Find a vector orthogonal to other two given and ends at a plane

Question

I am reviewing Calculus III using Mahavier, W. Ted's material and get stuck on one question in chapter 1. Here is the problem:

Assume $\vec{u},\vec{v}\in \mathbb{R}^3$. Find a vector $\vec{x}=(x,y,z)$ so that $\vec{x}\perp\vec{u}$ and $\vec{x}\perp\vec{v}$ and $x+y+z=1$.

My attempt: From the last condition, I know that $\vec{x}$ ends at the plane intersecting the $x-,y-,z-$axis at $(1,0,0),(0,1,0)$ and $(0,0,1)$. From the orthogonal conditions, $\vec{x}$ is perpendicular to the plane formed by $\vec{u},\vec{v}$ if they are distinct, otherwise, any plane that contains $\vec{u},\vec{v}$.

Am I on the right track? And how do I go from here? Thanks!

Edit: Thanks for all who responded! I do remember cross product. However, at this point of the book, the definition of cross product has not been introduced yet. I wonder whether there are other means to attack this problem without invoking a to-be-introduced concept?

Thanks again!

Answer 1

Hint: compute the cross-product of $\vec{u}$ and $\vec{v}$ and normalize the result (assuming that they are not collinear).

Answer 2

You get three equations in the three unknowns $x$, $y$, and $z$ from $x+y+z=1$, $\vec{x}\cdot\vec{u} = 0$, and $\vec{x}\cdot\vec{v} = 0$. If these three equations have a solution, that is the vector you are looking for. However, they do not always have a solution (for example, try $\vec{u} = (1,1,1)$ and $\vec{v} = (1,0,1)$). There are other cases where they will not have a unique solution (where $\vec{u}$ and $\vec{v}$ are collinear).

Answer 3

Do you remember the definition of the cross product? Given vectors $u = (u_1,u_2, u_3)$ and $v = (v_1,v_2,v_3)$ define $u\times v$ as the unique vector such that

$$(u\times v)\cdot a = \left| \matrix{a_1&a_2&a_3\\u_1&u_2&u_3\\v_1&v_2&v_3}\right|$$

where $a=(a_1,a_2,a_3)$, the dot represents the dot product and the expression on the right is the determinant of that matrix.

From the properties of the determinant it's easy to see that $(u\times v)\cdot u = (u\times v)\cdot v = 0$, hence the cross product is orthogonal to both vectors.

From here we have to assume that $u$ and $v$ are linearly independent. The magnitude of the cross product is $|u||v|\sin(\theta)$ where $\theta$ is the angle between $u$ and $v$. So if $\theta = 0$ then the cross product is $0$. In order to get $x+y+z = 1$ simply divide $u\times v$ by $z_1+z_2+z_3$ where $u\times v = (z_1,z_2,z_3)$.