If veca= hati+2 hatk, vecb= hati+ hatj+ hatk, vecc=7 hati-3 hatk+4 hatk, vec r × vec b + vec b × vec c = vec0 and vec r ⋅ vec a =0 then vec r ⋅ vec c is equal to:

Question

If veca= hati+2 hatk, vecb= hati+ hatj+ hatk, vecc=7 hati-3 hatk+4 hatk, vec r × vec b + vec b × vec c = vec0 and vec r ⋅ vec a =0 then vec r ⋅ vec c is equal to: (A) 34 (B) 12 (C) 36 (D) 30

Tardigrade Admin · Accepted Answer

vec r × vec b - vec c × vec b =0 ⇒( vec r - vec c ) × vec b =0 ⇒ vec r - vec c =λ vec b ⇒ vec r = vec c +λ vec b And given that vec r ⋅ vec a =0 ⇒( vec c +λ vec b ) ⋅ vec a =0 ⇒ vec c ⋅ vec a +λ vec b ⋅ vec a =0 ⇒ λ=(- vec c ⋅ vec a / vec b ⋅ vec a ) Now vec r ⋅ vec c =( vec c +λ vec b ) ⋅ vec c =( vec c -( vec c ⋅ vec a / vec b ⋅ vec a ) vec b ) ⋅ vec c =| vec c |-(( vec c ⋅ vec a / vec b ⋅ vec a ))( vec b ⋅ vec c ) =74-[(15/3)] 8 =74-40=34

Q. If a=i^+2k^,b=i^+j^​+k^,c=7i^−3k^+4k^, r×b+b×c=0 and r⋅a=0 then r⋅c is equal to:

34

12

36

30

r×b−c×b=0 ⇒(r−c)×b=0 ⇒r−c=λb ⇒r=c+λb And given that r⋅a=0 ⇒(c+λb)⋅a=0 ⇒c⋅a+λb⋅a=0 ⇒λ=b⋅a−c⋅a​ Now r⋅c=(c+λb)⋅c =(c−b⋅ac⋅a​b)⋅c =∣c∣−(b⋅ac⋅a​)(b⋅c) =74−[315​]8 =74−40=34

Q. If $a = \hat{i} + 2 \hat{k}, b = \hat{i} + \hat{j} + \hat{k}, c = 7 \hat{i} - 3 \hat{k} + 4 \hat{k}$ , $r \times b + b \times c = 0$ and $r \cdot a = 0$ then $r \cdot c$ is equal to: