Question

(a) Find parametric equations for the line through (4, 5, 6) that is perpendicular to the plane x − y + 3z = 2. (Use the parameter t.) (x(t), y(t), z(t)) = (b) In what points does this line intersect the coordinate planes? xy-plane (x, y, z) = yz-plane (x, y, z) = xz-plane (x, y, z) =

Answer 1

A. Therefore the required equation of the line is

x(t)= 4+t             y(t) = 5- t                    z(t) = 6+3t

B.

The intersection point  of xy plane and the line is(2,7,0)

The intersection point  of yz plane and the line is(0,9,-6)

The intersection point  of xz  plane and the line is(9,0,21)

Step-by-step explanation:

A.

Given that the line passes through (4,5,6) and is perpendicular to the plane

x- y+3z=2

Since the line is perpendicular to the given plane So the direction ratio of the line will be same with the direction ratio of the plane.

Therefore the direction ratio of line is (1,-1,3).

The parametric equation of a line is

$x(t)= x_\circ +t l_1$   ,      $y(t)= y_\circ +tm_1$    and      $z(t)= z_\circ +t n_1$

Here   $x_\circ= 4$  ,   $y_\circ= 5$    ,    $z_\circ= 6$ ,   $l_1 = 1$   ,  $m_1 =- 1$  ,    $n_1 = 3$

Therefore the required equation of the line is

x(t) = 4+t(1)        y(t) = 5+ t(-1)       and  z(t) = 6+ t(3)

x(t)= 4+t             y(t) = 5- t                    z(t) = 6+3t

B.

For xy plane , z(t)= 0

∴6+3t = 0 ⇔ t= -2

The intersection point  of xy plane and the line is (4-2,5+2,0) =(2,7,0)

For yz plane , x(t) = 0

∴4+t= 0⇔t = -4

The intersection point  of yz plane and the line is (0,5+4,6+3(-4)) =(0,9,-6)

For xz plane ,y(t)= 0

∴5-t=0 ⇔ t=5

The intersection point  of xz plane and the line is (4+5,0 , 6+3(5))=(9,0,21)