Find $y$ so that the vectors $\begin{pmatrix} 1 \\ -3 \\ -4 \end{pmatrix}$ and $\begin{pmatrix} -2 \\ y \\ -1 \end{pmatrix}$ are orthogonal.
For the vectors $\begin{pmatrix} 1 \\ -3 \\ -4 \end{pmatrix}$ and $\begin{pmatrix} -2 \\ y \\ -1 \end{pmatrix}$ to be orthogonal, their dot product should be 0:
\[(1)(-2) + (-3)(y) + (-4)(-1) = 0.\]Solving, we find $y = \boxed{\frac{2}{3}}.$
\frac{2}{3}
Level 2
Precalculus
Problem: Find $y$ so that the vectors $\begin{pmatrix} 1 \\ -3 \\ -4 \end{pmatrix}$ and $\begin{pmatrix} -2 \\ y \\ -1 \end{pmatrix}$ are orthogonal.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To determine the value of $y$ that makes the vectors orthogonal, we need to use the property of orthogonality, which states that the dot product of two orthogonal vectors is zero. 

Let the first vector be $\mathbf{a} = \begin{pmatrix} 1 \\ -3 \\ -4 \end{pmatrix}$ and the second vector be $\mathbf{b} = \begin{pmatrix} -2 \\ y \\ -1 \end{pmatrix}$. The dot product $\mathbf{a} \cdot \mathbf{b}$ is calculated as follows:

\[
\mathbf{a} \cdot \mathbf{b} = 1 \cdot (-2) + (-3) \cdot y + (-4) \cdot (-1)
\]

Calculating each term gives:

1. The first term: $1 \cdot (-2) = -2$.
2. The second term: $(-3) \cdot y = -3y$.
3. The third term: $(-4) \cdot (-1) = 4$.

Combining these results, we have:

\[
\mathbf{a} \cdot \mathbf{b} = -2 - 3y + 4
\]

Simplifying this expression results in:

\[
\mathbf{a} \cdot \mathbf{b} = 2 - 3y
\]

For the vectors to be orthogonal, we set the dot product equal to zero:

\[
2 - 3y = 0
\]

To solve for $y$, we isolate $y$:

\[
-3y = -2
\]

Dividing both sides by -3 gives:

\[
y = \frac{2}{3}
\]

Thus, the value of $y$ that ensures the vectors are orthogonal is $\frac{2}{3}$.<|END_OF_LATENT|> The final answer is $\frac{2}{3}$.