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Let  P and Q  be two vectors in space.  In the plane that is formed by P and Q vectors,  draw them so that their starting point is the same at O.  The angle between them is Ф  (Ф < 180 deg.)

In order to know the resultant vector of combining P and Q, we construct a parallelogram with sides OA and OB whose lengths are equal to the magnitudes of the vectors P and Q respectively.  The angle AOB is the angle Ф. 

Then the length of diagonal OC is equal to the magnitude of A + B.  The direction of the resultant is given by the direction of diagonal OC.
The magnitude of vector P = 100 units and that of Q = 200 units.  The angle between them be  Ф.  Then construct a parallelogram AOBC  such that OA = 5 cm and OB = 10 cm.  with the angle Ф between them.  Complete the parallelogram.  Measure the length d  of OC - diagonal thro O..    Then  20 * d will be the magnitude of the resultant vector.   Here 20 is the scale of drawing..
We can find the formula for the resultant vector magnitude as:
   Draw a perpendicular CE  from C on to OB (extend if required). 

   OC² = OE² + EC² = (OB + BC Cos Ф)² + (BC Sin Ф)²
         = OB² + BC² + 2 OB * BC Cos Ф
           = OB² + OA² + 2 OA * OB * Cos Ф

   Hence,  R² = P² + Q² + 2 P Q Cos Ф

triangle law of vector addition

   Let vectors P and Q be drawn in a different way.  First draw OA vector with its magnitude equal to vector P.  Then from the end point of the vector OA, draw the vector AB such that its magnitude is equal to that of vector Q.  The direction of OA is parallel to that of P.  Direction of AB is parallel to vector Q.  The angle between the two vectors  P and Q  is preserved between OA and AB.

   The complete the triangle  OAB by joining OB.  The vector OB gives the magnitude and direction of the resultant of the two vectors P and Q. 
         vector OB  = vector P + vector Q = vector OA + vector AB

          we can prove that  OB² = P² + Q² + 2 P Q Cos Ф

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