Given two non-collinear vectors a and b, show that a, axb, and (axb)xa are mutually perpendicular.

Also

Express the unit vectors i, j, and k as ordered triplets and show that
i x j = k
what are ordered triplets?

Using components show that
u x v = -v x u for any vectors u and v.

A cross product is always perpendicular to BOTH vectors, unless it the cross product is zero, which happens only for collinear vectors.

Therefore (a x b) is perpendicular to a.

My statement also proves that (axb)x a is perpendicular to both a and a x b

The "proof" they want you to do involves actually computing the cross product of a x b where a = i and b = j. You will end up proving that the cross product is k. Use these "ordered triplet" designations:
i = (1 0 0)
j = (0 1 0)
k = (0 0 1)

i x j = value of the determinant
|i j k|
|1 0 0|
|0 1 0| = 0i + 0j + k

Your last question <<Using components show that u x v = -v x u for any vectors u and v.>>

can be done similarly using arbitrary ordered triplet designations for u and v,
u = ai + bj + ck
v = di + ej + fk
where a, b, c, d, e and f are arbitrary constants and then using the determinant formula to compute the cross products. u x v and v x u

u x v =
|i j k|
|a b c|
|d e f|
= (bf - ce)i + (cd - af)j + (ae - bd)k
Now compare that to v x u =
|i j k|
|d e f|
|a b c| = (ec - bf)i, etc

Advertising, public

To show that the vectors a, axb, and (axb)xa are mutually perpendicular, we need to demonstrate that their dot products are zero.

1. Vector a and axb:
To find the dot product between these two vectors, we simply multiply their corresponding components and sum the results. The dot product of a and b is given by a · (axb) = |a| |(axb)| cos(θ), where |a| and |(axb)| represent the magnitudes of the vectors a and axb, respectively, and θ is the angle between them. Since a and b are non-collinear, the angle between them is 90 degrees (π/2 radians). Therefore, cos(θ) = 0, which implies that the dot product a · (axb) equals zero.

2. Vector axb and (axb)xa:
Similarly, we need to calculate the dot product between these two vectors. The dot product of (axb) and (axb)xa is given by (axb) · [(axb)xa] = |(axb)| |(axb)xa| cos(ϕ), where |(axb)| and |(axb)xa| represent the magnitudes of the vectors (axb) and (axb)xa, respectively, and ϕ is the angle between them. Again, since (axb) and (axb)xa are non-collinear, the angle between them is 90 degrees (π/2 radians), meaning cos(ϕ) = 0. This implies that the dot product (axb) · [(axb)xa] equals zero.

Therefore, we have shown that both a · (axb) and (axb) · [(axb)xa] equal zero, which proves that the vectors a, axb, and (axb)xa are mutually perpendicular.

Now, moving on to the next question. In linear algebra, the unit vectors i, j, and k are typically used to denote the standard basis vectors in a three-dimensional coordinate system. They represent the directions of the x, y, and z axes, respectively.

The ordered triplets for the unit vectors i, j, and k are:

i = (1, 0, 0)
j = (0, 1, 0)
k = (0, 0, 1)

By convention, the order of the components in an ordered triplet corresponds to the x, y, and z coordinates, respectively. So i has a 1 in the x-component and zeros in the y and z components. Similarly, j has a 1 in the y-component and zeros in the x and z components. Lastly, k has a 1 in the z-component and zeros in the x and y components.

Now, to show that i x j = k, we can use the cross product formula between two vectors. The cross product of i and j is given by:

i x j = (i2j3 - i3j2) i + (i3j1 - i1j3) j + (i1j2 - i2j1) k

Substituting the values for i, j, and k into this equation, we get:

i x j = (0)(0) i + (0)(0) j + (1)(1) k = 0i + 0j + 1k = k

Hence, we have shown that i x j equals k.

Lastly, we will use components to show that u x v = -v x u for any vectors u and v.

The cross product of two vectors u = (u1, u2, u3) and v = (v1, v2, v3) is given by:

u x v = (u2v3 - u3v2) i + (u3v1 - u1v3) j + (u1v2 - u2v1) k

Similarly, the cross product of v x u can be calculated as:

v x u = (v2u3 - v3u2) i + (v3u1 - v1u3) j + (v1u2 - v2u1) k

To prove that u x v = -v x u, we need to compare each component.

The i-component of u x v is (u2v3 - u3v2), while the i-component of -v x u is -(v2u3 - v3u2). Taking the negative of the i-component in -v x u gives (-v2u3 + v3u2). We can observe that these two expressions are equal.

Similarly, by comparing the j and k components, we find that they are equal as well.

Therefore, u x v = -v x u holds true for any vectors u and v.