Must two lines actually intersect in order to be perpendicular to one another?
The word “perpendicular” when referring to a pair of lines usually assumes that they lie in the same plane. So the question is: Is there a reasonable definition for noncoplanar lines to be perpendicular, and has that definition ever been used?Yes, there is such a definition. Consider this labelled cube:The lines [math]AB[/math] and [math]CG[/math] don’t lie in the same plane, so they’re skew lines, yet they seem to be perpendicular in some sense. The sense can be made precise in a way that works in arbitrary dimensions using linear algebra. Let [math]\mathbf v[/math] be the displacement vector from [math]A[/math] to [math]B[/math], and let [math]\mathbf w[/math] be the displacement vector from [math]C[/math] to [math]G[/math]. The angle [math]\theta[/math] between these vectors is [math]90^\circ[/math] if and only if their dot product (also called inner product) [math]\mathbf v\cdot\mathbf w[/math] is [math]0[/math]. That says the directions of the lines [math]AB[/math] and [math]CG[/math] are at right angles.That definition has been used for perpendicular lines on occasion, but it’s not standard. If you do use that definition, be sure to say so because otherwise readers will assume you’re using the standard definition which requires the lines to be coplanar.
Which is not true about parallel and perpendicular lines?
A.) Given any two distinct lines in the Cartesian plane, the two lines will either be parallel or perpendicular. B.) The ratios of the vertical rise to the horizontal run of any two distinct nonvertical parallel lines must be equal. C.) If two distinct nonvertical lines are parallel, then the two lines must have the same slope.
Show that the line: x =-5-4t, y=1-t,z=3+2t and plane: x+2y+3z-9=0 are parallel.?
The normal vector of a plane is perpendicular to any vector that lies in the plane. So we need to show that the directional vector v, of the line is perpendicular to the normal vector n, of the plane. If this is true then the dot product is zero. v • n = <-4, -1, 2> • <1, 2, 3> = -4 - 2 + 6 = 0 So the directional vector of the line is parallel to the plane. Now we show that a point on the line is not in the plane. Let t = 0 and we get the point P(-5, 1, 3). x + 2y + 3z - 9 = 0 -5 + 2*1 + 3*3 - 9 = -5 + 2 + 9 - 9 = -3 So the line does not lie in the plane. It is parallel to the plane.
If the Earth's axis is perpendicular to its orbital plane, then what impact would it have on the duration of days and nights?
If the Earth’s axis was changed from its current 23 degree angle to the earth’s orbital plane to being perpendicular, days and nights would be of equal length. Also the angle of the sun (hence it’s intensity) at different latitudes would not change with the seasons.Seasons would be very much milder but would still exist. The earth moves in an elliptical path around the sun. This would create a slight yearly cycle in temperature. Because of the influence of the other planets the eccentricity of the earth’s orbit varies over time. The distance between the maximum and minimum distance varies between about 1 percent to as much as 12% over a 92,000 year cycle.
How do you determine whether a line is parallel to a plane? What are some examples?
If a line is parallel to a plane, it will be perpendicular to the plane’s normal vector (just like any other line contained within the plane, or parallel to the plane).(Note that I’m using “perpendicular” here, not in the sense that they intersect, necessarily, but in the sense that their vectors would be at 90 degrees if they were placed next to one another)To find if two vectors are perpendicular, just take their dot product. If it equals 0, then they are perpendicular.So, for example, if we have the plane: 2x + 3y - 4z = 7 (normal vector here would be <2,3,-4>)And we want to find out if the line: x=2+t, y=3–2t, z=5-t, is parallel to it, we just need the dot product of the line’s vector (<1, -2, -1>) and the plane’s normal vector.<1, -2, -1> DOT <2, 3, -4> = 1*2 + -2*3 + -1*-4 = 2 - 6 + 4 = 0So in this case, the line & plane are parallel.If we want to use the same plane, but compare it to the line: x=4+2t, y=3+6t, z=5+9t, then we will get:<2, 6, 9> DOT <2, 3, -4> = 2*2 + 6*3 + 9*-4 = 4 + 18 - 36 = -14So we can see these two will not be parallel.
If a straight line does not intersect the x-axis, does it have any slope?
Maybe I'm too meticulous… But the answer to your question is: It depends, but not on whether or not it crosses the x-axis. You asked about a “straight line” of which there are an infinite amount of straight line orientations.There are horizontal lines, vertical lines, and then everything in between….including diagonal which is a term that could apply to all of the in-betweens.If you have a horizontal line, it cannot intersect the x-axis, as that axis is too, is a horizontal line, but it can fall directly on the x-axis. This would be called an overlap. In any case, the slope of a horizontal line is always zero because it doesn't increase vertically.If you have a vertical line, the slope is always undefined, because there are no horizontal increments and we as humans have not universally defined y/x, where x is zero. It doesn't matter if it crosses the x-axis. Slope is just a term that represents how much a line rises (y) divided by how much it carries (x). If a line has no carry (x=0) then we are left with y/0 for our slope, undefined, unless y is also zero is which case this is not a line at all and beyond the scope of this answer.If you have a diagonal line, then the slope is defined by calculating the change in y value divided by he change in x value between any two or more points on the line. This is true no matter if it crosses the x-axis or not.
