Question

A vector q which has magnitude 8 is added to another vector p which lies on the x axsis the resultant of these two vectors is the third vector r which lies on y axis and has magnitude twice of p the value of p is

Answer 1

Answer:

Explanation:

Let the vector with magnitude 8 m be denoted by B , then | B | = B = 8. Let O be the origin of the CO-ordinate plane and P be a point on x- axis such that vector OP represents the vector A. Choose a point Q on the y- axis such that vector PQ = B and vector OQ = 2 A. This way we get a rt. angled triangle OPQ with PQ as its diagonal. Therefore in the rt. angled triangle OPQ ;

PQ^2 = OP^2 + OQ^2. Or. B^2 = A^2. + ( 2 A )^2. ===> 8^2 = A^2 + 4 A^2 Or

64 = 5 A^2 which gives A = 8 / sqrt(5).

Answer 2

Answer: Let the vector with magnitude 8 m be denoted by B , then | B | = B = 8. Let O be the origin of the CO-ordinate plane and P be a point on x- axis such that vector OP represents the vector A. Choose a point Q on the y- axis such that vector PQ = B and vector OQ = 2 A. This way we get a rt. angled triangle OPQ with PQ as its diagonal. Therefore in the rt. angled triangle OPQ ;

PQ^2 = OP^2 + OQ^2. Or. B^2 = A^2. + ( 2 A )^2. ===> 8^2 = A^2 + 4 A^2 Or

64 = 5 A^2 which gives A = 8 / sqrt(5).

Explanation: