The plane 2x - y + z = 4 intersects the line segment joining the points A(a, -2, 4) and B(2, b, -3)

Q: Why do we reject the positive root for $b$?

The problem imposes a strict condition: $ab < 0$. When we solve the quadratic, we get two potential values for $b$. We must test both against the relationship $a = b + 2$. Only the value that results in a negative product for $a$ and $b$ is physically consistent with the problem's constraints. Always check your constraints at the end!

Q: Is there a faster way to find $C$ without the section formula?

You could use the parametric form of the line $AB$, where any point on the line is $P(t) = A + t(B-A)$. By setting the ratio to $2:1$, you are essentially choosing a specific parameter $t$. The section formula is just a shortcut for this parametric approach—it's the most efficient tool in your kit for this specific problem.

Q: Why did we need to calculate $P$ separately?

The point $P$ is defined by the parameters $a$ and $b$ we just discovered. It acts as a final test of your algebraic accuracy. Once you have the correct $a$ and $b$, $P$ becomes a fixed point in space, allowing us to calculate the final distance $CP^2$ with confidence. Think of it as the final destination of our journey.

Question

The plane $2x - y + z = 4$ intersects the line segment joining the points $A(a, -2, 4)$ and $B(2, b, -3)$ at the point $C$ in the ratio $2 : 1$ and the distance of the point $C$ from the origin is $\sqrt{5}$. If $ab < 0$ and $P$ is the point $(a - b, b, 2b - a)$ then $CP^2$ is equal to :

Options:

(A) $17/3$

(B) $16/3$

(C) $73/3$

(D) $97/3$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a problem; we are visualizing a collision in three-dimensional space.

Imagine a plane, defined by the equation $2x - y + z = 4$, slicing through the void. A line segment, anchored by points $A(a, -2, 4)$ and $B(2, b, -3)$, pierces this plane at a specific point $C$.

Our mission is to uncover the identity of this point $C$ and use it to unlock the final value of $CP^2$. Let us begin.

### The Bridge of the Section Formula

When a point $C$ divides a line segment $AB$ in a ratio $m:n$, it is essentially a weighted average of the two endpoints. We are given the ratio $2:1$.

Using the section formula, we can express the coordinates of $C$ as:

$$C = \left(\frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n}, \frac{m z_2 + n z_1}{m+n}\right)$$

Substituting our values $m=2$ and $n=1$, we find the coordinates of $C$ in terms of our unknowns $a$ and $b$:

$$C = \left(\frac{2(2) + 1(a)}{2+1}, \frac{2(b) + 1(-2)}{2+1}, \frac{2(-3) + 1(4)}{2+1}\right) = \left(\frac{a+4}{3}, \frac{2b-2}{3}, -\frac{2}{3}\right)$$

We have successfully reduced the mystery of point $C$ to two variables.

### The Constraint of the Plane

Point $C$ is not just floating in space; it is trapped on the plane $2x - y + z = 4$. This means the coordinates of $C$ must satisfy this equation perfectly.

Substituting our expressions for $x, y,$ and $z$ into the plane equation:

$$2\left(\frac{a+4}{3}\right) - \left(\frac{2b-2}{3}\right) + \left(-\frac{2}{3}\right) = 4$$

Multiply the entire equation by $3$ to clear the denominators:

$$2(a+4) - (2b-2) - 2 = 12$$

Expanding this, we find $2a + 8 - 2b + 2 - 2 = 12$, which simplifies to $2a - 2b = 4$, or simply $a - b = 2$. This is our golden key: $a = b + 2$.

### The Distance Filter

We are told the distance of $C$ from the origin is $\sqrt{5}$, which means $OC^2 = 5$. Using the distance formula $x^2 + y^2 + z^2 = 5$:

$$\left(\frac{a+4}{3}\right)^2 + \left(\frac{2b-2}{3}\right)^2 + \left(-\frac{2}{3}\right)^2 = 5$$

Substitute $a = b + 2$ into this equation to solve for $b$:

$$\left(\frac{b+6}{3}\right)^2 + \left(\frac{2b-2}{3}\right)^2 + \frac{4}{9} = 5$$

Multiplying by $9$ clears the path: $(b+6)^2 + (2b-2)^2 + 4 = 45$. Expanding these squares gives us $5b^2 + 4b - 1 = 0$.

Factoring this quadratic yields $(5b - 1)(b + 1) = 0$. We have two candidates: $b = -1$ or $b = 1/5$.

### The Final Victory

We must apply the final constraint: $ab < 0$.

If $b = 1/5$, then $a = 11/5$, and $ab > 0$ (rejected). If $b = -1$, then $a = 1$, and $ab = -1 < 0$ (accepted!).

With $a=1$ and $b=-1$, we find $C = (5/3, -4/3, -2/3)$ and $P = (2, -1, -3)$. Finally, the distance $CP^2$ is:

$$CP^2 = \left(2 - \frac{5}{3}\right)^2 + \left(-1 - \left(-\frac{4}{3}\right)\right)^2 + \left(-3 - \left(-\frac{2}{3}\right)\right)^2$$

$$CP^2 = \left(\frac{1}{3}\right)^2 + \left(\frac{1}{3}\right)^2 + \left(-\frac{7}{3}\right)^2 = \frac{1}{9} + \frac{1}{9} + \frac{49}{9} = \frac{51}{9} = \frac{17}{3}$$

The final result is **$\frac{17}{3}$**.

Select Answer:

Visualized Solution

Visualizing the Geometry

Applying Section Formula for $C$

Coordinates of $C$

Constraint 1: $C$ lies on the Plane

Simplifying the Plane Equation

Constraint 2: Distance from Origin

Substituting into Distance Formula

Solving the Quadratic Equation

Applying the Condition $ab < 0$

Coordinates of $C$ and $P$

Final Calculation: $C P^{2}$

The Sigma Insight: Intersection of a Line and a Plane

Analyzing the Setup

The Bridge of the Section Formula

The Constraint of the Plane

The Distance Filter

The Final Victory

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Select Answer:

Visualized Solution

Visualizing the Geometry

Applying Section Formula for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

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

Constraint 1: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C lies on the Plane