What is 4 plus 4?12234

Let p be the plane in space that intersects the x-axis at −5, the y-axis at −2, and the z-axis at 5. Find a vector v¯¯¯ that is

Lana71 [14]

Let P be the plane that intersects

x-axis at point (-5,0,0);
y-axis at point (0,-2,0);
z-axis at point (0,0,5).

Write the equation of the plane P:

$\left|\begin{array}{ccc}x-(-5)&y-0&z-0\\0-(-5)&-2-0&0-0\\0-(-5)&0-0&5-0\end{array}\right|=0\Rightarrow \left|\begin{array}{ccc}x+5&y&z\\5&-2&0\\5&0&5\end{array}\right|=0.$

Then

$-10(x+5)+10z-25y=0,\\ \\2(x+5)-2z+5y=0,\\ \\2x+5y-2z=-10.$

The coefficients at variables x, y and z are the coordinates of perpendicular vector to the plane. Thus

$\vec{v}=(2,5,-2)\perp P.$

Answer: $\vec{v}=(2,5,-2).$

7 0

4 years ago

Using the law of cosines, write an algebraic proof to show that the angles in an equilateral triangle must equal 60°. Use "∧" to

bekas [8.4K]

The algebraic proof shows that the angles in an equilateral triangle must equal 60° each

<h3>Laws of cosines </h3>

From the question, we are to use the law of cosines to write an algebraic proof that shows that the angles in an equilateral triangle must equal 60°.

Given any triangle ABC, the measures of angles A, B, and C by the law of cosines are

cos A = (b^2 + c^2 - a^2)/2bc

cos B= (a^2 + c^2 - b^2)/2ac

cos C = (a^2 + b^2 - c^2)/2ab

Now, given that the triangle is equilateral, with each of the side lengths equal to s

That is, a = b = c = s

Then, we can write that

cos A = (s^2 + s^2 - s^2)/(2s×s)

cos A = (s^2 )/(2s^2)

cos A = 1/2

cos A = 0.5

∴ A = cos⁻¹(0.5)

A = 60°

Also

cos B = (s^2 + s^2 - s^2)/(2s×s)

cos B = (s^2 )/(2s^2)

cos B = 1/2

cos B = 0.5

∴ B = cos⁻¹(0.5)

B = 60°

and

cos C = (s^2 + s^2 - s^2)/(2s×s)

cos C = (s^2 )/(2s^2)

cos C = 1/2

cos C = 0.5

∴ C = cos⁻¹(0.5)

C = 60°

Thus,

A = 60°, B = 60° and C = 60°

Hence, the algebraic proof above shows that the angles in an equilateral triangle must equal 60° each.

Learn more on The law of cosines here: brainly.com/question/2866347

#SPJ1

3 0

2 years ago

Direction: True or False

Goshia [24]

Answer:

<u>False </u>Any three points are always coplanar

<u>False </u>Two points are always collinear

<u>False </u>Two planes intersect at a point

Read explanations, as the answer to these questions, is subjective.

Step-by-step explanation:

The definition of coplanar is; figures that exist on the same plane. Any three points might not always be colinear, as some points might exist on one plane, but other points could exist on other planes. To visualize this phenomenon, refer to the attached image. Two points could be on the red plane, but the third could be on the green. Therefore, while there are three points, not all of them exist on the same plane. However, another plane can be constructed to connect these points, bear in mind certain problems might specifically indicate that three points are not coplanar, therefore, the answer is false.

The definition of collinear is existing on the same line. This statement is a little subject, it can be both true and false depending on the way one looks at it. A line can be drawn to connect any two points, one only needs two points to determine a line. So technically two points are always collinear. However, it also depends on the circumstance. Certain geometry problems might specifically indicate that two points are not collinear. Thus, it really depends on the context. Therefore, the answer is technically false.

Two planes usually intersect on a line. To understand this, please refer to the image attached. Tehcnailly, if a corner of one plane intersects another, then one can state that the plane intersects on a point, but typically, planes intersect on lines. Therefore, technically the answer to this problem is false, as two planes usually intersect on a line.

Images credits: Geogebra