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Question : 17 of 100
Marks:
+1,
-0
Solution:
b=− c=−− r×a=c×a ⇒r×a−c×a=0 ⇒(r−c)×a=0 ∴r−c=λa ⇒r=λa+c ⇒r⋅b=λa⋅b+c⋅b (taking dot
w)
⇒0=λa⋅b+c⋅b ⇒λ(+2−)⋅(−)+(−−)⋅(−)=0 ⇒λ(1−2)+2=0 ⇒λ=2 ∴r=2a+c ⇒r⋅a=2a⋅a+c⋅a [taking dot with
a ]
=2|a|2+a⋅c =2(1+4+1)+(1−2+1) r⋅a=12
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