26. Find the angle between the vectors A⃗ = 6î+8ĵ+ 10k̂ and B⃗⃗ = 6î+ 8ĵ- 10k̂.A. θ = 10°B. θ = 25°C. θ = 35°D. θ = 55°E. θ = 90°

The angle θ between two vectors A and B can be found using the dot product formula:

A.B = |A||B|cos(θ)

Where:

• A.B is the dot product of A and B
• |A| and |B| are the magnitudes of A and B
• θ is the angle between A and B

First, let's find the dot product of A and B:

A.B = (66) + (88) + (10*-10) = 36 + 64 - 100 = 0

Next, let's find the magnitudes of A and B:

|A| = sqrt((6^2) + (8^2) + (10^2)) = sqrt(36 + 64 + 100) = sqrt(200)

|B| = sqrt((6^2) + (8^2) + (-10^2)) = sqrt(36 + 64 + 100) = sqrt(200)

Now, we can find the angle θ:

0 = sqrt(200)*sqrt(200)*cos(θ)

0 = 200cos(θ)

cos(θ) = 0/200 = 0

Therefore, θ = cos^-1(0) = 90°

So, the answer is E. θ = 90°.