Vectors: Find all values of a and b such that $(1,2,a)$ and $(2,(1-b),3)$ are perpendicular

Question

I know that, for the vectors to be perpendicular, their cross product must equal $0$. i.e. $w = u \times v$, where $u = (1,2,a)$ and $v = (2,(1-b),3)$

From that, I got $w = ( (6-a+ab), (-3+2a), (-b-3) )$

When checking, $w \cdot u$ and $w \cdot v$ must equal $0$

This does occur, but I am no closer to finding what $a$ and $b$ are

Answer 1

Note that we can write $$u \cdot v = |u| |v| \cos \theta$$ where $\theta$ is the angle between $u$ and $v$ (where $u,v$ are vectors). So, if we want the vectors to be perpendicular, we want $\theta = \pi/2$ so that the RHS is zero. In particular, two vectors are perpendicular if $u \cdot v = 0$. For $u = (1,2,a)$ and $v = (2,(1-b),3)$ we have $$ u \cdot v = 2 + 2(1-b) + 3a$$ so that, setting this equal to zero, $$4 - 2b + 3a = 0 \implies 3a -2b = -4$$ for them to be perpendicular.

Answer 2

Scalar not cross/Vector product has to be zero for orthogonal vectors. Therefore the values of a and b has to satisfy $$ 3a-2b+4= 0$$

Answer 3

Noting that the dot product (which is a scalar, not the cross product, which is a vector) of any two perpendicular vectors must be $0$ (thanks to Macavity's comment on the question), we can solve for $a$ and $b$ using the definition of the dot product.

First of all, we know that, by the definition of the dot product, the dot product of any two vectors $(a_1, a_2, a_3)$ and $(b_1, b_2, b_3)$ $= a_1b_1 + a_2b_2 + a_3b_3$.

Here, substituting the respective values for $u_i$ and $v_i$ and simplifying the expression, we can find a "linear" equation representing all values for $a$ and $b$.

\begin{align} u \cdot v &= u_1v_1 + u_2v_2 + u_3v_3 \\ &= 1 \cdot 2 + 2(1-b) + a \cdot 3 \\ &= 2 + (2-2b) + 3a \\ &= 4 - 2b + 3a \end{align}

As we know that the dot product of two perpendicular vectors is $0$, we can substitute and solve:

\begin{align} 0 &= 4 - 2b + 3a \\ 2b - 3a &= 4 \quad && \text{In standard form} \\ a &= \frac{2b - 4} 3 \quad && \text{Solving for $a$} \\ b &= \frac{3a + 4} 3 \quad && \text{Solving for $b$} \end{align}

Therefore, given any value of $a$ or $b$, we should be able to solve for the other.

Note that there are, however, infinitely many solutions to the problem for $a$ and $b$ - essentially, the values are represented by a linear model where $a$ varies accordingly to $b$ (and vice versa). Below would be a graph that contains values of $a$ and $b$, graphed by WolframAlpha: