# Thread: Dot product question

1. ## Dot product question

Consider the triangle ABC in $\mathbb{R}^3$ 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 $\mathbb{R}^3$ 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 $\mathbf{AD}$ is perpendicular to $\mathbf{BC}$, then $\mathbf{AD}\cdot\mathbf{BC} = 0$.

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

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

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

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

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

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

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

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

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

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

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

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

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