How do you determine if a line lies on a plane?
Take the vector equation of a line:[math]\vec{r}(\lambda) = \vec{a} + \lambda \vec{b}[/math]For a given line to lie on a plane, it must be perpendicular to the normal vector of the plane. If we want to know whether a line lies on the plane or not, we need to look at the part which judges its direction - the vector [math]\vec{b}[/math] from the equation I quoted above. If our line lies on the plane, then this vector will be parallel to the plane, meaning it will be perpendicular to a normal vector of that plane. Thus, the dot product of [math]\vec{b}[/math] with the normal vector must be zero:[math]\vec{b} \cdot \vec{N} = 0[/math]Where [math]\vec{b}[/math] is the line’s directional vector, and [math]\vec{N}[/math] is a normal vector to the plane.It’s not enough that the line is parallel to the plane, though - a line can be parallel to the plane, yet still not in it. We must be able to take any point on the line, and any point on the plane, and have the vector between these points be parallel to the plane (and so perpendicular to the normal vector). We can write this in equation form as follows:[math]\Big(\vec{r}(\lambda) - \vec{P_0}\Big) \cdot \vec{N} = 0[/math]Where [math]\vec{r}(\lambda)[/math] is a vector to any given point on the line, [math]\vec{P_0}[/math] is a given point on the plane, and [math]\vec{N}[/math] is a normal vector to the plane.Substituting in the equation of the line:[math](\vec{a} + \lambda \vec{b} - \vec{P_0}) \cdot \vec{N} = 0[/math]If the line is parallel to the plane, you’ll only have to test the equation for one value of [math]\lambda[/math] - and to simplify things, you can choose [math]\lambda = 0[/math]. This would make the equation into:[math](\vec{a} - \vec{P_0}) \cdot \vec{N} = 0[/math]To conclude, if [math]\vec{b} \cdot \vec{N} = 0[/math] (the line is parallel to the plane) then [math](\vec{a} - \vec{P_0}) \cdot \vec{N} = 0[/math] must be true for the line to lie in the plane. If [math]\vec{b} \cdot \vec{N} \neq 0[/math], then the line does not lie in the plane.
Why doesn't the electric field of an infinitely large plane depend on the distance from the plane?
*HI5* EVEN I FOUND IT HARD TO ACCEPT THAT THE FIELD DUE TO AN INFINITE PLANE HAS A DISTANCE INVARIANT ELECTRIC FIELD.You might have proved this by gauss law in school, but if you find it hard to accept that then this is one of my hopefully satisfying explanation:Let us assume consider 2 points P and Q. P is closer to the plane than Q.Differential areas of the plane can be considered point charges of appropriate magnitude. Every point charge is closer to P than to Q and so P is expected to have a higher magnitude of field. The lower value of θ for Q means cosθ is higher for Q and thus the expected higher magnitude of P is compensated.If you're still not satisfied: This is the electric field expression at an axial point due to a disk.As R->∞ we get E=σ/2ɛ. :DTHANKS FOR THE A2A.
How do you determine if vectors are coplanar?
Two vectors are always co-planar. If the triple dot product [math]\vec{a}.(\vec{b}\times\vec{c}) [/math]of vectors [math]\vec{a}[/math], [math]\vec{b}[/math], [math]\vec{c}[/math] is zero, then the vectors are co-planar.
How to find the vector parallel to two planes?
A vector can't be parallel to a plane, since a plane has not a single direction, as a vector does. You can find a vector in the same direction as the intersection of two planes. That vector will be perpendicular to the normal vectors of both planes, and this will be true no matter if the planes intersect or not. The explanation to follow is for planes and vectors in R3. In a general vector space all of these formulas still hold, with the obvious modifications. Plane ax + by + cz = g has normal vector (a,b,c), and likewise dx + ey + fz = h has normal vector (d,e,f). Now find a vector (u,v,w) perpendicular to both of these, by setting (a,b,c) . (u,v,w) = 0 and (d,e,f) . (u,v,w) = 0. These can be combined by solving the matrix equation [a b c] [d e f] . (u,v,w) = 0. You will get a family of vectors in a particular direction as your solution. Any member of this family will be parallel to the direction of the intersection of the two planes, if any, and will be in a skew direction to the two planes, if they are parallel. In this latter case you will get a two-parameter family of vectors as a solution, of course.