© examsiri.com
Question : 33 of 160
Marks:
+1,
-0
Solution:
We have,
|a|=|b|=|c|=1a⋅b‌=|a||b|cos‌‌=(I)(I)(‌)=‌b⋅c‌=|b||c|cos‌‌=(I)(I)(‌)=‌And
c⋅a=|c||a|cos‌=(1)(1)(‌)=‌‌[abc]=a⋅(b×c)‌‌‌=|a||b×c|cos‌‌=(l)|b×c|(‌)=‌|b×c|Now,
|b×c|=|b||c|sin‌‌‌=(1)(1)(‌)=‌∴‌‌[abc]‌=‌×‌=‌Given,
xa+yb+zc=p(b×c)+q(c×a)+r(a×b)∴‌x(a⋅a)+y(b⋅a)+z(c⋅a)‌=p((b×c)⋅a)⇒x+y(‌)+z(‌)=p[abc]⇒2x+y+z=‌pSimilarly,
‌x(a⋅b)+y(b⋅b)+z(c⋅b)=q((c×a)⋅b)‌⇒‌‌x(‌)+y+z(‌)=q[abc]‌⇒‌‌x+2y+z=‌q‌‌...(ii)And
x(a⋅c)+y(b⋅c)+z(c⋅c)=r[abc]‌⇒‌‌x(‌)+y(‌)+z=r(‌)‌⇒‌‌x+y+2z=‌rOn adding Eqs. (i), (ii) and (iii), we get
‌4(x+y+z)=‌(p+q+r)⇒‌=‌
© examsiri.com
Go to Question: