There are 3 vectors:

p= 7i + 10j
q= 3i + 12j
r= -i +4j

Could you show me how to find the angle PQR. using the scalar product.

I will associate your vectors to have endpoints

P(7,10), Q(3,12) and R(-1,4)
then vector(PQ) = (-4,2)
vector(QR) = (-4,-8)

then PQ.QR = |PQ||QR|cos(theta), where theta is the angle between them
16-16 = |PQ||QR|cos(theta)
cos(theta) = 0
theta = 90 degrees

(notice the slope of lines PQ and QR are negative reciprocals of each other, so PQ and QR are perpendicular)

To find the angle between two vectors using the scalar product, you can use the formula:

cosθ = (p · q) / (|p| |q|)

where p · q represents the dot product of vectors p and q, and |p| and |q| represent the magnitudes of vectors p and q, respectively.

Let's calculate the dot products and magnitudes for vectors p, q, and r:

p · q = (7 * 3) + (10 * 12) = 21 + 120 = 141
|p| = √((7^2) + (10^2)) = √(49 + 100) = √149

q · r = (3 * -1) + (12 * 4) = -3 + 48 = 45
|q| = √((3^2) + (12^2)) = √(9 + 144) = √153

r · p = (-1 * 7) + (4 * 10) = -7 + 40 = 33
|r| = √((-1)^2 + (4^2)) = √(1 + 16) = √17

Now, substitute these values into the formula:

cosθ = (p · q) / (|p| |q|)
cosθ = 141 / (√149 √153)

Using a calculator, compute the value of cosθ. Then, find the angle by taking the inverse cosine (cos⁻¹) of the computed value.