1. ## Dot product question

Consider the triangle ABC in $\displaystyle \mathbb{R}^3$ formed by the points A(3,2,1), B(4,4,2) and C(6,1,0)

Find the coordinates of the point D on BC such that AD is perpendicular to BC

2. Originally Posted by acevipa
Consider the triangle ABC in $\displaystyle \mathbb{R}^3$ formed by the points A(3,2,1), B(4,4,2) and C(6,1,0)

Find the coordinates of the point D on BC such that AD is perpendicular to BC
If $\displaystyle \mathbf{AD}$ is perpendicular to $\displaystyle \mathbf{BC}$, then $\displaystyle \mathbf{AD}\cdot\mathbf{BC} = 0$.

If $\displaystyle D = (x, y, z)$, then

$\displaystyle \mathbf{AD} = \mathbf{AO} + \mathbf{OD}$

$\displaystyle = \mathbf{OD} - \mathbf{OA}$

$\displaystyle = (x\mathbf{i} + y\mathbf{j} + z\mathbf{k}) - (3\mathbf{i} + 2\mathbf{j} + \mathbf{k})$

$\displaystyle = (x - 3)\mathbf{i} + (y - 2)\mathbf{j} + (z - 1)\mathbf{k}$.

$\displaystyle \mathbf{BC} = \mathbf{BO} + \mathbf{OC}$

$\displaystyle = \mathbf{OC} - \mathbf{OB}$

$\displaystyle = (6\mathbf{i} + 1\mathbf{j} + 0\mathbf{k}) - (4\mathbf{i} + 4\mathbf{j} + 2\mathbf{k})$

$\displaystyle = 2\mathbf{i} - 3\mathbf{j} - 2\mathbf{k}$.

So $\displaystyle \mathbf{AD}\cdot\mathbf{BC} = 0$

$\displaystyle [(x - 3)\mathbf{i} + (y - 2)\mathbf{j} + (z - 1)\mathbf{k}]\cdot[2\mathbf{i} - 3\mathbf{j} - 2\mathbf{k}] = 0$

$\displaystyle 2(x - 3) - 3(y - 2) -2(z - 1) = 0$.

You are told that $\displaystyle (x, y, z)$ lies on the line $\displaystyle BC$, so I would advise finding the parametric or Cartestian equations of $\displaystyle BC$...