Question

If i, j, k are the unit vectors and mutually perpendicular, then the value of [i−j ​j−k​ k−i​] =
(a) 0
(b) 1
(c) −1
(d) none of these

Answer 1

Hint: To solve this problem, we should be aware about the basic concepts of scalar triple product of the vectors. The formula for scalar triple product of vectors is given by [a b c] = $\overrightarrow{a}.(\overrightarrow{b}\times \overrightarrow{c})$. Here, we make use of vector products for the vectors b and c, while we find the dot product of a with the vector product of the vectors b and c.

Complete step-by-step answer:
Before solving this problem, we should be aware that i, j and k are mutually perpendicular unit vectors. Thus, we should also know a few basic properties. For dot product, we have,
i.j = j.k = k.i = 0 -- (1)
i.i = j.j = k.k = 1 -- (2)
Also, in case of vector product, we have,
i $\times$ i = j $\times$ j = k $\times$ k = 0 -- (3)
i $\times$ j = k, j $\times$ k = i, k $\times$ i = j --(4)
Now, in the above problem, we have,
= [i−j ​j−k​ k−i​]
= (i-j). [(j-k) $\times$ (k-i)]

Now, we can the vector product of these vector term in brackets by individually performing vector product of each terms, we get,
= (i-j). [(j $\times$ k) + (j $\times$ (-i)) + (-k $\times$ k) + (-k $\times$ -i)]
= (i-j). [ i + (j $\times$ (-i)) + 0 + j] -- (A)

Now, to evaluate (j $\times$ (-i)), we use the property that $(\overrightarrow{b}\times \overrightarrow{c})=-(\overrightarrow{c}\times \overrightarrow{b})$, thus, we get,
(j $\times$ (-i)) = -(j $\times$ i) = (i $\times$ j) = k
Putting this is in (A), we get,
= (i-j). [i + k + j]
= (i.i) + (i.k) + (i.j) +(-j.i)+(-j.k)+(-j.j)
= 1 + 0 + 0 + 0 + 0 -1
= 0
Hence, the correct answer is (a) 0.

Note: While solving the problems on vectors involving three mutually perpendicular vectors (in this case, we have i, j and k), it is important to remember basic properties of these vectors in relation to dot product and vector product. This is because the results are much more simplified for perpendicular vectors and thus useful while solving.