I am battling to get my mind around some of the concepts involving vectors in $3$-space. This question asks me whether the following statements are True or False:
(A) The line $(a; b; c)$ is parallel to $V$
(B) The distance of the plane from the origin is $|d|$
For (A) I don't think I have enough to work with here. I know that if a line is parallel to a plane then it's direction vector must be perpendicular to the planes normal vector… I honestly do not know how to go about proving this and perhaps I am over thinking this.
For (B) I have said this is false as the distance between the plane from the origin is not $|d|$ and is given by the formula:
$$d=\frac{|Ax+By+Cz+D|}{|A^2+B^2+C^2|}$$
I would really appreciate any input on this.
Thanks!
Best Answer
For part (A) your reasoning is correct. A line with direction vector $\mathbf{v}$ is parallel to a plane with normal vector $\mathbf{n}$ if and only if $\mathbf{v}$ is orthogonal to $\mathbf{n}$, that is, if and only if $\mathbf{v}\cdot\mathbf{n} = 0$. In this case the direction vector of the line is the same as the normal vector of the plane and so they cannot be parallel since $\mathbf{v}\cdot\mathbf{n} = \mathbf{n}\cdot\mathbf{n} = 1 \neq 0$.
For part (B) you may use the following argument. Let $\mathbf{u}$ be any point in the plane. Then the distance from the origin to $\mathbf{u}$ is $\|\mathbf{u}\|$. Your goal is to find the minimum value of $\|\mathbf{u}\|$. By the Cauchy-Schwarz inequality we have that
$$ |d| = |\mathbf{n}\cdot\mathbf{u}| \leq \|\mathbf{n}\|\cdot\|\mathbf{u}\| = \|\mathbf{u}\| $$
since $\mathbf{n}$ is a unit vector. Therefore the minimum distance of the plane from the origin is bounded below by $|d|$. The last step is to show that there exists a vector in the plane whose norm is $|d|$. In this case such a vector is given by $-d\mathbf{n}$ since
$$ \mathbf{n}\cdot(-d\mathbf{n}) + d = -d\mathbf{n}\cdot\mathbf{n} +d = -d +d = 0 $$
and $\|-d\mathbf{n}\| = |d|\|\mathbf{n}\| = |d|$.