Where does the line through (1, 0, 1) and (4, −3, 3) intersect the plane x + y + z = 6?

Mathematics

1 answer:

Irina-Kira [14]3 years ago

7 0

Parametrically for real t the line is

$(x,y,z) = (1-t)(1,0,1) + t(4,-3,3)$

$(x,y,z) = (1,0,1) + t(4-1,-3-0,3-1)$

$(x,y,z) = (1,0,1) + t(3,-3,2)$

That's three equations; we substitute into

$x+y+z=6$

$(1+3t)+(0-3t) + ( 1+2t)=6$

$2t=4$

$t=2$

$(x,y,z) = (1,0,1) + 2(3,-3,2) = (7,-6,5)$

The intersection is a single point (7,-6,5).

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Examine the following system of inequalities.

lubasha [3.4K]

Answer:

Dotted linear inequality shaded above passes through (0, 4) and (4, 0). Solid exponential inequality shaded below passes through (negative 2,2) & (0,5)

Step-by-step explanation:

we have

$y > -x+4$ ----> inequality A

The solution of the inequality A is the shaded area above the dotted line $y=-x+4$

The dotted line passes through the points (0,4) and (4,0) (y and x-intercepts)

and

$y \leq -(1/2)^{x} +6$ -----> inequality B

The solution of the inequality B is the shaded area above the solid line $y=-(1/2)^{x} +6$

The solid line passes through the points (0,5) and (-2,2)

therefore

The solution of the system of inequalities is the shaded area between the dotted line and the solid line

see the attached figure

Dotted linear inequality shaded above passes through (0, 4) and (4, 0). Solid exponential inequality shaded below passes through (negative 2,2) & (0,5)