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}

asked 2021-01-27
If (v-ku) is orthogonal to v, then what is k??
\(u=\begin{bmatrix}1 \\1\\1 \end{bmatrix} \text{ and } v=\begin{bmatrix}2 \\-1\\2 \end{bmatrix}\)

Answers (1)

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 :
\(2\times (2-k)-1\times(-1-k)+2\times(2-k)=0\)
Which is required value of k and for this value of k (v-ku) is orthogonal to v.
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