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.
 
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