Assuming the plane $4x-3y+7z=0$ to be horizontal, find the equation of the line of greatest slope through the point $(2,1,1)$ in the plane $2x+y-5z=0$
[Math] Line of greatest slope
3d
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Well, there's a few different ways to define a plane. There's the way you will know which is a plane is defined by a point and two vectors in the plane $\mathbf r = \mathbf a + \lambda \mathbf b + \mu \mathbf c$. That's all you need to express everything we need to know about the plane in order to define it uniquely, but there's a way of simplifying this using the cross-product (if you haven't done the cross product, its an operation which combines two vectors to generate a third vector that is perpendicular to both).
If we take the cross product of the two lines in the plane, we get a (nearly) unique vector that's perpendicular to the plane. I say nearly because the vector could point the other way. Anyway, since this vector is perpendicular to the plane, we can say that the dot product of our vector with any vector in the plane is zero. So this gives us another way of expressing a plane. If $\mathbf r$ is the position vector of a general point in the plane, $\mathbf a$ is the position vector of a specific point in the plane, and $\mathbf v$ is a vector perpendicular to the plane, then:
$$ (\mathbf r - \mathbf a)\cdot \mathbf v = 0$$
We can link this back to our original equation by noting that $\mathbf v = \mathbf b \times \mathbf c$ is a vector perpendicular to the plane.
There's one final way of defining a plane - in Cartesian coordinates. It doesn't involve vectors though, and you'll probably come across it soon if you haven't already, so I won't explain it.
The diagram is a 3-D representation of a hill with:-
(1) $ABCD$ is a horizontal plane; (2) $ABQP$ is the slope of the hill; and (3) its vertical height $= QC = PD = h$.
To go up the hill from the bottom (any point on the line $AB$) to the top (any point on the line $PQ$), one has the choice of using routes $AP$ (equivalently $BQ$) or $AQ$. Obviously, following the route $AP$ is the hard way. The following explain why it is hard.
[Hope that I need not explain why those angles marked as right-angled are actually right angles.]
AP is inclined at an angle $\alpha$ to the horizontal and that of $AQ$ is $\beta$. Physically, there is no slope can have an inclination that is greater than $90^0$. Hence, we can safely assume that both $\alpha$ and $\beta$ are less than $90^0$.
In $⊿APD$, the gradient of $AP = \tan \alpha = h / x$ and in $⊿AQC$, that of $AQ$ is $\tan \beta = h / z$
By considering $⊿ABC$, $z > x$ because $z$ is the hypotenuse. This means $\tan \alpha > \tan \beta$ and further means $\alpha > \beta$ [This is because $\tan \theta$ is an increasing function when $\theta$ is acute] . Thus, we say $AP$ is a line having a greater slope.
But which line has the greatest slope? To explain it, we need :-
(1) "the line of intersection of two planes"
$AB$ serves as an example because it is the intersection of plane $ABCD$ and plane $ABQP$
(2) “the angle between two planes M and N, say” … Four conditions must be satisfied:-
(I) One of the two lines forming the angle is from plane M and the other from plane N.
(II) Each of these two lines must be perpendicular to “the “line of intersection of plane M and N”.
(III) These two lines must meet at the same point X, say.
(IV) X must be on “the “line of intersection of plane M and N”.
$\alpha$ happens to the one meeting all of the above requirements.
$AP$ is then the line of greatest slope because it has the properties of (i) a line on the slope and (ii) the angle it inclined to the horizontal plane is the greatest (than any other lines like $AQ$, as proved above).
Sometimes when the slope is so steep that ordinary vehicles cannot go from the bottom (point A) to the top (point P’) directly along the line of the greatest slope (APP’), we have to built Zigzag-type roads (from A to Q and then Q to P’) for them.
Best Answer
It is indeed a good idea to determine first the intersection of the two planes. So let us solve:
$$4x-3y+7z=0$$ $$2x+y-5z=0$$
Eliminating $x$ yields $5y = 17z$. Substituting this result back in either of the two equations, we obtain the following result for the straight line which is the intersection of the two planes:
$$(x,y,z) = C * (4,17,5)$$
We observe that this line defines the direction in the second plane with minimal slope (namely horizontal = zero slope) to the first plane. Therefore we must find the vector that is orthogonal to this line. This vector must also be orthogonal to the normal of the second plane. Hence the vector must satisfy these two equations:
$$4x+17y+5z=0$$ $$2x+y-5z=0$$
Eliminating $x$ tields $y = -z$. Substituting this back into our equations yields the vector $(3,-1,1)$. Therefore the line through $(2,1,1)$ with the greatest slope is given by:
$$(x,y,z) = (2,1,1) + L * (3,-1,1)$$