The distance of the point (1, 0, 2) from the point of intersection of the line (x-2)/3 = (y+1)/4 = (

Q: Why do we set the line equation equal to $r$?

Think of $r$ as a 'time' or 'position' variable. It allows us to describe every single point on the line using just one variable, which makes solving for the intersection point much easier.

Q: Is there any other way to solve this without parametric form?

Yes, you could use vector equations, but parametric form is the most intuitive for finding intersection points. It turns a geometry problem into a simple linear equation.

Q: What if the line was parallel to the plane?

Great question! If the line were parallel, the direction vector of the line would be perpendicular to the normal vector of the plane. In that case, the dot product would be zero, and you would find no unique solution for $r$.

Question

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

Options:

(A) $3\sqrt{21}$

(B) $13$

(C) $2\sqrt{14}$

(D) $8$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are tasked with finding the intersection point $P$ between a line and a plane, and subsequently calculating the distance from a given point $A(1, 0, 2)$ to $P$.

The line is defined by the symmetric equation:
$$\frac{x-2}{3} = \frac{y+1}{4} = \frac{z-2}{12}$$

The plane is defined by the linear equation:
$$x - y + z = 16$$

### The Parametric Bridge

To find the intersection, we define any point on the line using a parameter $r$. By setting the symmetric equations equal to $r$, we obtain:
$$\frac{x-2}{3} = \frac{y+1}{4} = \frac{z-2}{12} = r$$

This allows us to express the coordinates of any point on the line as functions of $r$:
$$x = 3r + 2, \quad y = 4r - 1, \quad z = 12r + 2$$

Thus, any point $P$ on the line can be represented as $(3r + 2, 4r - 1, 12r + 2)$.

### The Moment of Collision

At the point of intersection, the coordinates of $P$ must satisfy the plane equation $x - y + z = 16$. Substituting our parametric expressions into this equation yields:
$$(3r + 2) - (4r - 1) + (12r + 2) = 16$$

Expanding and grouping the terms:
$$3r + 2 - 4r + 1 + 12r + 2 = 16$$
$$11r + 5 = 16$$

Solving for $r$:
$$11r = 11 \Rightarrow r = 1$$

Substituting $r = 1$ back into our parametric equations, we find the coordinates of the intersection point $P$:
$$x = 3(1) + 2 = 5$$
$$y = 4(1) - 1 = 3$$
$$z = 12(1) + 2 = 14$$

The intersection point is $P(5, 3, 14)$.

### Final Calculation

We now calculate the distance between point $A(1, 0, 2)$ and point $P(5, 3, 14)$ using the 3D distance formula:
$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$$

Plugging in the coordinates:
$$d = \sqrt{(5 - 1)^2 + (3 - 0)^2 + (14 - 2)^2}$$
$$d = \sqrt{4^2 + 3^2 + 12^2}$$
$$d = \sqrt{16 + 9 + 144} = \sqrt{169}$$

The final distance is **13**.

Animated Solution for Mathematics - Three Dimensional Geometry: 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

Select Answer:

Visualized Solution

Visualizing the Geometry

Parametric Form of the Line

General Point $P$ on the Line

Intersection Condition

Solving for $r$

Coordinates of Intersection Point $P$

Distance Formula Setup

Calculating the Distance

The Sigma Insight: Intersection of a Line and a Plane

Analyzing the Setup

The Parametric Bridge

The Moment of Collision

Final Calculation

Similar Questions

The square of the distance of the point of intersection of the line $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z + 1}{6}$ and the plane $2 x - y + z = 6$ from the point $(- 1, - 1, 2)$ is .

The distance of the point $(- 1, 9, - 16)$ from the plane $2 x + 3 y - z = 5$ measured parallel to the line $\frac{x + 4}{3} = \frac{2 - y}{4} = \frac{z - 3}{12}$ 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:

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 :

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:

If the line, $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 2}{4}$ meets the plane, $x + 2 y + 3 z = 15$ at a point $P$ , then the distance of $P$ from the origin is

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 ___

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

The distance of the point $(1, - 2, 3)$ from the plane $x - y + z = 5$ measured parallel to a line, whose direction ratios are $2, 3, - 6$ is :

Select Answer:

Visualized Solution

Visualizing the Geometry

Parametric Form of the Line

General Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P on the Line

Intersection Condition

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

Coordinates of Intersection Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Distance Formula Setup

Calculating the Distance

The Sigma Insight: Intersection of a Line and a Plane

Analyzing the Setup

The Parametric Bridge

The Moment of Collision

Final Calculation

Similar Questions

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