Question

If\[\,\,a=i-j\]\[,b=j-k\]\[,c=k-i\]. If \[d\]be a unit vector such that\[\,a.d=0=\left[ b,c,d \right]\], then\[d\]is equal to\[?\]
A. \[\pm \dfrac{i+j-k}{\sqrt{3}}\]
B. \[\pm \dfrac{i+j-2k}{\sqrt{6}}\]
C. \[\pm \dfrac{i+j+k}{\sqrt{3}}\]
D. \[\pm k\]
E. \[i+j\]

Answer 1

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

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

So, the correct answer is “Option B”.

Note: Here in the given question it is written that \[\left[ b,c,d \right]=0\] that means \[b\]\[,c\]and \[d\]are 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 \[\,\,d\]that is \[\widehat{d}\] and use this method to solve similar problems.