Resolved: "The parallelogram shown here has vertices at A(2,..."

europaparksn

Answered question

2021-12-06

The parallelogram shown here has vertices at

A (2, - 1, 4), B (1, 0, - 1), C (1, 2, 3)

, and D. FInd the coordinates of D.

Answer & Explanation

Donald Proulx

Beginner2021-12-07Added 18 answers

Step 1
The parametric equations of line BA is,
$\frac{x - 1}{2 - t} = \frac{y - 0}{- 1 - 0} = \frac{z + 1}{4 + 1} = t$
$x = 1 + t, y = - t, z = - 1 + 5 t$
1) $L_{1} : x = 1 + t, y = 2 - t, z = 3 + 5 t$
Step 2
The parametric equations of line BC is as follows:
$\frac{x - 1}{1 - 1} = \frac{y - 0}{2 - 0} = \frac{z + 1}{3 + 1} = s$
$x = 1, y = 2 s, z = - 1 + 4 s$
2) $L_{2} : x = 2, y = - 1 + 2 s, z = 4 + 4 s$
To solve the point of intersection, solve equations (1) and (2) as follows:
$1 + t = 2 {Equate x}$
$t = 1$
$2 - t = - 1 + 2 s {Equate y}$
$2 - 1 = - 1 + 2 s {As t = 1}$
$2 s = 2$
$s = 1$
$x = 2$
$y = - 1 + 2 \times 1$
$y = 1$
$z = 4 + 4 \times 1$
$= 8$
Hence, the required coordinates of point D are $D (2, 1, 8)$

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