Let Q be the foot of the perpendicular from the point P(7, -2, 13) on the plane containing the lines

Q: Wait, why do we use the cross product of the direction vectors? Can't we just use the dot product?

That is a fantastic question! The dot product gives us a scalar value related to the angle between vectors, but it doesn't give us a new vector. We need a vector that is perpendicular to the plane, and the cross product is the only operation that produces a vector orthogonal to both input vectors. Since the lines lie in the plane, their direction vectors are parallel to the plane, making their cross product the perfect candidate for the plane's normal vector.

Q: Does it matter which point I pick from the lines to find the plane equation?

Not at all! Since both lines lie entirely within the plane, every single point on either line is also a point on the plane. You can pick any point from $L_1$ or $L_2$ to satisfy the plane equation. Just be careful to extract the coordinates correctly from the symmetric form of the line equations—remember that the form is $\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}$, so the point is $(x_0, y_0, z_0)$.

Q: Why do we simplify the normal vector $\vec{n}$ by dividing by 9? Is it necessary?

It is not strictly necessary, but it is a mark of an experienced problem solver! The normal vector $\vec{n}$ defines the orientation of the plane. Any scalar multiple of $\vec{n}$ (like $9\vec{n}$ or $\frac{1}{9}\vec{n}$) represents the same orientation. By dividing by 9, we work with smaller, cleaner numbers, which significantly reduces the chance of making a silly arithmetic error during the final distance calculation. Always look for ways to simplify your vectors before plugging them into complex formulas.

Question

Let $Q$ be the foot of the perpendicular from the point $P(7, -2, 13)$ on the plane containing the lines $\frac{x+1}{6} = \frac{y-1}{7} = \frac{z-3}{8}$ and $\frac{x-1}{3} = \frac{y-2}{5} = \frac{z-3}{7}$. Then $(PQ)^2$, is equal to .

Sigma Percentile · Accepted Answer

### The Geometry of 3D Space: A Journey to the Plane

Welcome, future engineer. Today, we are not just solving a problem; we are embarking on a journey through three-dimensional space. Imagine you are standing in a vast, empty room with a point $P$ hovering in the air and a flat, infinite plane defined by two lines that crisscross through it.

Our mission is to find the shortest path from point $P$ to that plane. This path is the perpendicular distance, and the point where it hits the plane is $Q$. We want to find $(PQ)^2$.

### Phase 1: Unlocking the Plane's Identity

To define a plane, we need a point on the plane and a normal vector—a vector that stands perfectly upright, perpendicular to the surface. We are given two lines, $L_1$ and $L_2$, that lie within this plane:

$$L_1: \frac{x+1}{6} = \frac{y-1}{7} = \frac{z-3}{8}$$

$$L_2: \frac{x-1}{3} = \frac{y-2}{5} = \frac{z-3}{7}$$

From these equations, we extract the direction vectors $\vec{b_1} = (6, 7, 8)$ and $\vec{b_2} = (3, 5, 7)$. Because these lines lie in the plane, their direction vectors are parallel to the plane.

### Phase 2: The Magic of the Cross Product

To find a vector perpendicular to both $\vec{b_1}$ and $\vec{b_2}$, we calculate the cross product $\vec{n} = \vec{b_1} 	imes \vec{b_2}$. We set up the determinant as follows:

$$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 6 & 7 & 8 \ 3 & 5 & 7 \end{vmatrix}$$

Expanding this, we get $\hat{i}(49 - 40) - \hat{j}(42 - 24) + \hat{k}(30 - 21)$, which simplifies to $\vec{n} = 9\hat{i} - 18\hat{j} + 9\hat{k}$.

For efficiency, we simplify our normal vector by dividing by 9 to obtain $\vec{n} = (1, -2, 1)$. This maintains the correct orientation while simplifying our arithmetic.

### Phase 3: Constructing the Plane Equation

We use the point $(1, 2, 3)$ from line $L_2$ and our normal vector $\vec{n} = (1, -2, 1)$. Applying the point-normal form $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$, we substitute our values:

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

Expanding this, we get $x - 1 - 2y + 4 + z - 3 = 0$. This simplifies beautifully to the plane equation:

$$x - 2y + z = 0$$

### Phase 4: The Final Leap to the Distance

We have point $P(7, -2, 13)$ and the plane $x - 2y + z = 0$. The perpendicular distance $PQ$ from a point $(x_1, y_1, z_1)$ to a plane $ax + by + cz + d = 0$ is:

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

Plugging in our values, we calculate:

$$PQ = \frac{|1(7) - 2(-2) + 1(13)|}{\sqrt{1^2 + (-2)^2 + 1^2}} = \frac{|7 + 4 + 13|}{\sqrt{1 + 4 + 1}} = \frac{24}{\sqrt{6}}$$

Rationalizing this, we find $PQ = 4\sqrt{6}$. Squaring our result, we get $(4\sqrt{6})^2 = 16 	imes 6 = 96$.

The final answer is **96**.

Enter Numerical Value:

Visualized Solution

Visualizing the Setup for $P$ and $Q$

Direction Vectors $b_{1}$ and $b_{2}$

Setting up the Normal Vector $n$

Calculating the Cross Product

Simplifying the Normal Vector $n$

Identifying a Point $(x_{0}, y_{0}, z_{0})$ on the Plane

Substituting into the Plane Equation

Final Equation of the Plane

Formula for Distance $P Q$

Substituting $P (7, - 2, 13)$ into Distance Formula

Calculating the Distance $P Q$

Finding the Final Value of $(P Q)^{2}$

The Sigma Insight: Equation of a Plane

The Geometry of 3D Space

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

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Enter Numerical Value:

Visualized Solution

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

Direction Vectors .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b1​​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b2​​

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