Question

An orthogonal set of vectors having a span that contains P, Q, R is , where P = (-10,-1,3)', Q = (-2,-5,9)' & R = (2,-7,12)' are three vectors.

• A. $\left[\begin{array}{c}-6\\ -3\\ 6\end{array}\right]\left[\begin{array}{c}4\\ -2\\ 3\end{array}\right]$
• B. $\left[\begin{array}{c}-4\\ 2\\ 4\end{array}\right]\left[\begin{array}{c}5\\ 7\\ -11\end{array}\right]\left[\begin{array}{c}6\\ 2\\ -3\end{array}\right]$
• C. $\left[\begin{array}{c}6\\ 7\\ -1\end{array}\right]\left[\begin{array}{c}-3\\ 2\\ -2\end{array}\right]\left[\begin{array}{c}3\\ 9\\ -4\end{array}\right]$
• D. $\left[\begin{array}{c}4\\ 3\\ 11\end{array}\right]\left[\begin{array}{c}1\\ 31\\ 3\end{array}\right]\left[\begin{array}{c}5\\ 3\\ 4\end{array}\right]$

Answer 1

The correct option is A $\left[\begin{array}{c}-6\\ -3\\ 6\end{array}\right]\left[\begin{array}{c}4\\ -2\\ 3\end{array}\right]$

We know that two vectors $(a_1,b_1,c_1)$ & $(a_2,b_2,c_2)$ are said to be orthogonal if $a_1{a}_2+b_1{b}_2+c_1{c}_2=0$ ...(i)
We are looking for an orthogonal set of vector having a span that contain P,Q & R

Take option (a)

$\left[\begin{array}{c}-6\\ -3\\ 6\end{array}\right]$ & $\left[\begin{array}{c}4\\ -2\\ 3\end{array}\right]$ = -6 $\times$ 4 + (-3) $\times$ (-2) + 6(3)

= -24 + 6 + 18 = 0

$\Rightarrow \left[\begin{array}{c}-6\\ -3\\ 6\end{array}\right]$ & $\left[\begin{array}{c}4\\ -2\\ -3\end{array}\right]$ are orthogonal.

The space spanned by these vectors is

$K_1\left[\begin{array}{c}-6\\ -3\\ 6\end{array}\right]+K_2\left[\begin{array}{c}4\\ -2\\ 3\end{array}\right]$

The span of $\left[\begin{array}{c}-6\\ -3\\ 6\end{array}\right]$ & $\left[\begin{array}{c}4\\ -2\\ 3\end{array}\right]$ contains P,Q & R. So, P.Q,R can be represented as linear combination of these two vectors by solving $K_1$ & $K_2$.

Option (b), (c) & (d) are not orthogonal.