Let a plane P pass through the point (3, 7, -7) and contain the line, (x-2)/-3 = (y-3)/2 = (z+2)/1.

Q: Why can't I just use the direction vector of the line as the normal vector?

The direction vector of the line is parallel to the line, which means it lies *in* the plane, not perpendicular to it. The normal vector must be perpendicular to *all* lines in the plane. By taking the cross product of the line's direction vector and another vector in the plane, we find a vector that is perpendicular to both, which is exactly what we need.

Q: Does it matter which point on the line I choose to construct the second vector?

No, it does not matter at all. The plane is uniquely defined by the line and the point $B$. Any point $A$ you choose on the line will result in a vector $\vec{AB}$ that lies in the plane. The cross product of the line's direction vector and $\vec{AB}$ will always yield a vector parallel to the same normal, so the final plane equation will be identical.

Q: Why is the distance formula $d = \frac{|D|}{\sqrt{A^2+B^2+C^2}}$?

This formula is derived from the projection of the vector from the origin to any point on the plane onto the unit normal vector. Since the origin is $(0,0,0)$, the distance simplifies to the absolute value of the constant term $D$ divided by the magnitude of the normal vector $\sqrt{A^2+B^2+C^2}$.

Question

Let a plane $P$ pass through the point $(3, 7, -7)$ and contain the line, $\frac{x-2}{-3} = \frac{y-3}{2} = \frac{z+2}{1}$. If distance of the plane $P$ from the origin is $d$, then $d^2$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, three-dimensional space. You have a single point $B(3, 7, -7)$ floating in the void, and a straight line $L$ defined by the equation:

$$\frac{x-2}{-3} = \frac{y-3}{2} = \frac{z+2}{1}$$

Your mission is to find the equation of the unique plane $P$ that contains both this point and this line. Once we have that, we will calculate the distance from the origin to this plane.

### Anchoring the Line

First, we need to understand our tools. From the symmetric form of the line $L$, we can extract a point $A$ that lies on the line and the direction vector $\vec{v}$ that defines its orientation.

By looking at the denominators, the direction ratios are $(-3, 2, 1)$, so our direction vector is $\vec{v} = -3\hat{i} + 2\hat{j} + \hat{k}$. From the numerators, we identify a point $A(2, 3, -2)$ that sits on the line.

Because the line $L$ lies entirely within our plane $P$, both the point $A$ and the vector $\vec{v}$ are contained within the plane.

### The Magic of the Cross Product

To define a plane, we need a normal vector $\vec{n}$ that stands perfectly perpendicular to the surface. We already have one vector in the plane: $\vec{v}$.

To find the normal, we need a second, non-parallel vector. We create this by connecting our known point $A$ on the line to the point $B$:

$$\vec{AB} = (3-2)\hat{i} + (7-3)\hat{j} + (-7 - (-2))\hat{k} = \hat{i} + 4\hat{j} - 5\hat{k}$$

Now, we have two vectors, $\vec{v}$ and $\vec{AB}$, both lying in the plane. The cross product $\vec{n} = \vec{v} 	imes \vec{AB}$ provides the normal vector:

$$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -3 & 2 & 1 \ 1 & 4 & -5 \end{vmatrix}$$

Expanding this determinant, we get:

$$\vec{n} = \hat{i}(-10 - 4) - \hat{j}(15 - 1) + \hat{k}(-12 - 2) = -14\hat{i} - 14\hat{j} - 14\hat{k}$$

Since we only care about the direction, we can scale this vector by dividing by $-14$, yielding the simplified normal vector $\vec{n} = \hat{i} + \hat{j} + \hat{k}$.

### The Equation and the Final Distance

With our normal vector $\vec{n} = (1, 1, 1)$ and point $A(2, 3, -2)$, we use the point-normal form:

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

Simplifying this, we arrive at the equation of the plane:

$$x + y + z - 3 = 0$$

The distance $d$ from the origin $(0, 0, 0)$ to the plane $Ax + By + Cz + D = 0$ is given by:

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

Substituting our values:

$$d = \frac{|-3|}{\sqrt{1^2+1^2+1^2}} = \frac{3}{\sqrt{3}} = \sqrt{3}$$

The final result, $d^2$, is:

**$d^2 = (\sqrt{3})^2 = 3$**

Animated Solution for Mathematics - Three Dimensional Geometry: Let a plane $P$ pass through the point $(3, 7, - 7)$ and contain the line, $\frac{x - 2}{- 3} = \frac{y - 3}{2} = \frac{z + 2}{1}$ . If distance of the plane $P$ from the origin is $d$ , then $d^{2}$ is equal to

Enter Numerical Value:

Visualized Solution

Problem Setup for Plane $P$

Extracting Information from Line $L$

Point $A$ and Vector $v$

Constructing Vector $A B$

Calculating Vector $A B$

The Normal Vector $n$

Setting up the Cross Product

Evaluating the Determinant

Simplifying the Normal Vector

Equation of the Plane $P$

Simplifying the Plane Equation

Distance $d$ from Origin

Substituting Values for Distance

Calculating Distance $d$

Final Answer: $d^{2}$

The Sigma Insight: Equation of a Plane

Analyzing the Setup

Anchoring the Line

The Magic of the Cross Product

The Equation and the Final Distance

Similar Questions

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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 $∣ d ∣$ .

Enter Numerical Value:

Visualized Solution

Problem Setup for Plane .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Extracting Information from Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L

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

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

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

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