Question

the magnitudes of two vectors P and Q differ by 1 the magnitude of their resultant makes an angle of tan inverse 3/4 with p the angle between P and Q

Answer 1

Answer:

If the value of tan = 3/4 then the value of perpendicular(p) is 3 and the value of base(b) is 3 + 1 = 4 according to the question hence the value of hypotenuse will be $\sqrt{3^2+4^2}$ which is 5.

Now, to find the angle between p and q we have to use the formulae of triangle law of vector.

5^2 = $\sqrt{p^2+q^2+2pqcosx}$  where x is the angle between  P and Q.

25 = 25 + 2pqcosx.

cosx=0.

x=90.

Answer 2

Answer:

Explanation:

as tan theta is 3/4

then perpendecular of triangle is 3

and base of the triangle is 4

using pytogorous therem h^2=s^2+s^2

then h^2=3^2+4^2,h=5.

Using formulae r^2=a^2+b^2+2ab cos theta

we get 5^2=3^2+4^2+2ab coos theta

we get 25=25+2ab cos theta

then 2ab cos theta=0,cos theta=o

as cos theta is 0 ,then theta =90