a) If vector u and vector v are non-collinear vectors show that vector u, vector u cross product vector v and (vector u cross product vector v) cross product vector u are mutually othogonal.
b) Verify this property using vectors collinear with the unit vector, i, j and k
c) Use tis property to detemine a set of three mutually orthaogonal vectors.
a) To show that vector u, vector u cross product vector v, and (vector u cross product vector v) cross product vector u are mutually orthogonal, we need to prove that the dot product of any two of these vectors is zero.
Let's start by calculating the dot product of vector u and vector u cross product vector v:
(u · (u x v))
The cross product of two vectors can be computed using the determinant formula:
(u x v) = (u2*v3 - u3*v2)i + (u3*v1 - u1*v3)j + (u1*v2 - u2*v1)k,
where u1, u2, u3 are the components of vector u, and v1, v2, v3 are the components of vector v.
Replacing (u x v) with its components, the dot product becomes:
(u · (u x v)) = u1(u2*v3 - u3*v2) + u2(u3*v1 - u1*v3) + u3(u1*v2 - u2*v1).
Now, let's compute the cross product between (u x v) and vector u:
((u x v) x u)
Again, using the determinant formula:
((u x v) x u) = ((u2*v3 - u3*v2)i + (u3*v1 - u1*v3)j + (u1*v2 - u2*v1)) x (u1i + u2j + u3k),
Expanding the cross product:
((u x v) x u) = ((u3*v1 - u1*v3)u3 - (u1*v2 - u2*v1)u2)i + ((u1*v2 - u2*v1)u1 - (u2*v3 - u3*v2)u3)j + ((u2*v3 - u3*v2)u2 - (u3*v1 - u1*v3)u1)k.
Finally, we will compute the dot product of (u x v) with (u x v) x u:
((u x v) · ((u x v) x u)).
((u x v) · ((u x v) x u)) = ((u2*v3 - u3*v2) + (u1*v2 - u2*v1) + (u3*v1 - u1*v3)) * ((u3*v1 - u1*v3)u3 - (u1*v2 - u2*v1)u2) +
(u1*v2 - u2*v1)(u1*v2 - u2*v1) +
(u2*v3 - u3*v2)(u2*v3 - u3*v2) = 0.
Therefore, vector u, vector u cross product vector v, and (vector u cross product vector v) cross product vector u are mutually orthogonal.
b) To verify this property using vectors collinear with the unit vector, i, j, and k, we need to substitute the components of these vectors into the dot product calculations from part a) and demonstrate that the result is zero. Let's consider the following cases:
1) Vector u is collinear with i:
Vector u = u1i, where u1 is any non-zero scalar.
(u · (u x v)) = u1(u2*v3 - u3*v2) = 0 (since u and v are non-collinear).
((u x v) x u) = ((u3*v1 - u1*v3)u3 - (u1*v2 - u2*v1)u2)i = 0 (since (u3*v1 - u1*v3) = 0 and (u1*v2 - u2*v1) = 0).
((u x v) · ((u x v) x u)) = 0 ((u1*v2 - u2*v1)(u1*v2 - u2*v1) + (u2*v3 - u3*v2)(u2*v3 - u3*v2) = 0).
2) Vector u is collinear with j:
Vector u = u2j, where u2 is any non-zero scalar.
(u · (u x v)) = u2(u3*v1 - u1*v3) = 0 (since u and v are non-collinear).
((u x v) x u) = ((u1*v2 - u2*v1)u1 - (u2*v3 - u3*v2)u3)j = 0 (since (u1*v2 - u2*v1) = 0 and (u2*v3 - u3*v2) = 0).
((u x v) · ((u x v) x u)) = 0 ((u2*v3 - u3*v2)(u2*v3 - u3*v2) + (u3*v1 - u1*v3)(u3*v1 - u1*v3) = 0).
3) Vector u is collinear with k:
Vector u = u3k, where u3 is any non-zero scalar.
(u · (u x v)) = u3(u1*v2 - u2*v1) = 0 (since u and v are non-collinear).
((u x v) x u) = ((u2*v3 - u3*v2)u2 - (u3*v1 - u1*v3)u1)k = 0 (since (u2*v3 - u3*v2) = 0 and (u3*v1 - u1*v3) = 0).
((u x v) · ((u x v) x u)) = 0 ((u3*v1 - u1*v3)(u3*v1 - u1*v3) + (u1*v2 - u2*v1)(u1*v2 - u2*v1) = 0).
Therefore, the property holds true for vectors collinear with i, j, and k.
c) To determine a set of three mutually orthogonal vectors, we can use the above property by choosing vectors for u and v that are non-collinear. For example:
Vector u = i
Vector v = j
Applying the property:
(u · (u x v)) = (i · (i x j)) = (i · k) = 0,
((u x v) x u) = ((i x j) x i) = (k x i) = -j,
Therefore, vector u = i, vector v = j, and (vector u cross product vector v) = -j form a set of mutually orthogonal vectors.