Evaluate |2x+y-3z| for x=-2, y=10, and z=3 How to solve?

Mathematics

2 answers:

Angelina_Jolie [31]3 years ago

4 0

|2x + y - 3z|
Plug in the values
|-4 + 10 - 9|
= |-3|
= 3

stiv31 [10]3 years ago

3 0

The answer is (3) because when all the letters are turned into numbers then solved it becomes [-4+10+(-9)] then you have to turn it into a positive because of the lines.

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Let T be the plane-2x-2y+z =-13. Find the shortest distance d from the point Po=(-5,-5,-3) to T, and the point Q in T that is cl

GaryK [48]

Answer:

d=10u

Q(5/3,5/3,-19/3)

Step-by-step explanation:

The shortest distance between the plane and Po is also the distance between Po and Q. To find that distance and the point Q you need the perpendicular line x to the plane that intersects Po, this line will have the direction of the normal of the plane $n=(-2,-2,1)$ , then r will have the next parametric equations:

$x=-5-2\lambda\\y=-5-2\lambda\\z=-3+\lambda$

To find Q, the intersection between r and the plane T, substitute the parametric equations of r in T

$-2x-2y+z =-13\\-2(-5-2\lambda)-2(-5-2\lambda)+(-3+\lambda) =-13\\10+4\lambda+10+4\lambda-3+\lambda=-13\\9\lambda+17=-13\\9\lambda=-13-17\\\lambda=-30/9=-10/3$

Substitute the value of $\lambda$ in the parametric equations:

$x=-5-2(-10/3)=-5+20/3=5/3\\y=-5-2(-10/3)=5/3\\z=-3+(-10/3)=-19/3\\$

Those values are the coordinates of Q

Q(5/3,5/3,-19/3)

The distance from Po to the plane

$d=\left| {\to} \atop {PoQ}} \right|=\sqrt{(\frac{5}{3}-(-5))^2+(\frac{5}{3}-(-5))^2+(\frac{-19}{3}-(-3))^2} \\d=\sqrt{(\frac{5}{3}+5))^2+(\frac{5}{3}+5)^2+(\frac{-19}{3}+3)^2} \\d=\sqrt{(\frac{20}{3})^2+(\frac{20}{3})^2+(\frac{-10}{3})^2}\\d=\sqrt{\frac{400}{9}+\frac{400}{9}+\frac{100}{9}}\\d=\sqrt{\frac{900}{9}}=\sqrt{100}\\d=10u$