The vector $\begin{pmatrix} k \\ 2 \end{pmatrix}$ is orthogonal to the vector $\begin{pmatrix} 3 \\ 5 \end{pmatrix}$. Find $k$.
I thought the answer should be 10/4/
You can't enter graphics here, sorry. I can't read what you intended.
To find the value of $k$ such that the vector $\begin{pmatrix} k \\ 2 \end{pmatrix}$ is orthogonal to the vector $\begin{pmatrix} 3 \\ 5 \end{pmatrix}$, we need to use the dot product.
The dot product of two vectors $\mathbf{u} = \begin{pmatrix} u_1 \\ u_2 \end{pmatrix}$ and $\mathbf{v} = \begin{pmatrix} v_1 \\ v_2 \end{pmatrix}$ is given by:
$\mathbf{u}\cdot\mathbf{v} = u_1v_1 + u_2v_2$.
In this case, we have:
$\begin{pmatrix} k \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ 5 \end{pmatrix} = k(3) + (2)(5) = 3k + 10$.
For the vectors to be orthogonal, their dot product must be zero. Therefore, we can set up the equation:
$3k + 10 = 0$.
To solve for $k$, we can subtract 10 from both sides:
$3k = -10$.
Lastly, we divide both sides by 3 to isolate $k$:
$k = -\frac{10}{3}$.
Therefore, the value of $k$ that makes the vectors orthogonal is $-\frac{10}{3}$.
To determine when two vectors are orthogonal, we can use the dot product. Two vectors $\vec{u} = \begin{pmatrix} u_1 \\ u_2 \end{pmatrix}$ and $\vec{v} = \begin{pmatrix} v_1 \\ v_2 \end{pmatrix}$ are orthogonal if their dot product is equal to zero, i.e., $\vec{u} \cdot \vec{v} = 0$.
In this case, we have the vector $\begin{pmatrix} k \\ 2 \end{pmatrix}$ as one vector, and the vector $\begin{pmatrix} 3 \\ 5 \end{pmatrix}$ as the other. We need to find the value of $k$ such that these two vectors are orthogonal.
Using the dot product, we have:
$\begin{pmatrix} k \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ 5 \end{pmatrix} = k(3) + (2)(5) = 3k + 10$
For these vectors to be orthogonal, the dot product should be equal to zero. Therefore, $3k + 10 = 0$.
Solving this equation for $k$, we get:
$3k + 10 = 0$
$3k = -10$
$k = -\dfrac{10}{3}$
Hence, the value of $k$ that makes the vectors orthogonal is $k = -\dfrac{10}{3}$.