Question

Vertices B and C of a ABC lie along the line
2 1 0
2 1 4
x  y  z 
  . Find the area of the triangle given
that A has Coordinates 1,1,2 and line segment BC has length 5 units

Answer 1

Answer on Question #44600 – Mathematics – Analytic Geometry

Question:

Vertices $B$ and $C$ of a $\Delta ABC$ lie along the line

210

214

xyz.

Find the area of the triangle given, that $A$ has coordinates $\{1, 1, 2\}$ and line segment $BC$ has length 5 units.

Solution:

To compute the area of a triangle formed by three points $A$, $B$ and $C$ in space (see fig.1) we use the vector product.

Fig. 1

By definition the area of this triangle is

$$S = \frac {1}{2} | \overrightarrow {B A} \times \overrightarrow {B C} | = \frac {1}{2} | \overrightarrow {B A} | \cdot | \overrightarrow {B C} | \sin \alpha ,$$

where $\alpha$ is the angle between the vectors $\overrightarrow{BA}$ and $\overrightarrow{BC}$. Taking $\overrightarrow{BC}$ to be the base of the triangle $ABC$, then the height of the triangle is $h = |\overrightarrow{BA}| \sin \alpha$. Therefore, we can rewrite (1) in the following form

$$S = \frac {1}{2} | \overrightarrow {B A} \times \overrightarrow {B C} | = \frac {1}{2} h \cdot B C,$$

where $BC = |\overrightarrow{BC}|$.

Let's find the height $h$ in two steps.

1) Write the canonical equation of the line $l$ passing through the two given points $M_1(2, 1, 0)$ and $M_2(2, 1, 4)$.

2) Determine the distance $h$ from the given point A (1, 1, 2) to the line $l$.

The canonical equation of the line passing through the two points in space is given by relation:

$$\frac {x - x _ {1}}{x _ {2} - x _ {1}} = \frac {y - y _ {1}}{y _ {2} - y _ {1}} = \frac {z - z _ {1}}{z _ {2} - z _ {1}}.$$

Substituting into (2) the coordinates of points $M_1$ and $M_2$ we obtain the equation of the line $l$

$$\frac {x - 2}{0} = \frac {y - 1}{0} = \frac {z}{4}.$$

From (2) and (3) it is easy to see that $\overrightarrow{M_1M_2} = \{x_2 - x_1, y_2 - y_1, z_2 - z_1\} = \{0, 0, 4\}$ is the directing vector of the line $l$ (fig.2).

Fig. 2

Hence, the distance between point $A$ and line $l$ can be found using the formula that follows from definition of parallelogram area:

$$h = \frac {\left| \overrightarrow {A M _ {1}} \times \overrightarrow {M _ {1} M _ {2}} \right|}{\left| \overrightarrow {M _ {1} M _ {2}} \right|}.$$

Let's calculate the numerator and denominator of (4) separately.

$$\overrightarrow {A M _ {1}} = \left\{x _ {1} - x _ {A}, y _ {1} - y _ {A}, z _ {1} - z _ {A} \right\} = \{1, 0, - 2 \},$$ $$\begin{array}{l} \overrightarrow {A M _ {1}} \times \overrightarrow {M _ {1} M _ {2}} = \left| \begin{array}{c c c} \vec {i} & \vec {j} & \vec {k} \\ 1 & 0 & - 2 \\ 0 & 0 & 4 \end{array} \right| = \big (0 \cdot 4 - 0 \cdot (- 2) \big) \vec {i} - \big (1 \cdot 4 - 0 \cdot (- 2) \big) \vec {j} + (1 \cdot 0 - 0 \cdot 0) \vec {k} = 4 \vec {j} = \\ = \{0, - 4, 0 \}, \\ \end{array}$$ $$\left| \overrightarrow {A M _ {1}} \times \overrightarrow {M _ {1} M _ {2}} \right| = \sqrt {0 ^ {2} + (- 4) ^ {2} + 0 ^ {2}} = 4,$$ $$\left| \overline {{M _ {1} M _ {2}}} ^ {\prime} \right| = \sqrt {0 ^ {2} + 0 ^ {2} + (4) ^ {2}} = 4.$$

From here we have: $h = \frac{4}{4} = 1$ unit.

Therefore, the area of the triangle $ABC$ is equal to

$$S = \frac {1}{2} h \cdot B C = \frac {1}{2} \cdot 1 \cdot 5 = 2.5 \text{ square units}.$$

Answer: the area of the triangle $ABC$ is equal to 2.5 square units.

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