1. ## Subspace

Again not sure what to do here.

Subspace of R3 spanned by the vectors u1 = (1,1,1) and u2 = (2, 0, -1) is a plane passing through the origin. Express w = (1, 2, 3) in the form w = w1 + w2 where w1 lies in the plane and w2 is perpendicular to the plane.

2. Hi
Again not sure what to do here.

Subspace of R3 spanned by the vectors u1 = (1,1,1) and u2 = (2, 0, -1) is a plane passing through the origin. Express w = (1, 2, 3) in the form w = w1 + w2 where w1 lies in the plane and w2 is perpendicular to the plane.
Let $u_3=u_1\times u_2$ and let $P$ be the plane spanned by $u_1$ and $u_2$ which passes through the origin.

$(u_1,u_2,u_3)$ is a basis of $\mathbb{R}^3$ hence $w=\alpha u_1+\beta u_2+\gamma u_3$ for some real numbers $\alpha,\,\beta$ and $\gamma$. As $\alpha u_1+\beta u_2\in P$ and $\gamma u_3 \perp P$, one can also write $w=w_1+w_2$ with $\begin{cases}w_1=\alpha u_1+\beta u_2, & w_1\in P \\w_2=\gamma u_3, & w_2\perp P\end{cases}$

Hence solving this problem boils down to finding the values of the three unknowns $\alpha,\,\beta$ and $\gamma$.

Does it help ?

3. The normal to the given plane is $N = u_2 \times u_1$
Now you want $w = w_1 + tN \Rightarrow \quad w_1 = w - tN\,\& \,w_1 \cdot N = 0$.
The last bit is to insure $w_1$ is in the plane.
Now solve for $t$.

4. Hi. I did it this way, which is a bit tedious but at least I understand what I'm doing.

First find the equation of the plane. Any point (x,y,z) on it is of the form $au_1+bu_2$, $a,b\in\mathbb{R}$. So

$x=a+2b$
$y=a$
$z=a-b$

whence eliminating a and b gives $x-3y+2z=0$.

Now let $w=w_1+w_2$ where $w=\left(\begin{array}{r}1\\2\\3\end{array}\right)$ and $w_1=\left(\begin{array}{c}3y-2z\\y\\z\end{array}\right)$.

Note that $u_1$ and $u_2$ lie in the plane. So take the dot product of w with $u_1$ and $u_2$, noting that $u_1\cdot w_2=u_2\cdot w_2=0$:

$w\cdot u_1=w_1\cdot u_1\ \implies\ 6=4y-z$
$w\cdot u_2=w_1\cdot u_2\ \implies\ -1=6y-5z$

Solving gives $y=\frac{31}{14}$, $z=\frac{40}{14}$

Hence

$\left(\begin{array}{r}1\\2\\3\end{array}\right)\ =\ \left(\begin{array}{r}\frac{13}{14}\\\\\frac{31}{1 4}\\\\\frac{40}{14}\end{array}\right)+\left(\begin {array}{r}\frac{1}{14}\\\\-\frac{3}{14}\\\\\frac{2}{14}\end{array}\right)$

,

,

### Let R3 have the Euclidean inner product. The subspace of spanned by the vectors and is aplane passing through the origin. Express in the form , where lies in the plane and is perpendicular to the plane

Click on a term to search for related topics.