Show that the xy plane W=(x,y,0) in R3 is generated by (i) u=[120] and v=[010] (ii) u=[2−10] and v=[130]

Check the solution to "Show that the xy plane W=(x,y,0) in R3..."

College
Algebra
Linear algebra
Vectors and spaces

Linda Seales

Answered question

2022-01-05

Show that the xy plane

W = (x, y, 0)

R^{3}

is generated by
(i)

u = [\begin{matrix} 1 \\ 2 \\ 0 \end{matrix}]

and

v = [\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}]

(ii)

u = [\begin{matrix} 2 \\ - 1 \\ 0 \end{matrix}]

and

v = [\begin{matrix} 1 \\ 3 \\ 0 \end{matrix}]

Answer & Explanation

Louis Page

Beginner2022-01-06Added 34 answers

Let $(x, y, 0)$ be any element from W. For a and b in $R$ , we have
$(x, y, 0) = a (1, 2, 0) + b (0, 1, 0)$
$= (a, 2 a + b, 0)$
Comparing both side we get
$x = a$ and $y = 2 a + b$
Now from $y = 2 a + b$
$b = y - 2 a$
$= - y - 2 x$
Therfore, for any $(x, y, 0)$ in W, we can write
$(x, y, 0) = x (1, 2, 0) + (y - 2 x) (0, 1, 0)$
$\Rightarrow (x, y, 0) {(1, 2, 0), (0, 1, 0)}$
Therefore, W is generated by u and v.
Let $(x, y, 0)$ be any element from W. For a and b in $R$ , we have
$(x, y, 0) = a (2, - 1, 0) + b (1, 3, 0)$
$= (2 a + b, - a + 3 b, 0)$
Comparing both side we get
$x = 2 a + b$ ...(1)
and
$y = - a + 3 b$ ...(2)
Now we solve equationn (1) and (2) to find the value of a and b. From (1) and (2), we get
$x + 2 y = 2 a + b - 2 a + 6 b = 7 b$
$\Rightarrow b = \frac{x + 2 y}{7}$
Now from (2)
$a = y + 3 b$
$= y + 3 (\frac{x + 2 y}{7})$
$= y + \frac{3 x + 6 y}{7}$
$= \frac{7 y + 3 x + 6 y}{7}$
$= \frac{3 x + 13 y}{7}$
$\Rightarrow a = \frac{3 x + 13 y}{7}$
Therfore, for any $(x, y, 0)$ in W, we can write
$(x, y, 0) = a (2, - 1, 0) + b (1, 3, 0)$
$= \frac{3 x + 13 y}{7} (2, - 1, 0) + \frac{x + 2 y}{7} (1, 3, 0)$
$\Rightarrow (x, y, 0) {(2, - 1, 0), (1, 3, 0)}$
Therefore, W is generated by u and v.

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Linear algebra

Didn't find what you were looking for?