Q 48

Asked by lekhakarthikeyan | 16th Jul, 2018, 03:58: PM

Expert Answer:

begin mathsize 16px style According space to space the space question comma
Vector space passing space through space top enclose straight r subscript 1 equals top enclose straight a subscript 1 plus straight lambda top enclose straight b subscript 1 space and space top enclose straight r subscript 2 equals top enclose straight a subscript 2 plus straight mu top enclose straight b subscript 2
open parentheses top enclose straight r minus top enclose straight a close parentheses times open parentheses top enclose straight b subscript 1 cross times top enclose straight b subscript 2 close parentheses equals 0
top enclose straight a equals straight i with hat on top plus 2 straight j with hat on top minus 4 straight k with hat on top
top enclose straight b subscript 1 equals 2 straight i with hat on top plus 3 straight j with hat on top plus 6 straight k with hat on top
top enclose top enclose straight b end enclose subscript 2 equals straight i with hat on top plus straight j with hat on top plus straight k with hat on top
Find space top enclose straight b subscript 1 cross times top enclose top enclose straight b end enclose subscript 2 space we space get space minus 3 straight i with hat on top minus 4 straight j with hat on top minus straight k with hat on top
Find space vector space equation space of space straight a space plane
open square brackets top enclose straight r minus open parentheses straight i with hat on top plus 2 straight j with hat on top minus 4 straight k with hat on top close parentheses close square brackets times open parentheses negative 3 straight i with hat on top minus 4 straight j with hat on top minus straight k with hat on top close parentheses equals 0
rightwards double arrow top enclose straight r times open parentheses 3 straight i with hat on top plus 4 straight j with hat on top plus straight k with hat on top close parentheses equals 7
Cartesian space form
open parentheses straight x straight i with hat on top plus straight y straight j with hat on top plus zk with hat on top close parentheses times open parentheses 3 straight i with hat on top plus 4 straight j with hat on top plus straight k with hat on top close parentheses equals 7
3 straight x plus 4 straight y plus straight z minus 7 equals 0
Find space distance space of space the space point space left parenthesis 9 comma negative 8 comma negative 10 right parenthesis space from space the space plane space we space get space straight p space equals space fraction numerator 22 over denominator square root of 26 end fraction units
end style

Answered by Sneha shidid | 17th Jul, 2018, 09:51: AM