Question

Find all points \(C\) on the line through \(A(1,-1,2)\) and \(B=(2,0,1)\) such that \(\|\overrightarrow{A C}\|=2\|\overrightarrow{B C}\|\).

Short answer

The points are \(C(3, 1, 0)\) and \(C\left( \frac{5}{3}, \frac{-1}{3}, \frac{4}{3} \right)\).

Step-by-step solution

Parameterize the Line

The line through points \(A(1, -1, 2)\) and \(B(2, 0, 1)\) can be expressed in vector form as \(\overrightarrow{r}(t) = (1, -1, 2) + t(2-1, 0-(-1), 1-2)\). Simplifying, we have \(\overrightarrow{r}(t) = (1 + t, -1 + t, 2 - t)\). Any point \(C\) on this line is \(C(1+t, -1+t, 2-t)\).

Write Vectors \(\overrightarrow{AC}\) and \(\overrightarrow{BC}\)

The vector \(\overrightarrow{AC}\) is obtained by \((1+t - 1, -1+t + 1, 2-t - 2) = (t, t, -t)\). The vector \(\overrightarrow{BC}\) is \((1+t - 2, -1+t - 0, 2-t - 1) = (t-1, t-1, -t-1)\).

Express the Norms

The norm \(\|\overrightarrow{AC}\|\) is \(\sqrt{t^2 + t^2 + (-t)^2} = \sqrt{3t^2}\). The norm \(\|\overrightarrow{BC}\|\) is \(\sqrt{(t-1)^2 + (t-1)^2 + (-t-1)^2} = \sqrt{3t^2 - 6t + 3}\).

Set Up the Equation

From the problem, we know that \(\|\overrightarrow{AC}\| = 2\|\overrightarrow{BC}\|\). Substituting the expressions, we have \(\sqrt{3t^2} = 2\sqrt{3t^2 - 6t + 3}\).

Square Both Sides to Eliminate the Square Roots

Squaring both sides gives \(3t^2 = 4(3t^2 - 6t + 3)\). Expanding the right side, we have \(3t^2 = 12t^2 - 24t + 12\).

Simplify and Solve for \(t\)

Rearranging terms gives \(9t^2 - 24t + 12 = 0\). Dividing through by 3, we have \(3t^2 - 8t + 4 = 0\). Using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3, b = -8, c = 4\), we find \(t = \frac{8 \pm \sqrt{64 - 48}}{6}\), or \(t = 2\) or \(t = \frac{2}{3}\).

Find Points \(C\) Using Values of \(t\)

Substitute \(t = 2\) and \(t = \frac{2}{3}\) into the parameterization \((1+t, -1+t, 2-t)\):- For \(t = 2\), \(C = (3, 1, 0)\).- For \(t = \frac{2}{3}\), \(C = \left( \frac{5}{3}, \frac{-1}{3}, \frac{4}{3} \right)\).

Key concepts

Parametric Equations

Parametric equations are a way of defining a line or curve using parameters, which are variables that represent point locations on a given geometric shape. For instance, when working with lines in a three-dimensional space, you often define the line using a parameter, often denoted as \(t\). This allows you to express points along the line.

In the solved example, the line passing through points \(A(1, -1, 2)\) and \(B(2, 0, 1)\) is described using parametric equations. Here, a vector equation \(\overrightarrow{r}(t) = (1, -1, 2) + t(1, 1, -1)\) outlines the line. Therefore, any point \(C\) on this line can be specified by the parametric formula \((1+t, -1+t, 2-t)\). This simplifies the task of identifying the coordinates of point \(C\) by simply plugging in the desired value of \(t\).

Using parametric equations gives flexibility, allowing the description of the entire line rather than fixed points, thus aiding in solving complex geometric problems more effectively.

Vector Operations

Vector operations are fundamental in vector algebra and include operations such as addition, subtraction, and finding magnitude (norms). These operations are crucial while dealing with problems involving lines and points in space.

In the context of the solved problem, you encounter such vector operations. The vectors \(\overrightarrow{AC}\) and \(\overrightarrow{BC}\) are derived based on the positions of the points relative to \(C\). Here's how it's done:

• \(\overrightarrow{AC}\) is calculated as \((t, t, -t)\) by subtracting \((1, -1, 2)\) from \(C(1+t, -1+t, 2-t)\).
• \(\overrightarrow{BC}\) is computed as \((t-1, t-1, -t-1)\) by subtracting \((2, 0, 1)\).

The norms of these vectors were important to establishing the condition \(\|\overrightarrow{AC}\| = 2\|\overrightarrow{BC}\|\), showing that understanding basic vector operations is essential for progressing through such exercises.

Quadratic Equations

Quadratic equations frequently appear in algebra and are critical when solving systems that involve squared terms. The general form of a quadratic equation is \(ax^2 + bx + c = 0\), which can be solved by various techniques, such as factoring, completing the square, or using the quadratic formula.

In this specific problem, after setting up the condition \(\|\overrightarrow{AC}\| = 2\|\overrightarrow{BC}\|\), it leads to a quadratic equation in terms of \(t\): \(3t^2 - 8t + 4 = 0\). By using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), and identifying \(a=3\), \(b=-8\), and \(c=4\), the roots \(t=2\) and \(t=\frac{2}{3}\) are found.

These solutions for \(t\) help find the exact points \(C\) on the line, namely \((3, 1, 0)\) and \(\left( \frac{5}{3}, \frac{-1}{3}, \frac{4}{3} \right)\). Thus, understanding how to manipulate and solve quadratic equations is vital for successfully resolving this task.