A proton moves with a velocity of vector v = (6ihat - 1jhat + khat) m/s in a region in which the magnetic field is vector B = (ihat + 2jhat - 3khat) T. What is the magnitude of the magnetic force this charge experiences?

answer in Newtons.

I tried F=qvb but the vectors are just stumping me. Any help is welcome!

Compute the determinant:

|i j k|
|6 -1 1|
|1 2 -3|
Then calculate the magnitude of the resulting vector. Then multiply that by e.

Why do you have an unnecessary 1 in front of the velocity j component? Is that a typing error?

For the determinant, I get
(4-2)i + (1+19)k + (12+1)k
= 2i +20j +12k

The force would be 1.6*10^-19*23.4 N

I am still not understanding it the way you explained it but your answer was correct.

You need to learn the method of taking the cross product of two vectors with determinants. That is what I did.

For a brief summary see
http://www.ucl.ac.uk/mathematics/geomath/level2/mat/mat121.html

For the determinant, I get

(4-2)i + (1+19)j + (12+1)k
(a typo error was corrected)
= 2i +20j +12k

The magnitude of that vector is
sqrt[2^2 + 20^2 + 12^2] = sqrt548 = 23.4

The force would be 1.6*10^-19*23.4 N
= 3.74*10^-18 N

To find the magnitude of the magnetic force experienced by the proton, you can use the equation F = qvB sinθ, where q is the charge of the particle, v is its velocity vector, B is the magnetic field vector, and θ is the angle between the velocity vector and the magnetic field vector.

In this case, the velocity vector is given as v = 6i - j + k m/s, and the magnetic field vector is given as B = i + 2j - 3k T.

To find θ, we need to take the dot product of the velocity vector and the magnetic field vector:

v · B = (6i - j + k) · (i + 2j - 3k)
= 6i·i + 6i·j - 6i·k - j·i - j·2j + j·3k + k·i + k·2j - k·3k
= 6 + 6(0) - 6(0) - 0 - 2 + 0 + 0 + 0 - 3
= 1

Since the dot product of two vectors is equal to the product of their magnitudes multiplied by the cosine of the angle between them, we can rearrange the equation to solve for cosθ:

cosθ = (v · B) / (|v| |B|)
cosθ = 1 / (sqrt(6^2 + (-1)^2 + 1^2) * sqrt(1^2 + 2^2 + (-3)^2))
cosθ = 1 / (sqrt(38) * sqrt(14))
cosθ = 1 / (2.912)
cosθ = 0.343

Now, to find sinθ, we can use the fact that cos^2θ + sin^2θ = 1:

sin^2θ = 1 - cos^2θ
sin^2θ = 1 - (0.343)^2
sin^2θ = 0.882
sinθ = sqrt(0.882)
sinθ ≈ 0.939

Let's substitute the values back into the equation F = qvB sinθ:

F = (q)(|v|)(|B|)(sinθ)
F = (1.6 x 10^(-19) C)(sqrt(6^2 + (-1)^2 + 1^2) m/s)(sqrt(1^2 + 2^2 + (-3)^2) T)(0.939)
F = (1.6 x 10^(-19) C)(sqrt(38) m/s)(sqrt(14) T)(0.939)
F ≈ 1.098 x 10^(-18) N

Therefore, the magnitude of the magnetic force experienced by the proton is approximately 1.098 x 10^(-18) Newtons.