Brennan Flores

2021-01-27

If (v-ku) is orthogonal to v, then what is k??

$u=\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]\text{and}v=\left[\begin{array}{c}2\\ -1\\ 2\end{array}\right]$

wheezym

Skilled2021-01-28Added 103 answers

Step 1

We have$u=\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]\text{and}v=\left[\begin{array}{c}2\\ -1\\ 2\end{array}\right]$ 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$\left[\begin{array}{c}2-k\\ -1-k\\ 2-k\end{array}\right]$ and according to the given problem ,this is orthogonal to

$v=\left[\begin{array}{c}2\\ -1\\ 2\end{array}\right]$

So,$\left[\begin{array}{c}2-k\\ -1-k\\ 2-k\end{array}\right]\left[\begin{array}{c}2\\ -1\\ 2\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

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.

We have

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

Now we know that two vectors are orthogonal if their dot product is 0.

Now (v-ku) gives the vector

So,

Since this is simply a vector dot product ,so we can write :

Which is required value of k and for this value of k (v-ku) is orthogonal to v.

Jeffrey Jordon

Expert2022-01-24Added 2607 answers

Answer is given below (on video)

Jeffrey Jordon

Expert2022-08-23Added 2607 answers

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