a,b and c are mutually perpendicular vectors of equal magnitude. Find the angle between vector a and a+b+c

1 answer

To find the angle between vector a and a+b+c, we can first find the dot product of these two vectors and then use the properties of the dot product to find the angle.

Let's start by finding the dot product of vector a and a+b+c:

a · (a+b+c) = a · a + a · b + a · c

Since a, b, and c are mutually perpendicular vectors, their dot products with each other will be zero:

a · a = |a|^2 = a^2 (magnitude of a squared)
a · b = 0
a · c = 0

So we are left with:

a · (a+b+c) = a^2 + 0 + 0
= a^2

Now, let's find the magnitudes of vector a and vector (a+b+c):

|a| = |a+b+c| (since their magnitudes are equal)

Using the definition of the dot product, we know that:

a · (a+b+c) = |a| * |a+b+c| * cosθ

where θ is the angle between vector a and vector (a+b+c).

From above, we have:

a^2 = |a| * |a+b+c| * cosθ

Dividing both sides by |a|:

a = |a+b+c| * cosθ

Dividing both sides by |a+b+c|:

a / |a+b+c| = cosθ

We can now take the inverse cosine of both sides to solve for θ:

θ = cos^(-1) (a / |a+b+c|)

So, the angle between vector a and vector (a+b+c) is given by the inverse cosine of a divided by the magnitude of (a+b+c).

This equation holds true as long as a, b, and c are mutually perpendicular vectors of equal magnitude.