Question

1. find the area of the triangle with vertices ( a, b, ) and ( c ) :
2. ( a = (2,3,4) ), ( b = (4,5,6) ), ( c = (7,7,7) ).
3. ( a = (2,3,4) ), ( b = (6,5,4) ), ( c = (8,8,8) ).
4. find a vector perpendicular to the plane containing the points ( a, b, ) and ( c ) :
5. ( a = (2,3,4) ), ( b = (4,5,6) ), ( c = (7,7,7) ).
6. ( a = (2,3,4) ), ( b = (6,5,4) ), ( c = (8,8,8) ).

Explanation:

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Problem 6.1

Step1: Find vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$

$\overrightarrow{AB} = (4-2, 5-3, 6-4) = (2,2,2)$
$\overrightarrow{AC} = (7-2, 7-3, 7-4) = (5,4,3)$

Step2: Compute cross product $\overrightarrow{AB} \times \overrightarrow{AC}$

$$ \overrightarrow{AB} \times \overrightarrow{AC} =
LATEXBLOCK0
= \mathbf{i}(2 \cdot 3 - 2 \cdot 4) - \mathbf{j}(2 \cdot 3 - 2 \cdot 5) + \mathbf{k}(2 \cdot 4 - 2 \cdot 5) = (-2, 4, -2) $$

Step3: Calculate magnitude of cross product

$|\overrightarrow{AB} \times \overrightarrow{AC}| = \sqrt{(-2)^2 + 4^2 + (-2)^2} = \sqrt{4+16+4} = \sqrt{24} = 2\sqrt{6}$

Step4: Find half the magnitude (area)

$\text{Area} = \frac{1}{2}|\overrightarrow{AB} \times \overrightarrow{AC}|$

Step1: Find vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$

$\overrightarrow{AB} = (6-2, 5-3, 4-4) = (4,2,0)$
$\overrightarrow{AC} = (8-2, 8-3, 8-4) = (6,5,4)$

Step2: Compute cross product $\overrightarrow{AB} \times \overrightarrow{AC}$

$$ \overrightarrow{AB} \times \overrightarrow{AC} =
LATEXBLOCK0
= \mathbf{i}(2 \cdot 4 - 0 \cdot 5) - \mathbf{j}(4 \cdot 4 - 0 \cdot 6) + \mathbf{k}(4 \cdot 5 - 2 \cdot 6) = (8, -16, 8) $$

Step3: Calculate magnitude of cross product

$|\overrightarrow{AB} \times \overrightarrow{AC}| = \sqrt{8^2 + (-16)^2 + 8^2} = \sqrt{64+256+64} = \sqrt{384} = 8\sqrt{6}$

Step4: Find half the magnitude (area)

$\text{Area} = \frac{1}{2}|\overrightarrow{AB} \times \overrightarrow{AC}|$

Step1: Find vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$

$\overrightarrow{AB} = (2,2,2)$, $\overrightarrow{AC} = (5,4,3)$

Step2: Compute cross product (perpendicular vector)

$$ \overrightarrow{AB} \times \overrightarrow{AC} =
LATEXBLOCK0
= (-2, 4, -2) $$

Answer:

$\sqrt{6}$

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Problem 6.2