   Chapter 12.5, Problem 5E

Chapter
Section
Textbook Problem

# Find a vector equation and parametric equations for the line.5. The line through the point (1, 0, 6) and perpendicular to the plane x + 3y + z = 5

To determine

To find: A vector equation and the parametric equations for a line through the point (1,0,6) and perpendicular to the plane x+3y+z=5.

Explanation

Formula used:

The expression to find the vector equation (r) is,

r=r0+tv (1)

The expressions to find the parametric equations for a line through the point (x0,y0,z0) and parallel to the direction vector ai+bjck.

x=x0+at,y=y0+bt,z=z0+ct (2)

Here,

r0 is the position vector of the point P0(1,0,6) and v is the parallel vector to the line.

Calculation:

Write the position vector (r0) using the point P0(1,0,6) as follows.

r0=1i+(0)j+6k=i+6k

Write the parallel vector (v) from the line which is perpendicular to the plane x+3y+z=5 as follows.

If the line is perpendicular to a certain plane, the direction of line is same as the direction of normal vector of that plane.

The normal vector of the plane x+3y+z=5 is 1,3,1, which is written as i+3j+k

Therefore, the parallel vector is v=i+3j+k.

In equation (1), substitute i+6k for r0 and i+3j+k for v

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