Question # If (v-ku) is orthogonal to v, then what is k?? u=begin{bmatrix}1 11 end{bmatrix} text{ and } v=begin{bmatrix}2 -12 end{bmatrix}

Matrices
ANSWERED If (v-ku) is orthogonal to v, then what is k??
$$u=\begin{bmatrix}1 \\1\\1 \end{bmatrix} \text{ and } v=\begin{bmatrix}2 \\-1\\2 \end{bmatrix}$$ 2021-01-28
Step 1
We have $$u=\begin{bmatrix}1 \\1\\1 \end{bmatrix} \text{ and } v=\begin{bmatrix}2 \\-1\\2 \end{bmatrix}$$ are two column matrices .
We have to find the value of k so that (v-ku) is orthogonal to v.
Step 2
First observe that we are given two column matrices u and v. We know that column matrices simply represents a vector in the said space(here $$\mathbb{R}^3$$) .So here u and v are vector in $$\mathbb{R}^3$$.
Now we know that two vectors are orthogonal if their dot product is 0.
Now (v-ku) gives the vector $$\begin{bmatrix}2-k \\-1-k\\2-k \end{bmatrix}$$ and according to the given problem ,this is orthogonal to
$$v=\begin{bmatrix}2 \\-1\\2 \end{bmatrix}$$
So, $$\begin{bmatrix}2-k \\-1-k\\2-k \end{bmatrix}\begin{bmatrix}2 \\-1\\2 \end{bmatrix}=\begin{bmatrix}0 \\0\\0 \end{bmatrix}$$
Since this is simply a vector dot product ,so we can write :
$$2\times (2-k)-1\times(-1-k)+2\times(2-k)=0$$
$$4-2k+1+k+4-2k=0$$
$$-3k+9=0$$
$$-3k=-9$$
$$3k=9$$
$$k=3$$
Which is required value of k and for this value of k (v-ku) is orthogonal to v.