# Find the line perpendicular to these two vectors

A line goes through point $$A = (9, -7, 31)$$ and the line is perpendicular to vectors $$[7, 1, 2]$$ and $$[3, 0, 1]$$.

What is the equation of the line?

The cross product of $$[7, 1, 2]$$ and $$[3, 0, 1]$$ is $$[1, -1, -3]$$.

I believe the line can be written as $$x = [9, -7, 31] + t [1, -1, -3]$$, where $$t$$ is a real number.

I'm somewhat skeptical of my own work. If you find the cross product (which will be perpendicular to the two given vectors); why does that mean it is automatically parallel with the line that is also perpendicular to these two vectors?

Given the vectors are in three-dimensional space; is it possible to produce a cross product of these two vectors and that vector not to be parallel with the line I found?

• The answer to your last question is "no." Two vectors determine a plane. There is only one direction normal to that plane and the cross product gives it. – B. Goddard Jan 18 at 13:00
• The people first think how to formulate a vector to be perpendicular to other two vectors in 3-dimensional space, then they invented the cross product. – kelvin hong 方 Jan 18 at 13:05
• I believe two non-parallel vectors make a plane. That's true. That makes it easy to visualise. The cross product makes a vector that pierces the plane at right angles from above and below said plane. That's my understanding. – Madison Jan 18 at 13:15
• I would slightly amend @B.Goddard's assertion to say that there are two opposite directions that are normal (perpendicular) to the plane, and if your vectors are $u$ and $v$, then these two directions are given by $u \times v$ and $v \times u$. – John Hughes Jan 18 at 13:20
• Yes, above and below the plane. I believe. The opposite direction being the anti-commutative. – Madison Jan 18 at 13:27

## 2 Answers

When you find a cross product,

$$A=\begin{bmatrix}a&b&c\\7&1&2\\ 3& 0& 1\end{bmatrix}$$

you are finding a linearly independent vector, namely $$\vec{v}=(a,b,c)$$, to the other two vectors, namely $$\vec{u}=(7,1,2),\vec{w}=(3,0,1)$$ (and it is like that because you are finding a vector which gives you $$\det(A)\neq 0$$, hence $$\dim(A)=3$$, hence $$\vec{u},\vec{v},\vec{w}$$ generate $$\mathbb{R}^3$$).

That means that you will have a basis of the $$3$$-dimensional space (since you have $$3$$ linearly independent vectors). An that is why if you have a fourth vector, namely $$\vec{\alpha}$$ which is perpendicular to the first two vectors, it has to be proportional to the one you have find with the cross product.

Hence, what you have done is OK.

• Okay, this is good. I was basically taught to just follow that it's true; but, it was lacking intuition. Soon to study linear independence. – Madison Jan 18 at 13:20

Consider the plane containing the point $$A=(9,-7,31)$$ and perpendicular to the vector $$[7,1,2]$$. Its equation is $$7x + 1y + 2z = 7(9) + 1(-7) + 2(31)$$, which simplifies to $$7x + y + 2z = 118$$.

Similarly the plane containing the point $$A=(9,-7,31)$$ and perpendicular to the vector $$[3,0,1]$$ has equation $$3x+z=58$$.

The intersection of these two planes will be the line that contains the point $$A$$ and is perpendicular to both vectors. It is not hard to check that the line $$\ell(t) = [9+t, -7-t, 31-3t]$$ is that line:

$$7x + y + 2z = 7(9+t)+(-7-t)+2(31-3t) = 118$$

and $$3x + z = 3(9+t)+(31-3t) = 58$$