a square is defined by the unit vectors i(vector) and j(vector). FInd the projections of i(vector) and j(vector) on each of the diagonals of the square.

To find the projections of the unit vectors i and j on each of the diagonals of the square, we need to use the dot product.

Let's start with the first diagonal of the square, which connects the opposite corners of the square. We can call one of the corners A and the other corner B. The vector AB represents the diagonal.

To find the projection of a vector on another vector, we can use the formula:

Projection of vector A on vector B = (A dot B) / (|B|)

Similarly, the projection of vector B on vector A = (B dot A) / (|A|)

Now, let's find the projection of vector i on the diagonal AB:

To find the dot product of i and AB, we can represent i as a unit vector in terms of i and j vectors.

i = i * (i dot i) + j * (i dot j)

Since i is a unit vector, the dot product of i with itself is 1, and the dot product of i with j is 0, because i and j are perpendicular.

i = i * 1 + j * 0

So, i = i

Now, we can find the projection of i on the diagonal AB:

Projection of i on AB = (i dot AB) / (|AB|)

But since i = i, the projection of i on AB simplifies to:

Projection of i on AB = (i dot AB) / (|AB|) = (i dot i) / (|AB|) = 1 / (|AB|)

Similarly, we can find the projection of vector j on the diagonal AB:

To find the dot product of j and AB, we can represent j as a unit vector in terms of i and j vectors.

j = i * (j dot i) + j * (j dot j)

Since j is a unit vector, the dot product of j with itself is 1, and the dot product of j with i is 0, because i and j are perpendicular.

j = i * 0 + j * 1

So, j = j

Now, we can find the projection of j on the diagonal AB:

Projection of j on AB = (j dot AB) / (|AB|)

But since j = j, the projection of j on AB simplifies to:

Projection of j on AB = (j dot AB) / (|AB|) = (j dot j) / (|AB|) = 1 / (|AB|)

This means that both i and j have equal projections of 1 / (|AB|) on the diagonal AB.

To find the projections on the other diagonal of the square, we can follow a similar approach as above.