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Linear Algebra: Orthogonal Columns
Let
M be

1
3
1

2
-2

a

1
1

b

and tet
a and
b be such that the columns of
M are orthogonat. Compute
a
Pick ONE option

-1//2

-1

1//2
2
Ciear Selection

$7$ . Linear Algebra: Orthogonal Columns $\newline$ Let $M$ be $\newline$ \begin{tabular}{|l|l|l|} $\newline$ \hline $1$ & $3$ & $1$ \\ $\newline$ \hline $2$ & $-2$ & $a$ \\ $\newline$ \hline $1$ & $1$ & $b$ \\ $\newline$ \hline $\newline$ \end{tabular} $\newline$ and tet $\mathrm{a}$ and $\mathrm{b}$ be such that the columns of $\mathrm{M}$ are orthogonat. Compute $\mathrm{a}$ $\newline$ Pick ONE option $\newline$ $-1 / 2$ $\newline$ $-1$ $\newline$ $1 / 2$ $\newline$ $2$ $\newline$ Ciear Selection

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Find the magnitude of a vector

Full solution

Q.

7

. Linear Algebra: Orthogonal Columns

\newline

Let

M

be

\newline

\begin{tabular}{|l|l|l|}

\newline

\hline

1

&

3

&

1

\\

\newline

\hline

2

&

-2

&

a

\\

\newline

\hline

1

&

1

&

b

\\

\newline

\hline

\newline

\end{tabular}

\newline

and tet

\mathrm{a}

and

\mathrm{b}

be such that the columns of

\mathrm{M}

are orthogonat. Compute

\mathrm{a}

\newline

Pick ONE option

\newline

-1 / 2

\newline

-1

\newline

1 / 2

\newline

2

\newline

Ciear Selection

Determine value of $a$ : To determine the value of $a$ , we need to ensure that the dot product of the first and second columns of matrix $M$ is zero, since orthogonal columns have a dot product of zero. $\newline$ The first column of $M$ is $[1, 2, 1]$ and the second column is $[3, -2, a]$ . $\newline$ The dot product of these two columns is calculated as follows: $\newline$ $(1 \times 3) + (2 \times -2) + (1 \times a) = 0$ $\newline$ $3 - 4 + a = 0$ $\newline$ $a = 1$
Calculate dot product: Now, we need to check if the third column $[1, b]$ is orthogonal to the other two columns. Since we only have a value for $a$ and not for $b$ , we can only check the orthogonality with respect to $a$ . The dot product of the first and third columns is: $(1 \times 1) + (2 \times b) + (1 \times 1) = 0$ $1 + 2b + 1 = 0$ $2b = -2$ $b = -1$
Check orthogonality: However, we made a mistake in the previous step. We were only asked to compute $a$ , not $b$ . Therefore, we should not have calculated the value of $b$ . We will disregard the previous step's calculation for $b$ and focus on the value of $a$ that we found.

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