MHB How to know if plane is perpendicular to another plane?

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To determine if a plane is perpendicular to the z-axis, one must use the normal vector, which for the z-axis is <0, 0, 1>. A plane perpendicular to this axis can be expressed as 0x + 0y + 1z = C, simplifying to z = C for any constant C. For a line perpendicular to the equation 2x + 3y = 6, first identify the slope, which is -2/3; the perpendicular slope will be 3/2. The resulting line equation can be expressed as y = (3/2)x + 2 or transformed into standard form as 2y - 3x = 2. Understanding normal vectors and slopes is crucial for establishing perpendicular relationships in both planes and lines.
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The question that I'm trying to answer states "Make a vector equation of a plane that is perpendicular to the z axis." How do i ensure its perpendicular? How do i start this equation?

Another question similar to this that i am also struggling states "What is the vector equation of a 2D line that is perpendicular to this line: 2x + 3y = 6 though the lines y intercept?"

Any help for both of these questions would be greatly appreciated!
 
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nicole said:
The question that I'm trying to answer states "Make a vector equation of a plane that is perpendicular to the z axis." How do i ensure its perpendicular? How do i start this equation?

Another question similar to this that i am also struggling states "What is the vector equation of a 2D line that is perpendicular to this line: 2x + 3y = 6 though the lines y intercept?"

Any help for both of these questions would be greatly appreciated!

Find a NORMAL VECTOR to the Planes. If they are orthogonal, you're done.

Rather the same deal for the lines. Compare the slopes.
 
Any plane can be written in the form Ax+ By+ Cz= D for constants A, B, C, and D. And then <A, B, C> is a vector normal (perpendicular) to that plane.

The z-axis has "unit direction vector" <0, 0, 1>. A plane perpendicular to it must have that as "normal vector" and so the equation of the plane can be taken as 0x+ 0y+ 1z= z= C for some constant C.

The line 2x+ 3y= 6 has y-intercept (0, 2) (the y-intercept of a line is the point where it crosses the y-axis so x= 0). Further, if we were to "solve" that equation for y, 3y= 6- 2x, so y= 2- (2/3)x. That shows, again, that the y-intercept is 2 and that the slope is -2/3. A line perpendicular to the given line must have slope 3/2. Any (non-vertical) line can be written in the form y= mx+ b where m is the slope and b is the y-intercept. A line with slope 3/2 and y-intercept 2 is y= (3/2)x+ 2. If you don't like fractions, multiply by 2 to get 2y= 3x+ 2 which can be written in the same form as the line was originally given as 2y- 3x= 2.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.

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