MHB Cartesian Equation of the Plane Passing Through Two Vectors

[Guest2]

Could someone please help with me with part $(b)$? By "the other set of vectors" they mean $R$, and the linear combination is

$(1,2,9) = -3(1,0,-1)+2(2,1,3)$.​

https://i.imgsafe.org/b80212f.png

 

[I like Serena]

Could someone please help with me with part $(b)$? By "the other set of vectors" they mean $R$, and the linear combination is

$(1,2,9) = -3(1,0,-1)+2(2,1,3)$.​

Hi Guest,

If we write the 3rd vector as a linear combination of the first two, we have:
$$(1,2,9) = a(1,0,-1)+b(2,1,3)$$
Can you solve it? (Wondering)

 

[Guest2]

Hi Guest,

If we write the 3rd vector as a linear combination of the first two, we have:
$$(1,2,9) = a(1,0,-1)+b(2,1,3)$$
Can you solve it? (Wondering)

So $(1,2,9) = (a, 0,-a)+(2b, b, 3b) = (a+2b, b, 3b-a) \implies b = 2 \implies a = -3. $

 

[Guest2]

Woah, I've solved it I think (Happy)

Since $v_1 = (1,0,-1)$ and $v_2 = (2,1,3)$ span the plane, we can write any point in the plane as:

$\begin{align*}
\begin{pmatrix}x\\y\\z\end{pmatrix}=tv_1+rv_2=t\begin{pmatrix}1\\0\\-1\end{pmatrix}+r\begin{pmatrix}2\\1\\3\end{pmatrix}
=\begin{pmatrix}1&2\\0&1\\-1&3\end{pmatrix}\begin{pmatrix} t \\ r\end{pmatrix}
\end{align*}$, call this (1). Now, row reducing:

$\begin{pmatrix}1&2&|&x\\0&1&|&y\\-1&3&|&z\end{pmatrix} \to \begin{pmatrix}1&2&|&x\\0&1&|&y\\0&5&|&x+z\end{pmatrix} \to \begin{pmatrix}1&0&|&x-2y\\0&1&|&y\\0&5&|&x+z\end{pmatrix} \to \begin{pmatrix}1&0&|&x-2y\\0&1&|&y\\0&0&|&x-5y+z\end{pmatrix}$

We find that (1) has a solution if and only if $x-5y+z = 0$, and this is the equation for the plane $\Pi$.