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

The vector equation for the line through the point (1,0,6) and perpendicular to the plane x+3y+z=5 is v =(1+t)i + (3t)j + (6+t)k

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

The parametric equation of a line through the point A(x, y, z) perpendicular to the plane ax+by+cz= d is expressed generally as:

A + vt where:

A = (x, y, z)

v = (a, b, c) (normal vector)

This can then be expressed as:

s = A + vt

s = (x, y, z) + (a, b, c)t

Given the point

(x, y, z) = (1,0,6)

(a, b, c) = (1, 3, 1)

Substitute the given coordinate into the equation above:

s = (1,0,6) + (1, 3, 1)t

s = (1+t) + (0+3t) + (6+t)

The parametric equations from the equation above are:

x(t) = 1+t

y(t) = 3t

z(t) = 6+t

The vector equation will be expressed as v = xi + yj + zk

v =(1+t)i + (3t)j + (6+t)k

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The normal vector to the plane x + 3y + z = 5 is n = (1, 3, 1). The line we want is parallel to this normal vector.

Scale this normal vector by any real number t to get the equation of the line through the point (1, 3, 1) and the origin, then translate it by the vector (1, 0, 6) to get the equation of the line we want:

(1, 0, 6) + (1, 3, 1)t = (1 + t, 3t, 6 + t)

This is the vector equation; getting the parametric form is just a matter of delineating

x(t) = 1 + t

y(t) = 3t

z(t) = 6 + t

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