Given that vector c and vector d are non zero vectors such that vector c =x1i + y1j + z1k and vector d =(y1z2-y2z1)+(x1y2 -x2y1)show that the two vectors are perpendicular.

To show that two vectors are perpendicular, we need to demonstrate that their dot product is zero.

Let's calculate the dot product of vector c and vector d:

c · d = (x1i + y1j + z1k) · ((y1z2 - y2z1) + (x1y2 - x2y1))

To calculate the dot product, we distribute and multiply each component:

c · d = (x1 * (y1z2 - y2z1)) + (x1 * (x1y2 - x2y1)) + (y1j * (y1z2 - y2z1)) + (y1j * (x1y2 - x2y1)) + (z1k * (y1z2 - y2z1)) + (z1k * (x1y2 - x2y1))

Now, since i, j, and k are orthogonal unit vectors (perpendicular to each other), their dot products with each other are zero. This means only the terms of the form i · j, j · k, and k · i will yield non-zero values.

Let's simplify the expression:

c · d = 0 + x1(x1y2 - x2y1) + 0 + y1(y1z2 - y2z1) + 0 + z1(y1z2 - y2z1)

c · d = x1²y2 - x1x2y1 + y1²z2 - y1y2z1 + z1y1z2 - z1y2z1

Now, let's rearrange the terms:

c · d = (-x1x2y1 + x1²y2) + (-y1y2z1 + y1²z2) + (z1y1z2 - z1y2z1)

Notice that the first two terms are identical to the second two terms, but with opposite signs. This results in the cancellation of those terms:

c · d = 0

Since the dot product of vector c and vector d is zero, we can conclude that the two vectors are perpendicular.

To summarize, we showed that the dot product of the two vectors, c · d, is zero, which proves that vector c and vector d are perpendicular.