Given points $A(0, 1, 1)$, $B(2, 0, 2)$, and $C(3, -1, -1)$ in a right-handed orthonormal coordinate system. a. Calculate the cross product $\vec{AB} \times \vec{AC}$ and deduce that points $A$, $B$, and $C$ are not collinear. b. Find the equation of the plane $P$ passing through points $A$, $B$, and $C$. c. Find the parametric equations of the line $L$ passing through point $D(1, 1, 2)$ and perpendicular to plane $P$. Then find the coordinates of point $M$, the intersection of plane $P$ and line $L$. d. Find the equation of the sphere passing through points $A$, $B$, $C$, and $D$.

1. Problem Description

Given points $A(0, 1, 1)$, $B(2, 0, 2)$, and $C(3, -1, -1)$ in a right-handed orthonormal coordinate system.

a. Calculate the cross product $\vec{AB} \times \vec{AC}$ and deduce that points $A$, $B$, and $C$ are not collinear.

b. Find the equation of the plane $P$ passing through points $A$, $B$, and $C$.

c. Find the parametric equations of the line $L$ passing through point $D(1, 1, 2)$ and perpendicular to plane $P$. Then find the coordinates of point $M$, the intersection of plane $P$ and line $L$.

d. Find the equation of the sphere passing through points $A$, $B$, $C$, and $D$.

2. Solution Steps

a.

First, we find the vectors $\vec{AB}$ and $\vec{AC}$.

$\vec{AB} = B - A = (2 - 0, 0 - 1, 2 - 1) = (2, -1, 1)$

$\vec{AC} = C - A = (3 - 0, -1 - 1, -1 - 1) = (3, -2, -2)$

Next, we compute the cross product $\vec{AB} \times \vec{AC}$:

$\vec{AB} \times \vec{AC} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 2 & -1 & 1 \\ 3 & -2 & -2 \end{vmatrix} = \vec{i}((-1)(-2) - (1)(-2)) - \vec{j}((2)(-2) - (1)(3)) + \vec{k}((2)(-2) - (-1)(3)) = (2 + 2)\vec{i} - (-4 - 3)\vec{j} + (-4 + 3)\vec{k} = 4\vec{i} + 7\vec{j} - \vec{k} = (4, 7, -1)$

Since $\vec{AB} \times \vec{AC} \neq (0, 0, 0)$, the vectors $\vec{AB}$ and $\vec{AC}$ are not parallel. Thus, the points $A$, $B$, and $C$ are not collinear.

b.

The equation of the plane $P$ can be written as $ax + by + cz = d$, where the normal vector to the plane is $\vec{n} = (a, b, c)$.

From part a, we know that $\vec{AB} \times \vec{AC} = (4, 7, -1)$ is a normal vector to the plane passing through $A$, $B$, and $C$. So, we can take $\vec{n} = (4, 7, -1)$.

The equation of the plane is of the form $4x + 7y - z = d$.

Since the plane passes through $A(0, 1, 1)$, we have $4(0) + 7(1) - 1 = d$, so $d = 6$.

The equation of the plane $P$ is $4x + 7y - z = 6$.

c.

The line $L$ passes through $D(1, 1, 2)$ and is perpendicular to plane $P$. This means that the direction vector of the line $L$ is parallel to the normal vector of plane $P$, which is $(4, 7, -1)$.

The parametric equations of line $L$ are:

$x = 1 + 4t$

$y = 1 + 7t$

$z = 2 - t$

To find the intersection point $M$ of line $L$ and plane $P$, we substitute the parametric equations of $L$ into the equation of $P$:

$4(1 + 4t) + 7(1 + 7t) - (2 - t) = 6$

$4 + 16t + 7 + 49t - 2 + t = 6$

$9 + 66t = 6$

$66t = -3$

$t = -\frac{3}{66} = -\frac{1}{22}$

Now we substitute $t = -\frac{1}{22}$ into the parametric equations of line $L$ to find the coordinates of point $M$:

$x = 1 + 4(-\frac{1}{22}) = 1 - \frac{4}{22} = 1 - \frac{2}{11} = \frac{9}{11}$

$y = 1 + 7(-\frac{1}{22}) = 1 - \frac{7}{22} = \frac{15}{22}$

$z = 2 - (-\frac{1}{22}) = 2 + \frac{1}{22} = \frac{45}{22}$

So the coordinates of point $M$ are $(\frac{9}{11}, \frac{15}{22}, \frac{45}{22})$.

d.

Let the equation of the sphere be $(x - a)^2 + (y - b)^2 + (z - c)^2 = r^2$.

Since $A(0, 1, 1)$, $B(2, 0, 2)$, $C(3, -1, -1)$, and $D(1, 1, 2)$ lie on the sphere, we have:

$(0 - a)^2 + (1 - b)^2 + (1 - c)^2 = r^2$ (1)

$(2 - a)^2 + (0 - b)^2 + (2 - c)^2 = r^2$ (2)

$(3 - a)^2 + (-1 - b)^2 + (-1 - c)^2 = r^2$ (3)

$(1 - a)^2 + (1 - b)^2 + (2 - c)^2 = r^2$ (4)

Subtracting (1) from (2), (3), and (4) gives:

$(2 - a)^2 - a^2 + b^2 - (1 - b)^2 + (2 - c)^2 - (1 - c)^2 = 0$

$(3 - a)^2 - a^2 + (-1 - b)^2 - (1 - b)^2 + (-1 - c)^2 - (1 - c)^2 = 0$

$(1 - a)^2 - a^2 + (2 - c)^2 - (1 - c)^2 = 0$

Simplifying:

$4 - 4a + a^2 - a^2 + b^2 - (1 - 2b + b^2) + 4 - 4c + c^2 - (1 - 2c + c^2) = 0 \implies 4 - 4a - 1 + 2b + 4 - 4c - 1 + 2c = 0 \implies -4a + 2b - 2c = -6 \implies 2a - b + c = 3$

$9 - 6a + a^2 - a^2 + (1 + 2b + b^2) - (1 - 2b + b^2) + (1 + 2c + c^2) - (1 - 2c + c^2) = 0 \implies 9 - 6a + 4b + 4c = 0 \implies 6a - 4b - 4c = 9$

$1 - 2a + a^2 - a^2 + 4 - 4c + c^2 - (1 - 2c + c^2) = 0 \implies 1 - 2a + 4 - 4c - 1 + 2c = 0 \implies -2a - 2c = -4 \implies a + c = 2$

From $a + c = 2$, $c = 2 - a$. Substituting into the first equation: $2a - b + 2 - a = 3 \implies a - b = 1 \implies b = a - 1$.

Substituting into the second equation: $6a - 4(a - 1) - 4(2 - a) = 9 \implies 6a - 4a + 4 - 8 + 4a = 9 \implies 6a = 13 \implies a = \frac{13}{6}$

Then $b = a - 1 = \frac{13}{6} - 1 = \frac{7}{6}$ and $c = 2 - a = 2 - \frac{13}{6} = -\frac{1}{6}$.

So the center of the sphere is $(\frac{13}{6}, \frac{7}{6}, -\frac{1}{6})$.

Now, we find $r^2$ using point $A(0, 1, 1)$:

$r^2 = (\frac{13}{6})^2 + (1 - \frac{7}{6})^2 + (1 + \frac{1}{6})^2 = (\frac{13}{6})^2 + (-\frac{1}{6})^2 + (\frac{7}{6})^2 = \frac{169 + 1 + 49}{36} = \frac{219}{36} = \frac{73}{12}$

Thus, the equation of the sphere is $(x - \frac{13}{6})^2 + (y - \frac{7}{6})^2 + (z + \frac{1}{6})^2 = \frac{73}{12}$.

3. Final Answer

a. $\vec{AB} \times \vec{AC} = (4, 7, -1)$. $A, B, C$ are not collinear.

b. $4x + 7y - z = 6$

c. $x = 1 + 4t$, $y = 1 + 7t$, $z = 2 - t$. $M(\frac{9}{11}, \frac{15}{22}, \frac{45}{22})$

d. $(x - \frac{13}{6})^2 + (y - \frac{7}{6})^2 + (z + \frac{1}{6})^2 = \frac{73}{12}$