Question

Determine the values of \(x\) and \(y\) such that the points \((1,2,3),(4,7,1),\) and \((x, y, 2)\) are collinear (lie on a line).

Short answer

Answer: The points will be collinear when x = 5/2 and y = 9/2.

Step-by-step solution

Find the vectors between the points

First, find the vectors from point \((1,2,3)\) to the other two given points. The vectors are \(\overrightarrow{AB} = \begin{pmatrix}4-1\\7-2\\1-3\end{pmatrix} = \begin{pmatrix}3\\5\\-2\end{pmatrix}\) and \(\overrightarrow{AC} = \begin{pmatrix}x-1\\y-2\\2-3\end{pmatrix} = \begin{pmatrix}x-1\\y-2\\-1\end{pmatrix}\).

Check if the vectors are parallel

We need to see if there exists a scalar \(k\) such that \(\overrightarrow{AB} = k \cdot \overrightarrow{AC}\). This would imply that: \(\begin{pmatrix}3\\5\\-2\end{pmatrix} = k\begin{pmatrix}x-1\\y-2\\-1\end{pmatrix}\).

Find the scalar

Now we need to find the scalar \(k\) by setting up and solving equations from the components of both vectors: - For the x-components: \(3 = k(x-1)\) - For the y-components: \(5 = k(y-2)\) - For the z-components: \(-2 = k(-1)\)

Solve the scalar equation for z-components

By solving the equation for the z-components, we obtain the value of \(k\): \(-2 = k(-1) \Rightarrow k = 2\).

Substitute the scalar value in the other equations and solve for x and y

Now, substitute the value of \(k\) into the other two equations: - For the x-components: \(3 = 2(x-1) \Rightarrow x= 3/2 + 1 \Rightarrow x=5/2\) - For the y-components: \(5 = 2(y-2) \Rightarrow y = 5/2 + 2 \Rightarrow y= 9/2\) Therefore, the values of \(x\) and \(y\) that make the points collinear are \(x = \frac{5}{2}\) and \(y = \frac{9}{2}\).