find equation of the line containing the origin and is perpendicular to the another line So, I'm given a line with parametric equations


*

*$x = 2t + 3$

*$y = 1 - t$

*$z = 5t$


and I'm supposed to find  symmetric and parametric equations of another line containing the origin and is perpendicular to the given line
i think a way to answer this would be to find a plane containing the given line because the normal vector can be used as a parallel vector of the missing line but i don't know how to get that normal vector. tips and help would be highly appreciated :)
 A: the line $$l:(3, 1, 0) + t(2, -1, 5)$$ is through $(3, 1, 0)$ and parallel to $n = (2, -1, 5)$ therefore a plane through the origin and orthogonal to this line is $$2x - y + 5z = 0$$  the line $l$ cuts this plane at the $t$ value given by $$2 \times 3 - 1 + t(2 \times 2 + 1+ 5 \times 5)= 0 \to t = -\frac 16$$ the line you are looking for is $$6s(8/3,7/6,-5/6)=s(16, 7,-5),\text{ where  $s$ is real number.}$$
A: one approach is to find the value of $t$ which minimizes the quadratic expression:
$$
(2t+3)^2 + (1-t)^2 + (5t)^2
$$
this gives the parameter of the point $T$ on the line which is closest to the origin $O$.
if the co-ordinates of $T$ are $(a,b,c)$ then a parametric form of the line required is:
$$
x=as \\
y=bs \\
z=cs
$$
A: Let's call the given line $L$, and let's take a line $M$ that passes through the origin and the point $(2t+3,1-t,5t)$ on $L$. We're going to adjust the value of $t$ to make $M$ perpendicular to $L$.
The direction vector of $L$ is $(2,-1,5)$, and the direction vector of $M$ is $(2t+3,1-t,5t)$. So, in order for $M$ to be perpendicular to $L$, we need
$$
(2t+3,1-t,5t) \cdot (2,-1,5) = 0
$$
This gives $30t+5=0$, so $t=-\tfrac16$. Putting $t=-\tfrac16$ in the equation of $L$, we get the point $P=\tfrac16(16,7,-5)$. The desired line $M$ passes through the origin and $P$, so its equation is $M(s) = \tfrac{s}{6}(16,7,-5)$.
You can confirm that $L$ and $M$ are perpendicular, because
$$
(2,-1,5) \cdot (16,7,-5) = 0
$$
A: There is a straightforward calculation using vector operations in $\mathbb R^3$,
specifically the "dot product" (aka inner product):
$$u \cdot v = 
\begin{pmatrix}u_1\\u_2\\u_3\end{pmatrix} \cdot \begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix}
= u_1 v_1 + u_2 v_2 + u_3 v_3.
$$
Suppose we have two vectors, $u$ and $v$.
The idea is to find two components of $u$ with respect to $v$. 
The first component is the parallel projection of $u$ onto $v$,
which we can write as
$$u_\parallel = \frac{u \cdot v}{v \cdot v} \, v.$$
The second component is the component of $u$ perpendicular to $v$:
$$u_\perp = u - u_\parallel = u - \frac{u \cdot v}{v \cdot v} \, v.$$
To verify that $u_\perp$ really is perpendicular to $v$, observe that
$$u_\perp \cdot v = \left(u - \frac{u \cdot v}{v \cdot v} \, v \right) \cdot v
= u \cdot v - \frac{u \cdot v}{v \cdot v} \, v \cdot v
 = u \cdot v - u \cdot v = 0.$$
Applying this to  your problem, let
$$u = \begin{pmatrix}3\\1\\0\end{pmatrix} \quad \mbox{and}
\quad v = \begin{pmatrix}2\\-1\\5\end{pmatrix}.$$
The line $su$ through the origin intersects your given line, $u + tv,$ at $u$.
Two intersecting lines define a plane. The plane in this particular case
contains (for example) $u$ and $u+v$, so it also contains $v$ and
any linear combination of $u$ and $v$.
One such linear combination is $u_\perp$.
So the solution is to set vectors $u$ and $v$ so your given line is
parameterized by $u + tv$, and compute 
$u_\perp = u - \frac{u \cdot v}{v \cdot v} \, v.$
The result, $u_\perp,$ lies in the plane containing the origin and your given line,
and it is perpendicular to that line.
