Find all unit vectors in the plane determined by $u = (3, 0, 1)$ and $v= (1, -1, 1)$ that are orthogonal to the vector $w= (1, 2, 0)$.
Answer: $± \frac{1}{\sqrt {61}} (6, -3, 4)$
Can someone explain the process for me? I understand where the former part $± \frac{1}{\sqrt {61}}$ is from, but I don't understand how you get $(6, -3, 4)$.
Best Answer
This is an answer based on Dhanvi's comment.
Any vector in the plane can be represented as a linear combination of $u$ and $v$. This means that for arbitrary reals $a, b$, the vector $a (3,0,1) + b (1, -1, 1) = (3a+b, -b, a+b)$ is in the plane.
The dot product with $w$ is $(3a+b) \cdot 1 + (-b) \cdot 2 + (a+b) \cdot 0 = 3a-b$. From the definition of the dot product, $3a - b =0$.
Since the vector must be a unit vector, $(3a+b)^2 + (-b)^2 + (a+b)^2 = 1$ (why?). Solve this system of equations to find $a$ and $b$:
$$3a-b = 0 \tag{1}$$ $$(3a+b)^2 + (-b)^2 + (a+b)^2 = 1 \tag{2}$$
which does give $a = ±\frac{1}{\sqrt{61}}$, $b = ±\frac{3}{\sqrt{61}}$. Substituting $a$ and $b$ and taking out a factor of $±\frac{1}{\sqrt{61}}$ gives the book's answer.