$ and $L_2$.

Your mission is to find the perpendicular distance between these two parallel worlds. This is not just a calculation; it is a journey into the heart of 3D geometry.

### The Strategic Shortcut

Many students immediately panic, thinking they need to find the equation of the second plane. While that is a valid path, it is like taking the long way around a mountain.

Remember, the distance between two parallel planes is constant. This means the distance from *any* point on the second plane to the first plane is the same.

The most elegant point to choose is the intersection of the two lines, $L_1$ and $L_2$. Let us call this point $P$. By finding $P$, we reduce a complex plane-to-plane problem into a simple point-to-plane problem.

### The Parametric Dance

To find the intersection point $P$, we must express the lines in their parametric forms. Let us equate the first line to a parameter $p$ and the second to a parameter $q$:

For $L_1$: 
$$\frac{x+1}{2} = \frac{y-3}{4} = \frac{z+1}{3} = p \Rightarrow (2p - 1, 4p + 3, 3p - 1)$$

For $L_2$: 
$$\frac{x+3}{2} = \frac{y+2}{6} = \frac{z-1}{\lambda} = q \Rightarrow (2q - 3, 6q - 2, \lambda q + 1)$$

At the intersection point, the $x$ and $y$ coordinates must be identical. This gives us a system of two linear equations:

1) $2p - 1 = 2q - 3 \Rightarrow p - q = -1$

2) $4p + 3 = 6q - 2 \Rightarrow 4p - 6q = -5$

Solving this system is like solving a puzzle. Substituting $p = q - 1$ into the second equation, we get $4(q - 1) - 6q = -5$, which simplifies to $-2q = -1$, or $q = \frac{1}{2}$.

Consequently, $p = -\frac{1}{2}$.

### The Final Destination

With our parameters in hand, we can find the coordinates of $P$. Substituting $p = -\frac{1}{2}$ into the parametric form of $L_1$, we get:

$x = 2(-\frac{1}{2}) - 1 = -2$

$y = 4(-\frac{1}{2}) + 3 = 1$

$z = 3(-\frac{1}{2}) - 1 = -\frac{5}{2}$

Our point $P$ is $(-2, 1, -\frac{5}{2})$.

Now, we use the classic distance formula for a point $(x_1, y_1, z_1)$ to a plane $ax + by + cz + d = 0$:

$$d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}$$

Plugging in our values:

$$d = \frac{|23(-2) - 10(1) - 2(-\frac{5}{2}) + 48|}{\sqrt{23^2 + (-10)^2 + (-2)^2}}$$

Calculating the numerator: $|-46 - 10 + 5 + 48| = |-3| = 3$.

The denominator is $\sqrt{529 + 100 + 4} = \sqrt{633}$.

Thus, the distance is $d = \frac{3}{\sqrt{633}}$. Comparing this to the given form $\frac{k}{\sqrt{633}}$, we find that **$k = 3$**.

Enter Numerical Value:

Visualized Solution

The Geometric Setup

Strategy: Point to Plane Distance

Parametric Coordinates of Lines

Equating $x$ and $y$ Coordinates

Solving for Parameters $p$ and $q$

The Intersection Point $P$

Distance Formula

Substituting Values

Evaluating the Numerator

Evaluating the Denominator

Finding $k$

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Strategic Shortcut

The Parametric Dance

The Final Destination

Similar Questions

If the distance between the plane, $23 x - 10 y - 2 z + 48 = 0$ and the plane containing the lines $\frac{x + 1}{2} = \frac{y - 3}{4} = \frac{z + 1}{3}$ and $\frac{x + 3}{2} = \frac{y + 2}{6} = \frac{z - 1}{λ}$ ( $λ \in R$ ) is equal to $\frac{k}{633}$ , then $k$ equal to

If the distance between the plane $A x - 2 y + z = d$ and the plane containing the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 4}{5}$ is $6$ , then find $∣ d ∣$ .

If the lines $r = (\hat{i} - \hat{j} + \hat{k}) + λ (3 \hat{j} - \hat{k})$ and $r = (α \hat{i} - \hat{j}) + μ (2 \hat{i} - 3 \hat{k})$ are co-planar, then distance of the plane containing these two lines from the point $(α, 0, 0)$ is :

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

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

Let the plane containing the line of intersection of the planes $P 1 : x + (λ + 4) y + z = 1$ and $P 2 : 2 x + y + z = 2$ pass through the points $(0, 1, 0)$ and $(1, 0, 1)$ . Then the distance of the point $(2 λ, λ, - λ)$ from the plane $P 2$ is

Let the equation of the plane $P$ containing the line $\frac{x + 10}{1} = \frac{8 - y}{2} = z$ be $a x + b y + 3 z = 2 (a + b)$ and the distance of the plane $P$ from the point $(1, 27, 7)$ be $c$ . Then $a^{2} + b^{2} + c^{2}$ is equal to ______.

A plane which is perpendicular to two planes $2 x - 2 y + z = 0$ and $x - y + 2 z = 4$ , passes through $(1, - 2, 1)$ . The distance of the plane from the point $(1, 2, 2)$ is

If the plane $2 x - y + 2 z + 3 = 0$ has the distances $1/3$ and $2/3$ units from the planes $4 x - 2 y + 4 z + λ = 0$ and $2 x - y + 2 z + μ = 0$ , respectively, then the maximum value of $λ + μ$ is equal to :

The distance of the point $(1, 3, - 7)$ from the plane passing through the point $(1, - 1, - 1)$ , having normal perpendicular to both the lines $\frac{x - 1}{1} = \frac{y + 2}{- 2} = \frac{z - 4}{3}$ and $\frac{x - 2}{2} = \frac{y + 1}{- 1} = \frac{z + 7}{- 1}$ , is: