Values of x and y such that $a$ and $b$ are orthogonal

Question

So I have 2 vectors, $a=(x+y)i+2j+yk$, and $b=3i+(4x+y+1)j+4k$. I need to find the values for $x$ and $y$ such that the two vectors and orthogonal, and parallel. I know that two vectors are orthogonal when their dot product is zero, but solving it doesn't leave me with any way to solve for $x$ and $y$ (i.e. no quadratic or anything of the nature). It is a similar story for the cross product. Any insight would be much appreciated.

Answer 1

Suppose your vectors are given as

$$ a = (x+y)i + 2j + yk \\ b = 3i + (4x+y+1)j+k \tag{1} $$

Their dot product is

$$ a \cdot b = 3 \cdot (x+y) + 2\cdot (4x+y+1) + y \tag{2} $$

Then the objective like you said is when we set the dot product to zero

$$ 3 \cdot (x+y) + 2\cdot (4x+y+1) + y =0 \\ 3x + 3y + 8x +2y+2 +y=0 \tag{3} $$

Now you need to combine like terms

$$ \\ 11x + 6y +2 =0 \tag{4} $$

Your objective is to figure out when this is zero. Let's suppose instead you want to use the cross product.

$$ a \times b = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ (x+y) & 2 & y \\ 3 & (4x+y+1) & 1 \end{vmatrix} \tag{5}$$

Now when we reduce this we get

$$ a \times b = \begin{vmatrix} 2 & y \\ (4x+y+1) & 1 \end{vmatrix} \hat{i} + \begin{vmatrix} (x+y) & y \\ 3 & 1 \end{vmatrix}\hat{j} + \begin{vmatrix} (x+y) & 2 \\3 & (4x+y+1) \end{vmatrix}\hat{k} \tag{6}$$

Now the last step here is to compute the determinants. The $2 \times 2$ determinant is $ad-bc$

$$ a\times b = (2 - y(4x+y+1) )\hat{i} + ((x+y)-3y )\hat{j} + ((x+y)(4x+y+1) -6) \hat{k}\tag{7} $$

Now this vector $ a \times b$ will be orthogonal (perpendicular) unless the following is true. If $ a ,b $ are parallel. This implies

$$ \| a \times b\| =0 \tag{8}$$

This implies

$$ a\times b = \hat{0} \tag{9} $$

Which is saying, find where each of those vector entries are zero. That is solve for the zeros of.

$$ -y^{2} -4xy -y +2 =0 \\ x-2y =0 \\ 4x^{2}+5xy+x+y^{2}+y-6 =0 \tag{10} $$

Which is a much worse problem. The vector should be orthogonal in all other cases..