Question

. Explain triangle method of addition of two vectors.​

Answer 1

Answer:

hope it help you.................

Answer 2

Tʀɪᴀɴɢʟᴇ Lᴀᴡ ᴏғ Vᴇᴄᴛᴏʀ Aᴅᴅɪᴛɪᴏɴ:-

Sᴛᴀᴛᴇᴍᴇɴᴛ ᴏғ Tʀɪᴀɴɢʟᴇ Lᴀᴡ:-

If 2 vectors acting simultaneously on a body are represented both in magnitude and direction by 2 sides of a triangle taken in an order then the resultant(both magnitude and direction) of these vectors is given by 3side of that triangle taken in opposite order.

Dᴇʀɪᴠᴀᴛɪᴏɴ ᴏғ ᴛʜᴇ ʟᴀᴡ:-

Consider two vectors P and Q acting on a body and represented both in magnitude and direction by sides OA and AB respectively of a triangle OAB. Let θ be the angle between P and Q. Let R be the resultant of vectors P and Q. Then, according to triangle law of vector addition, side OB represents the resultant of P and Q.

So, we have

R = P + Q

Now, expand A to C and draw BC perpendicular to OC.

From triangle OCB,

${OB }^{2} = {OC}^{2} + {BC}^{2} \\ or \\ {OB }^{2} = {(OA+ AC)}^{2} + {BC}^{2}......(i)$

in triangle ACB

$\cos θ= \frac{AC}{AB} \\or\\ BC=ABsinθ =Qsinθ$

Magnitude of resultant:

Substituting value of AC and BC in (i), we get

${R}^{2} = {(P+ Q \: cosθ)}^{2} + {(Q \sinθ)}^{2} \\ or\\{R}^{2} ={P}^{2} +2PQ\:cosθ+{Q}^{2}{cos}^{2}θ+{Q}^{2} {sin}^{2} θ\\or\\{R}^{2}={P}^{2}+2PQ\:cosθ+{Q}^{2}\\ therefore \\ R=\sqrt{ {P}^{2}+2PQ\:cosθ+{Q}^{2}}$

which is the magnitude of resultant.

Direction of resultant: Let ø be the angle made by resultant R with P. Then,

From triangle OBC,

$\tan∅ = \frac{BC}{OC} =\frac{BC}{OA+AC} \\or,\\tan∅=\frac{Q\sinθ}{P+Q\cosθ}\\therefore\\∅=tan-1(\frac{Q\sinθ}{P+Q\cosθ})$

which is the direction of resultant.