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Question: If\[\,\,a=i-j\]\[,b=j-k\]\[,c=k-i\]. If \[d\]be a unit vector such that\[\,a.d=0=\left[ b,c,d \right...

If\,\,a=i-j$$$$,b=j-k$$$$,c=k-i. If ddbe a unit vector such that a.d=0=[b,c,d]\,a.d=0=\left[ b,c,d \right], thenddis equal to??
A. ±i+j−k3\pm \dfrac{i+j-k}{\sqrt{3}}
B. ±i+j−2k6\pm \dfrac{i+j-2k}{\sqrt{6}}
C. ±i+j+k3\pm \dfrac{i+j+k}{\sqrt{3}}
D. ±k\pm k
E. i+ji+j

Explanation

Solution

Here in the question it is given that [b,c,d]=0\left[ b,c,d \right]=0that means the vector b,c\,b,candddare the coplanar of the vector that is d=b+λc\,d=b+\lambda c. After solving this equation we will get the value of ddin terms ofλ\lambda .
And using this condition is given in the question is  a.d=0\,\,a.d=0and find the unit vector of d\,d that is d^=d∣d∣\,\widehat{d}=\dfrac{d}{\left| d \right|}.

Complete step by step answer:
According to the question \,\,a=i-j$$$$,b=j-kand  c=k−i\,c=k-i
According to the given conditions [ b,c,d]=0\left[ \,b,c,d \right]=0that is a vector and is the coplanar of the vector.
Then the equation is given by,
d=b+λc−−−−(1)d=b+\lambda c----(1)
After substituting the values of  b=j−k\,b=j-k and c=k−i\,c=k-iin equation (1)(1)
d=(j−k)+λ(−i+k)d=(j-k)+\lambda (-i+k)
After further simplifying this equation you will get:
d=(−λ)i+j+(λ−1)k−−−−(2)d=(-\lambda )i+j+(\lambda -1)k----(2)
So in this equation you will get the values of  d\,din terms of λ\,\lambda . To find the value of λ\,\lambda
By using the condition which is given in the question.
a.d=0−−−−(3)a.d=0----(3)
After substituting the values of  a=i−j\,\,a=i-jand equation (2)(2) in equation (3)(3)
(i−j).((−λ)i+j+(λ−1)k)=0(i-j).((-\lambda )i+j+(\lambda -1)k)=0
After simplifying this equation you will get:
(i−j+0k).((−λ)i+j+(λ−1)k)=0(i-j+0k).((-\lambda )i+j+(\lambda -1)k)=0
By using the multiplication property of dot product you will get:
−λ−1=0-\lambda -1=0
λ=−1   −−−−(4)\lambda =-1\,\,\,----(4)
After substituting the value of equation (4)(4)in equation (2)(2) you will get:
d=i+j−2k−−−(5)d=i+j-2k---(5)
∣d∣=6−−−−(6)\left| d \right|=\sqrt{6}----(6)
To find the unit vector of ddis represented as d^\widehat{d}
By using this formula is given by
d^=d∣d∣−−−−(7)\widehat{d}=\dfrac{d}{\left| d \right|}----(7)
After substituting the values of equation(5)(5) and equation (6)(6) in equation(7)(7)
Unit vector of ddthat is d^\widehat{d}is equal to:
d^=i+j−2k6\widehat{d}=\dfrac{i+j-2k}{\sqrt{6}}

So, the correct answer is “Option B”.

Note: Here in the given question it is written that [b,c,d]=0\left[ b,c,d \right]=0 that means b$$$$,cand ddare the coplanar vectors. Coplanar vectors are defined as vectors which are lying on the same in a three-dimensional plane. The vectors are parallel to the same plane. Any two random vectors on a plane can always be found to be coplanar. When the scalar product of three vectors equals zero, they are said to be coplanar. And remember the formula for the unit vector   d\,\,dthat is d^\widehat{d} and use this method to solve similar problems.