show that the line joining (1, 10) and (-1, 11) is perpendicular to (3, 5) and (5, 9).
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show that the line joining (1, 10) and (-1, 11) is perpendicular to (3, 5) and (5, 9).
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Answer:
Step-by-step explanation:
Two lines are perpendicular[1] if they intersect at right angles. (Some textbooks have a more relaxed definition of perpendicular in which two lines can be “perpendicular” if one of them is normal to a plane containing the other.[2] However these texts seem to be in the minority as there is a perfectly cromulent word, “orthogonal”,[3] which describes this condition.)
So, can we agree that we need to check whether the lines are orthogonal and that they intersect?
Okay, then. The easy part is verifying they are orthogonal using the dot product.
(B⃗ −A⃗ )⋅(D⃗ −C⃗ )=⟨2,5,−4⟩⋅⟨3,2,4⟩=0
To check whether the two lines intersect, just see if they lie in the same plane and aren’t parallel, in which case they must intersect. So I’ll check whether A, B, C, and D are coplanar.
Let’s start by finding the equation of a plane that contains the first three points:
A (1,−1,2),
B (3,4,−2), and
C (0,3,2).
We can do this by letting vector R⃗ =B⃗ −A⃗ , and vector S⃗ =C⃗ −A⃗ , and then the cross product N⃗ =R⃗ ×S⃗ is normal to the plane.
R⃗ =B⃗ −A⃗ =⟨2,5,−4⟩
S⃗ =C⃗ −A⃗ =⟨−1,4,0⟩
N⃗ =R⃗ ×S⃗ =⟨16,4,13⟩
From the normal vector N⃗ , we get the equation of the plane N⃗ ⋅⟨x,y,z⟩=d, where ⟨x,y,z⟩ represents an arbitrary point of the plane. So we can pick any of our three original points to calculate d=N⃗ ⋅A⃗ = N⃗ ⋅B⃗ = N⃗ ⋅C⃗ . Just picking X⃗ at random, we calculate D as follows:
d=N⃗ ⋅A⃗ =38.
Now we know the values of N⃗ and d, so we can write N⃗ ⋅⟨x,y,z⟩=d as
16x+4y+13z=38,
which is the equation of the plane containing the three given points. Now, plug point D into this equation, and — uh oh,
N⃗ ⋅D⃗ =⟨16,4,13⟩⋅⟨3,5,6⟩=146.
So point D doesn’t lie in the same plane as A, B, and C, so the two lines do not intersect. They are orthogonal skew lines, but not perpendicular by the most commonly accepted definition of the word.
An easier way to check whether four points, A, B, C, and D are coplanar is to make a 3×3 matrix from the x, y, and z coordinates of B⃗ −A⃗ , C⃗ −A⃗ , and D⃗ −A⃗ , and then check the determinant of that matrix. If the determinant is zero, then each of these difference vectors is a linear combination of the other two, which means each pair of vectors determined the same plane, which is another way of saying all three of the difference vectors lie in the same plane.
In this case, the matrix is
∣∣∣∣ 2−12546−404 ∣∣∣∣=108
The determinant is not zero, so the four points are not coplanar.
Mathworld suggests a clever approach to determining whether four points are coplanar: calculate the volume of the tetrahedron[4] defined by the four points, which is a sixth of the absolute value of the determinant of the following matrix. If and only if that volume is zero, the points are coplanar.[5]
V = 13! ∣∣∣∣∣∣ x1x2x3x4y1y2y3y4z1z2z3z41111 ∣∣∣∣∣∣
For our four points,
V = 13! ∣∣∣∣∣∣ 1303−14352−2261111 ∣∣∣∣∣∣=−18
So, again, this method shows us that our points are not coplanar.