© examsiri.com
Question : 10 of 38
Marks:
+1,
-0
Solution:
OB×OC=‌OB×(OB−λOA) =‌(OA×OB)....(i)
Since,
OAâ‹…OB=0 therfore
OA⟂rOB ∴|OB×OC|=|‌(OA×OB)|  from(i)
=‌|OA|OB|sin‌ ⇒‌=‌×3×3 ⇒λ=1( given
λ>0) So,
=‌ =‌(−−4+) (a) Projection
OCon
OA =‌=‌=−‌ (b) Area of
△DAB =‌|OA×OB|=‌ (b) Area of
△DAB =‌|OA×OB|=‌ (c) Area of
△ABC =‌|×|=|| =‌|6−3−6|=‌ Let acute angle between diagonals is
θ then
∴‌‌θ≠‌
© examsiri.com
Go to Question: