Competency-Based
Assessment Made Simple.

1. Find the equations of the sides:

The equation of a line passing through two points with position vectors $\vec{a}$ and $\vec{b}$ is given by $\vec{r} = \vec{a} + \lambda(\vec{b} - \vec{a})$.

Side AB:

$\vec{a} = 4\hat{i} + 7\hat{j} + 8\hat{k}$

$\vec{b} = 2\hat{i} + 3\hat{j} + 4\hat{k}$

$\vec{b} - \vec{a} = (2-4)\hat{i} + (3-7)\hat{j} + (4-8)\hat{k} = -2\hat{i} - 4\hat{j} - 4\hat{k}$

Equation of AB: $\vec{r} = (4\hat{i} + 7\hat{j} + 8\hat{k}) + \lambda(-2\hat{i} - 4\hat{j} - 4\hat{k})$

In Cartesian form: $\frac{x-4}{-2} = \frac{y-7}{-4} = \frac{z-8}{-4}$ or $\frac{x-4}{1} = \frac{y-7}{2} = \frac{z-8}{2}$

Side BC:

$\vec{b} = 2\hat{i} + 3\hat{j} + 4\hat{k}$

$\vec{c} = -1\hat{i} - 2\hat{j} + 1\hat{k}$

$\vec{c} - \vec{b} = (-1-2)\hat{i} + (-2-3)\hat{j} + (1-4)\hat{k} = -3\hat{i} - 5\hat{j} - 3\hat{k}$

Equation of BC: $\vec{r} = (2\hat{i} + 3\hat{j} + 4\hat{k}) + \mu(-3\hat{i} - 5\hat{j} - 3\hat{k})$

In Cartesian form: $\frac{x-2}{-3} = \frac{y-3}{-5} = \frac{z-4}{-3}$

Side CD:

$\vec{c} = -1\hat{i} - 2\hat{j} + 1\hat{k}$

$\vec{d} = 1\hat{i} + 2\hat{j} + 5\hat{k}$

$\vec{d} - \vec{c} = (1-(-1))\hat{i} + (2-(-2))\hat{j} + (5-1)\hat{k} = 2\hat{i} + 4\hat{j} + 4\hat{k}$

Equation of CD: $\vec{r} = (-1\hat{i} - 2\hat{j} + 1\hat{k}) + \nu(2\hat{i} + 4\hat{j} + 4\hat{k})$

In Cartesian form: $\frac{x+1}{2} = \frac{y+2}{4} = \frac{z-1}{4}$ or $\frac{x+1}{1} = \frac{y+2}{2} = \frac{z-1}{2}$

Side DA:

$\vec{d} = 1\hat{i} + 2\hat{j} + 5\hat{k}$

$\vec{a} = 4\hat{i} + 7\hat{j} + 8\hat{k}$

$\vec{a} - \vec{d} = (4-1)\hat{i} + (7-2)\hat{j} + (8-5)\hat{k} = 3\hat{i} + 5\hat{j} + 3\hat{k}$

Equation of DA: $\vec{r} = (1\hat{i} + 2\hat{j} + 5\hat{k}) + \tau(3\hat{i} + 5\hat{j} + 3\hat{k})$

In Cartesian form: $\frac{x-1}{3} = \frac{y-2}{5} = \frac{z-5}{3}$

2. Find the foot of the perpendicular from A to CD:

Let the coordinates of the foot of the perpendicular be P.

The equation of line CD is $\frac{x+1}{1} = \frac{y+2}{2} = \frac{z-1}{2} = k$

So, the coordinates of any point on CD are $(k-1, 2k-2, 2k+1)$.

Let P be $(k-1, 2k-2, 2k+1)$.

Direction ratios of AP are $(k-1-4, 2k-2-7, 2k+1-8) = (k-5, 2k-9, 2k-7)$.

Direction ratios of CD are $(1, 2, 2)$.

Since AP is perpendicular to CD, the dot product of their direction ratios is zero.

$1(k-5) + 2(2k-9) + 2(2k-7) = 0$

$k - 5 + 4k - 18 + 4k - 14 = 0$

$9k - 37 = 0$

$k = \frac{37}{9}$

Coordinates of P are $(\frac{37}{9}-1, 2(\frac{37}{9})-2, 2(\frac{37}{9})+1) = (\frac{28}{9}, \frac{56}{9}, \frac{83}{9})$.