(a) Find, correct to one decimal place, the angle between (p = begin{pmatrix} 3 \ -1 end{pmatrix}) and (q = begin{pmatrix} 3 \ 4 end{pmatrix}).
(b) ABCD is a square with vertices at A(0, 0), B(2, 0), C(2, 2) and D(0, 2). Forces of magnitude 10 N, 15 N, 20 N and 5 N act along (overrightarrow{BA}, overrightarrow{BC}, overrightarrow{DC}) and (overrightarrow{AD}) respectively. Find the (i) magnitude (ii) direction; of the resultant.
Explanation
(a) (p = begin{pmatrix} 3 \ -1 end{pmatrix} = 3i – j ; q = begin{pmatrix} 3 \ 4 end{pmatrix} = 3i + 4j)
Let the angle be (theta),
(p . q = |p||q| cos theta)
((3i – j).(3i + 4j) = 9 – 4 = 5)
(|3i – j| = sqrt{3^{2} + (-1)^{2}} = sqrt{10})
(|3i + 4j| = sqrt{3^{2} + 4^{2}} = sqrt{25} = 5)
(5 = (sqrt{10})(5) cos theta implies cos theta = frac{sqrt{10}}{10} = 0.3162)
(theta = cos^{-1} 0.3162 = 71.567° approxeq 71.6°)
(b)
(i) Resolving the forces along the x and y axes, we obtain
Resultant (R) = (10i + 20j) N.
Magnitude = (sqrt{10^{2} + 20^{2}} = sqrt{500})
= (10sqrt{5} approxeq 22.4 N)
(ii) Direction : (tan theta = frac{20}{10} = 2)
(theta = tan^{-1} 2 = 63.4°).