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Find a vector equation and parametric equations for the line. (Use the parameter t.) The line through the point (1, 0, 9) and perpendicular to the plane

x + 3y + z = 5

1 Answer

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Answer:

vector equation and parametric equations for the line are;

r(t) = < 1, 0, 9 > + t< 1, 3, 1 >

x(t),y(t),z(t) = < 1 + t, 3t, 9 + t >

Explanation:

Given the data in the question;

Plane = x + 3y + z = 5

Normal of pane ⇒ <a> = < 1, 3, 1 >

As line is perpendicular to pane, Line is parallel to normal of plane.

⇒ Direction ratio of line < 1, 3, 1 >

Also line passes through < 1, 0, 9 >

so

r(t) = < 1, 0, 9 > + t< 1, 3, 1 >

Also,

(x-1)/1 = (y-0)/3 = (z-9)/1 = t

⇒ (x-1)/1 =t, (y-0)/3 = t, (z-9)/1 = t

⇒ x = 1 + t, y = 3t, z = 9 + t

⇒ x(t),y(t),z(t) = < 1 + t, 3t, 9 + t >

Therefore, vector equation and parametric equations for the line are;

r(t) = < 1, 0, 9 > + t< 1, 3, 1 >

x(t),y(t),z(t) = < 1 + t, 3t, 9 + t >

User Carlos Pascual
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