Crossing an intersection

2.2.8.85] Show that the curve

intersects the plane

at a right angle.

My intuitive solution is to use the tangent to the curve and set it perpendicular to the given plane, as given in the problem. As such, there are two perpendicular vectors to the plane: the normal of the plane and the tangent of the curve.

The normal of the plane is immediately obvious from the equation of the plane.

The tangent vector is actually more complicated in form because of the scalar denominator representing the length or magnitude of the tangent vector. It will be seen later that this scalar part of the tangent looks complex because of the presence of variable t. Good thing, this scalar is obviously non-zero, regardless of the value of t.

These two vectors, being both perpendicular to the same plane, must be parallel to each other. And so, the cross product of these two vectors must be zero.

Getting the cross product doesn't give us a clear answer immediately because of the presence of the variable t. If the cross product is zero, the scalar part must be zero or the resulting vector is in itself zero. Or both. In this case, the scalar is non-zero which leads us to set up the individual vector components (of the resulting cross product) to be zero. This way, we can solve for t which gives us zero vector components.

At this value of t, we can get the point x,y,z at which the curve intersects the plane perpendicularly. This turns out to be (2,8,8).

The other solution to this is to use the velocity vector of the curve, instead of the tangent vector. They are very similar but the velocity vector is simpler in form. I did in fact use the velocity vector the first time I solved this problem a long time ago. When I came back to this problem this morning, I spontaneously used the tangent vector.

What is the difference between the velocity and tangent approaches? Not much, actually. In my view however, the velocity vector is simpler but the tangent vector approach is more intuitive. Tangent is a concept that has a long history of supporting drills from lower levels of mathematics so I suppose most students will be more comfortable with the tangent concept. Velocity is generally regarded as a physics concept rather than a math concept, unless otherwise a given math problem has expressly used the term velocity. In this particular case, velocity isn't mentioned at all.

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Reference
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Distance education

Here are two problems about distance from a point and a line in space.

1.12.13.52] Find the distance from the point (5,7,14) to the line passing through (2,3,8) and (3,6,12).

1.12.14.52] Determine the shortest distance from the point (3,4,5) to the line through the origin parallel to the vector

Both of these problems can be solved in at least two ways. One is by using pythagorean theorem and the other is by using vector properties, particularly parallel-perpendicular decomposition.

The answers are:





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For solution, email me at agriengineering at yahoo.com

Reference(s):
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What's the point?

If the vector
represents the segment AB, and the midpoint of AB is (2,1), find A and B.

This is one of the problems which is easier solved by utilizing the properties of vectors.

The other solution is the conventional distance formula to solve for the coordinates of A and B but this actually takes longer because there are four unknowns in this case. At least two simultaneous equations may be required.

On the other hand, the vector method hinges on the vector property that its components represent the difference between the coordinates at endpoints. With this, the coordinates of A and B can be solved using a simple linear equation, one unknown at a time.

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Reference:
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