Question

Given points A(3;4;6),B(3;−2;−3),C(6;3;2),D(5;2;2). Make up: a) canonical and parametric equations of the straight line passing through points A and B

Answer 1

The parametric equations of the line passing through points A and B are x = 3, y = 4 - 6t, and z = 6 - 9t. The canonical equation is x = 3, and y/(-6) = z/(-9), demonstrating that the line passes through point A and that y and z are proportional to the direction vector components.

Step-by-step explanation:

To find the canonical and parametric equations of the straight line passing through points A(3;4;6) and B(3;−2;−3), we first need to determine the direction vector of the line, which is given by the difference between the coordinates of B and A:

• Direction vector, Δ = B - A = (3 - 3, -2 - 4, -3 - 6) = (0, -6, -9).

Now we can write the parametric equations of the line using point A and the direction vector Δ:

• x = x1 + tΔx = 3 + 0t
• y = y1 + tΔy = 4 + t(-6)
• z = z1 + tΔz = 6 + t(-9)

Thus, the parametric equations are x = 3, y = 4 - 6t, and z = 6 - 9t, where t is the parameter.

For the canonical equation of the line, we can express t in terms of each variable and then eliminate t:

• t = (x - 3)/0 (which is undefined for x - in this case, we simply see that x remains constant at 3)
• t = (y - 4)/(-6)
• t = (z - 6)/(-9)

Since 't' cannot be determined from x because of the division by zero, we express t in terms of y and z. The canonical equation assuming y and z change while x is constant:

• x = 3
• y/(-6) = z /(-9)

This shows that the line is passing through point A, and the variables y and z are changing in the ratio of their direction vector components.