Find all the vectors that are perpendicular to (1,4,4,1) and (2,9,8,2).
Two vectors are orthogonal to each other when their inner product equals zero.
For example,
(1,-3) and (3,1) are orthogonal because
(1,-3).(3,1)=1*3+(-3)*1=0
If we are looking for a vector, (p,q,r,s) which is orthogonal to the given vectors (1,4,4,1) and (2,9,8,2), we just have to form the dot products and equate them to zero, and subsequently solve for the values of p,q,r and s.
So
(1,4,4,1).(p,q,r,s)=0
=> p+4q+4r+s = 0 ....(1)
(2,9,8,2).(p,q,r,s)=0
=> 2p+9q+4r+s = 0 ....(2)
Now solving the system of 2 equations in 4 unknowns, we arrive at the reduced echelon form:
p 0 4r s=0 ... (1A)
0 q 0 0=0 ...(2A)
which gives us q=0, and p=-4r-s
q and r are free variables (can take on any value) while p is a dependent variable.
Thus the solution vector is
(-4r-s, 0, r, s)
Check:
(1,4,4,1).(-4r-s,0,r,s)
=-4r-s +4*0 +4r +s
=0
(2,9,8,2).(-4r-s,0,r,s)
=-8r-2s +9*0 + 8r + 2s
=0
So the solution vector is orthogonal to the two given vectors.
Note: r and s can take on any finite value.
Well, finding the vectors that are perpendicular to two given vectors involves some mathematical calculations. But let's take a slightly humorous approach, shall we?
Imagine the first vector as a super-speedy rabbit hopping diagonally, and the second vector as a slow-moving turtle trying to catch up. Clearly, they're not going in the same direction. So, we need a third vector that is perpendicular to both of them, to keep things fair.
Let's call this third vector the "Clown Vector" - it adds a touch of humor to the mix. The Clown Vector will be perpendicular to both the rabbit and the turtle, doing its own funny dance in the third dimension.
Now, let's calculate this comical Clown Vector using some math!
To find the vectors that are perpendicular to two given vectors, we can take their cross product. The cross product of two vectors will result in a vector that is perpendicular to both of the original vectors.
Let's find the cross product of the vectors (1,4,4,1) and (2,9,8,2):
1. Take the cross product of the vectors:
(1, 4, 4, 1) × (2, 9, 8, 2)
This can be done by using the formula:
(a, b, c) × (d, e, f) = (bf - ce, cd - af, ae - bd)
2. Apply the formula to find the cross product:
(4*8 - 4*9, 4*2 - 1*8, 1*9 - 1*2)
Simplifying this, we get:
(-4, -4, 7)
3. Therefore, the vector (-4, -4, 7) is perpendicular to both (1,4,4,1) and (2,9,8,2).
So, that is one vector perpendicular to both (1,4,4,1) and (2,9,8,2).
To find the vectors that are perpendicular to two given vectors, you can use the cross product.
Let's call the given vectors A = (1, 4, 4, 1) and B = (2, 9, 8, 2).
1. Calculate the cross product of A and B:
- For a 3D cross product, we can extend A and B to four-dimensional vectors by setting the fourth component to 0.
A = (1, 4, 4, 1, 0) and B = (2, 9, 8, 2, 0)
- Now, calculate the cross product of A and B: A × B = (A2B3 - A3B2, A3B1 - A1B3, A1B2 - A2B1)
2. Calculate the dot product of the resulting cross product and any vector perpendicular to A and B:
- Let's call the unknown vector (x, y, z, w).
- Calculate the dot product of A × B and (x, y, z, w): (x(A2B3 - A3B2) + y(A3B1 - A1B3) + z(A1B2 - A2B1) + w(0)) = 0
- Simplify the equation to find the relationship between x, y, z, and w.
3. Solve for the unknowns to find all possible vectors:
- Based on the equation obtained from the dot product, you can solve for the unknowns x, y, z, and w.
- There will be infinitely many solutions, as there are four unknowns. So, you'll find a family of vectors that are perpendicular to both (1, 4, 4, 1) and (2, 9, 8, 2).
Note: It's important to note that the resulting vectors will not be unique. There will be an infinite number of vectors that are perpendicular to the given vectors.