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If `vec a=3 hat i- hat j` and `vec b=2 hat i+hat j-3 hat k`, then express `vec b` in the from `vec b=vec b_1+vec b_2`, where `vec b_1|| vec a` and `vec b_2 _|_ vec a`. |
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Answer» Correct Answer - `vec(B)=(vec(b_(1))+vec(b_(2)))`, where `vec(b_(1))=(3/2 hat(i)- 1/2 hat(j))` and `vec(b_(2))=(1/2 hat(i)+3/2 hat(j)-3 hat(k))` Let `vec(b)=vec(b_(1))+vec(b_(2))`, where `vec(b_(1))||vec(a)` and `vec(b_(2)) bot vec(a)`. Let `vec(b_(1))=lambda (3 hat(i)-hat(j))` and let `vec(b_(2))=x hat(i) + y hat(i) + z hat(k)`. Then, `vec(b_(2)) bot vec(a) implies (x hat(i)+ y hat(j)+z hat(k)). (3 hat(i)-hat(j))=0 implies 3x-y =0`. `:. vec(b)=(vec(b)_(1)+vec(b)_(2)) implies (2 hat(i)+hat(j)-3hat(k))=lambda (3hat(i)-hat(j))+(x hat(i)+y hat(j)+ z hat(k))` `implies (2 hat(i)+hat(j)-3hat(k))=(3 lambda + x) hat(i)+(y-lambda) hat(j)+z hat(k)` `implies z=-3, y- lambda =1` and `3 lambda +x=2` `implies z=-3, lambda =(y-1)` and `3(y-1)+x=2` `implies z=-3, lambda =(y-1)` and `3y+x=5`. On solving `3x-y=0` and `x+3y=5`, we get `x=1/2` and `y=3/2, z=-3` and `lambda =1/2`. `:. vec(b_(1))=1/2 (3hat(i)-hat(j))` and `vec(b_(2))=1/2 hat(i) +3/2 hat(j) - 3hat(k)`. |
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