If the distance of the point P(1, -2, 1) from the plane x + 2y - 2z = α, where α 0, is 5, then the

Q: Wait, why do we reject the negative root for $\alpha$?

Great catch! The problem explicitly states that $\alpha > 0$. When we solve the absolute value equation $|5 + \alpha| = 15$, we get two potential solutions: $5 + \alpha = 15$ (which gives $\alpha = 10$) and $5 + \alpha = -15$ (which gives $\alpha = -20$). Since the problem imposes the constraint $\alpha > 0$, we must discard the negative value. Always check your constraints at the end of your algebraic steps!

Q: Can I use the parametric form of the line instead of the foot formula?

Absolutely! In fact, using the parametric form is often safer and more intuitive. By writing the line passing through $P$ and perpendicular to the plane as $\frac{x-1}{1} = \frac{y+2}{2} = \frac{z-1}{-2} = k$, you are essentially saying that any point on this line can be represented as $(1+k, -2+2k, 1-2k)$. Substituting this into the plane equation $x + 2y - 2z = 10$ allows you to solve for $k$ directly. It is a robust method that reduces the chance of memorization errors.

Q: Why is the denominator in the distance formula $\sqrt{a^2 + b^2 + c^2}$?

Think of the plane equation $ax + by + cz = d$ as a dot product: $\vec{n} \cdot \vec{r} = d$, where $\vec{n} = (a, b, c)$ is the normal vector. The distance formula is essentially the projection of the vector connecting a point on the plane to our point $P$ onto the unit normal vector $\hat{n} = \frac{\vec{n}}{|\vec{n}|}$. The term $\sqrt{a^2 + b^2 + c^2}$ is simply the magnitude of the normal vector, which we use to normalize the projection.

Question

If the distance of the point $P(1, -2, 1)$ from the plane $x + 2y - 2z = \alpha$, where $\alpha > 0$, is 5, then the foot of the perpendicular from $P$ to the plane is

Options:

(A) $(\frac{8}{3}, \frac{4}{3}, -\frac{7}{3})$

(B) $(\frac{4}{3}, \frac{4}{3}, \frac{1}{3})$

(C) $(\frac{1}{3}, \frac{2}{3}, \frac{10}{3})$

(D) $(\frac{2}{3}, \frac{1}{3}, \frac{5}{2})$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, three-dimensional room. Before you, a flat, infinite plane stretches out, defined by the equation $x + 2y - 2z = \alpha$.

You are holding a point $P$ at coordinates $(1, -2, 1)$, suspended in the air. The problem asks us to find the 'foot of the perpendicular'—the exact spot on the plane where a laser pointer from $P$, aimed perfectly perpendicular to the surface, would strike.

### The Alpha Mystery

Before we can find the landing spot, we must define the plane itself. We are told the perpendicular distance from $P$ to the plane is $5$.

We reach for our most reliable tool: the perpendicular distance formula:

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

Here, $a=1, b=2, c=-2$. Substituting our point $P(1, -2, 1)$, the numerator becomes:

$$|1(1) + 2(-2) - 2(1) - \alpha| = |1 - 4 - 2 - \alpha| = |-5 - \alpha|$$

The denominator is $\sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{1 + 4 + 4} = 3$. Setting this equal to $5$, we get:

$$\frac{|-5 - \alpha|}{3} = 5 \Rightarrow |5 + \alpha| = 15$$

This yields two possibilities: $\alpha = 10$ or $\alpha = -20$. Since the problem statement specifies $\alpha > 0$, we reject $-20$ and embrace **$\alpha = 10$**. Our plane is now fully revealed: $x + 2y - 2z = 10$.

### The Path to the Foot

Now, we need to find the foot of the perpendicular, $F(x, y, z)$. The line connecting $P$ to $F$ must be perpendicular to the plane, meaning it is parallel to the plane's normal vector $\vec{n} = (1, 2, -2)$.

We can express any point on this line using a parameter $k$:

$$x = 1 + k, \quad y = -2 + 2k, \quad z = 1 - 2k$$

Since $F$ lies on the plane, these coordinates must satisfy the plane equation $x + 2y - 2z = 10$. Substituting our parametric expressions:

$$(1 + k) + 2(-2 + 2k) - 2(1 - 2k) = 10$$

Expanding this, we get $1 + k - 4 + 4k - 2 + 4k = 10$. Combining the $k$ terms, we have $9k - 5 = 10$, which leads to $9k = 15$, or **$k = \frac{5}{3}$**.

### Final Calculation

With $k = \frac{5}{3}$ in hand, the coordinates of $F$ are within our grasp.

For $x$: $1 + \frac{5}{3} = \frac{8}{3}$.

For $y$: $-2 + 2(\frac{5}{3}) = -2 + \frac{10}{3} = \frac{4}{3}$.

For $z$: $1 - 2(\frac{5}{3}) = 1 - \frac{10}{3} = -\frac{7}{3}$.

The foot of the perpendicular is **$(\frac{8}{3}, \frac{4}{3}, -\frac{7}{3})$**. This result is the precise intersection of a line and a plane in 3D space.

Animated Solution for Mathematics - Three Dimensional Geometry: If the distance of the point $P (1, - 2, 1)$ from the plane $x + 2 y - 2 z = α$ , where $α > 0$ , is 5, then the foot of the perpendicular from $P$ to the plane is

Select Answer:

Visualized Solution

Visualizing the Geometry

The Distance Formula Tool

Substituting the Values

Solving for $α$

The Foot of Perpendicular Formula

Calculating the Ratio

Finding the $x$ -coordinate

Finding the $y$ -coordinate

Finding the $z$ -coordinate

Summary and Key Takeaway

The Sigma Insight: Intersection of a Line and a Plane

Analyzing the Setup

The Alpha Mystery

The Path to the Foot

Final Calculation

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