find vector equation and parametric equation for line thru (1,0,6) and perpendicular to the plane x+3y+z=5.?

Answer 1

The direction of the line is the same as the normal vector of the plane x + 3y + z = 5. The normal vector of this plane is:

v = <1,3,1>, so the parametric equation of the line is:

L1:

x = t + 1
y = 3t
z = t + 6

The vector equation is:

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

Hope it helps

Answer 2

Since the line is perpendicular to the plane, the directional vector of the line is the same as the normal vector of the plane. The directional vector v of the line is taken from the coefficients of the the variables in the equation of the plane.

v = <1, 3, 1>

Withe the directional vector and a point P(1,0,6) we can write the vector equation of the line L.

L(t) = P + tv
L(t) = <1, 0, 6> + t<1, 3, 1>
where t is a scalar that ranges over the real numbers

The parametric equation of the line can be taken directly from the vector equation.

L(t):
x = 1 + t
y = 3t
z = 6 + t

Answer 3

Use the formula y = ax + b the position a is the slope... the slope of the given line is -a million/3 and for a perpendicular the slope should be 3 Then that is y = 3x+b ... now use the point to discover b ... that is 7 = 3*2 + b ==> b =a million Then the equation is y = 3x + a million ok!

Answer 4

7-up+s*sq/=5on t-_+or n3