Question

find parametric equations of the line that passes through the point P =(2, 0,-1) and is parallel to the vector n =<2, 1, 3>

1.2 find paramedic equations of the line that passes through the points A= (1, 2, - 3) and B =(7, 2, - 4).

1.3 find paramedic equations for the line of intersection of the planes - 5x + y - 2z =3 and 2x - 3y + 5z =-7

Answer 1

1.1)

P = (2, 0,-1)

vector n =<2, 1, 3>

Let the parameter of the line be t . Since the line is parallel to vector n, so the direction ratios of the line to be found will also be parallel to the vector n

The equation of the line is given by

$\dfrac{x - x_1}{a} = \dfrac{y-y_1}{b} = \dfrac{z-z_1}{c} = t$

So, on substituting the values in the above equation, we have

$\dfrac{x - 2}{2} = \dfrac{y-0}{1} = \dfrac{z+1}{3} = t$

From the above equation we get the parametric equation of line as

x = 2t + 2 , y = t, z = 3t - 1

1.2)

A= (1, 2, - 3) and B =(7, 2, - 4).

The equation of the line is given by

$\dfrac{x - x_1}{a} = \dfrac{y-y_1}{b} = \dfrac{z-z_1}{c} = t$

Here,

a = x₂ - x₁ = 7 - 1 = 6

b = y₂ - y₁ = 2 - 2 = 0

c = z₂ - z₁ = -4 + 3 = -1

Hence, on substituting the values we have

$\dfrac{x - 1}{6} = \dfrac{y-2}{0} = \dfrac{z+3}{-1} = t$

We get the parametric equation of line as

x = 6t + 1, y = 2, z = - t - 3

1.3)

- 5x + y - 2z = 3

and

2x - 3y + 5z = - 7

To find the parametric equation of line we will solve the given system of planes.

Let z = t, where t is our free parameter.

So, the equation of the plane takes the form

-5x + y = 3 + 2t

2x - 3y = -7 - 5t

On solving the above equations we get

x = $\dfrac{-1}{13}(2+t)$

y = $\dfrac{1}{13}(29 + 2t)$

So, the parametric equation of the line of intersection of the given planes is

x = $\dfrac{-1}{13}(2+t)$, y = $\dfrac{1}{13}(29 + 2t)$ , z = t