Select the following lines that would be perpendicular to y = 3x + 1. (Select ALL correct)

Mathematics

1 answer:

max2010maxim [7]3 years ago

6 0

Answer:

The line passing through points (1,4) and (4,3).

The line passing through points (1,-4) and (4,-5).

Step-by-step explanation:

Since the equation is y=3x+1

If it was perpendicular, it means the slope would be the negative reciprocal.

So the slope would be -1/3

9-8/6-3=1/3 (not negative) No

3-4/4-1=-1/3 Yes

-5+4/4-1=-1/3 Yes

6-3/5-4= 3/1 No

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Find equations of the tangent plane and the normal line to the given surface at the specified point. x + y + z = 8exyz, (0, 0, 8

Dima020 [189]

Let $f(x,y,z)=x+y+z-8e^{xyz}$ . The tangent plane to the surface at (0, 0, 8) is

$\nabla f(0,0,8)\cdot(x,y,z-8)=0$

The gradient is

$\nabla f(x,y,z)=\left(1-8yze^{xyz},1-8xze^{xyz},1-8xye^{xyz}\right)$

so the tangent plane's equation is

$(1,1,1)\cdot(x,y,z-8)=0\implies x+y+(z-8)=0\implies x+y+z=8$

The normal vector to the plane at (0, 0, 8) is the same as the gradient of the surface at this point, (1, 1, 1). We can get all points along the line containing this vector by scaling the vector by $t$ , then ensure it passes through (0, 0, 8) by translating the line so that it does. Then the line has parametric equation

$(1,1,1)t+(0,0,8)=(t,t,t+8)$

or $x(t)=t$ , $y(t)=t$ , and $z(t)=t+8$ .

(See the attached plot; the given surface is orange, (0, 0, 8) is the black point, the tangent plane is blue, and the red line is the normal at this point)