When the angle between forces P and Q is `alpha`, magnitude of the resultant is `5sqrt(P^2+Q^2)`. When th angle between them changes to `(90^@-alpha)`, magnitude of the resultant changes to `3sqrt(P^2+Q^2)`. Prove thet `tan alpha=1/3`.

The resultant of two forces P and Q is 5 sqrt(P^2+Q^2) when the angle between them is alpha . When the angle changes to (90^@-alpha) , the resultant becomes 3sqrt(P^2+Q^2) . Show that, tan alpha =1/3 .

The resultant of two forces of magnitude (x+y) and (x-y) is sqrt(x^2+y^2) . The angle between them is

What change takes place in the value of the resultant of two vectors when the angle between them is increased from 0 to 90^@ ?

The magnitude of the resultant of two forces P and Q acting at a point is (2m+1) sqrt(P^2+Q^2) When the angle between them is alpha , and is (2m-1) sqrt(P^2+Q^2) when the angle is ((pi)/(2)-alpha) . Prove that, tan alpha=(m-1)/(m+1) .

Magnitude of each of two vectors is P. Magnitude of the resultant of the two is also P. angle between the vectors is

When will the magnitude of the resultant of two equal vectors be (i) sqrt(2) times and

The maximum magnitude of the resultant of two vectors, vec(P) and vec(Q) ("where" Pgt Q) is x times the minimum magnitude ot the resultant. When the angle between vec(P) and vec(Q)" ""is" " "theta , the magnitude of the resultant is equal to half the sum of the magnitudes of the two vectors. Prove that, cos theta =(x^2+2)/(2(1-x^2))*

The magnitude of each of the two vectors are P the magnitude of resultant vector of them is also P. The angle between two vectors are:

Resultant of two forces P and Q is sqrt(3) Q . The resultant is inclined at 30^@ with P . Show that, either P =Q or P=2Q.

When will the magnitude of the resultant of two equal vectors be (ii) sqrt(3) times the magnitude of each of them ?