© examsiri.com
Question : 71 of 150
Marks:
+1,
-0
Solution:
‌‌ Given, ‌b×c=b×a‌⇒b×(c−a)=0‌⇒b‌ is parallel to ‌(c−a)‌⇒c−a=λb‌ for some scalar ‌λ‌⇒c=a+λb...(i)
‌⇒c⋅a=a⋅a+λ(b⋅a)‌⇒0=|...|2+λ(b⋅a)‌‌...[∵c⋅a=0‌ (given) ‌]‌⇒0=6+4λ‌⇒λ=−‌ Substituting the value of
λ in (i), we get
c‌=(+2−)−‌(+−)‌=−‌(−−)∴‌‌c⋅b‌=−‌(−−)⋅(+−)‌=−‌(1−1+1)=−‌
© examsiri.com
Go to Question: