The problem consists of two parts: (c) Given the position vectors of points $A(8, 4, -3)$, $B(6, 3, -4)$, and $C(7, 5, -5)$, find the area of triangle $ABC$. (d) Find the volume of the parallelepiped whose co-terminous edges are given by the vectors $a = 2i - 3j + 4k$, $b = i + 2j - k$, and $c = 2i - j + 2k$.

1. Problem Description

The problem consists of two parts:

(c) Given the position vectors of points $A(8, 4, -3)$, $B(6, 3, -4)$, and $C(7, 5, -5)$, find the area of triangle $ABC$.

(d) Find the volume of the parallelepiped whose co-terminous edges are given by the vectors $a = 2i - 3j + 4k$, $b = i + 2j - k$, and $c = 2i - j + 2k$.

2. Solution Steps

(c) To find the area of triangle $ABC$, we first find the vectors $AB$ and $AC$.

$AB = B - A = (6-8, 3-4, -4-(-3)) = (-2, -1, -1)$

$AC = C - A = (7-8, 5-4, -5-(-3)) = (-1, 1, -2)$

Next, we find the cross product of $AB$ and $AC$:

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

The area of the triangle $ABC$ is given by half the magnitude of the cross product $AB \times AC$:

Area $= \frac{1}{2} |AB \times AC| = \frac{1}{2} \sqrt{(3)^2 + (-3)^2 + (-3)^2} = \frac{1}{2} \sqrt{9 + 9 + 9} = \frac{1}{2} \sqrt{27} = \frac{1}{2} \cdot 3\sqrt{3} = \frac{3\sqrt{3}}{2}$

(d) The volume of the parallelepiped formed by vectors $a$, $b$, and $c$ is given by the absolute value of the scalar triple product $a \cdot (b \times c)$, which is the determinant of the matrix formed by the components of the vectors.

$a = (2, -3, 4)$, $b = (1, 2, -1)$, $c = (2, -1, 2)$

$V = |a \cdot (b \times c)| = \left| \begin{vmatrix} 2 & -3 & 4 \\ 1 & 2 & -1 \\ 2 & -1 & 2 \end{vmatrix} \right|$

$V = |2(2(2) - (-1)(-1)) - (-3)(1(2) - (-1)(2)) + 4(1(-1) - 2(2))| = |2(4-1) + 3(2+2) + 4(-1-4)| = |2(3) + 3(4) + 4(-5)| = |6 + 12 - 20| = |18 - 20| = |-2| = 2$

3. Final Answer

(c) Area of triangle $ABC = \frac{3\sqrt{3}}{2}$

(d) Volume of the parallelepiped = 2