Question

Write the value of \[\left( {\hat k \times \hat j} \right) \cdot \hat i + \hat j \cdot \hat k\].

Answer 1

Hint: To solve the question we must have an idea about the properties of unit vectors along the X, Y, and Z axes. Here we have to apply the formula dot product and cross product of vectors. The obtained value of the given expression must be a scalar.

Complete step-by-step solution:
We know that the cross product of two vectors \[\vec A\] and \[\vec B\] is defined by
\[\vec A \times \vec B = \left| {\vec A} \right|\left| {\vec B} \right|\sin \theta \;\; \hat n\] …………………………………….. (1)
Where,
 \[\theta \] is the angle made by \[\vec A\] with respect to \[\vec B\].
\[\hat n\] is the unit vector perpendicular to \[\vec A\] and \[\vec B\].
Again we know that the magnitude of a unit vector is always unity. As the vectors \[\hat i,\hat j\] and \[\hat k\] are unit vectors along X, Y and Z axes respectively, then
\[\left| {\hat i} \right| = \left| {\hat j} \right| = \left| {\hat k} \right| = 1\] …………………………………….. (2)
As the unit vectors \[\hat i,\hat j\]and\[\hat k\] are mutually perpendicular to each other, \[\hat j\]makes an angle \[\dfrac{\pi }{2}\] with respect to \[\hat k\] but \[\hat k\] makes an angle\[2\pi - \dfrac{\pi }{2} = \dfrac{{3\pi }}{2}\] with respect to \[\hat j\]. Let’s evaluate \[\left( {\hat k \times \hat j} \right)\]. Now applying the formula from eq. (1) we have,
\[\left( {\hat k \times \hat j} \right) = \left| {\hat k} \right|\left| {\hat j} \right|\sin \dfrac{{3\pi }}{2}\hat n\] …………………………………….. (3)
Now substituting the value of eq. (2) in eq. (3), we will get
\[ \left( {\hat k \times \hat j} \right) = \sin \dfrac{{3\pi }}{2}\hat n \\
   = - \hat n \] ……………………………….. (4)
But we know that the \[\hat i\] is mutually perpendicular to both \[\hat j\] and \[\hat k\]. Hence we can replace \[\hat n\] with \[\hat i\] in eq. (4). Thus we will obtain
\[\left( {\hat k \times \hat j} \right) = - \hat i\] ………………………………… (5)
We know that the dot product of two vectors \[\vec A\] and \[\vec B\] is defined by
\[\vec A \cdot \vec B = \left| {\vec A} \right|\left| {\vec B} \right|\cos \theta \] ………………………………… (6)
Now let’s substitute the value of eq. (5) in the expression \[\left( {\hat k \times \hat j} \right) \cdot \hat i + \hat j \cdot \hat k\], we will get
\[\left( {\hat k \times \hat j} \right) \cdot \hat i + \hat j \cdot \hat k = - \hat i \cdot \hat i + \hat j \cdot \hat k\] ……………………………… (7)
Now applying the formulae from eq. (6) the dot product of \[\hat i\] and \[\hat i\] is given by,
\[ \hat i \cdot \hat i = \left| {\hat i} \right|\left| {\hat i} \right|\cos \theta \\
   = 1 \times 1 \times \cos \theta \\
   = \cos \theta \\ \] ……………………………… (8)
But we know that the angle between two equal vectors is \[{0^ \circ }\], so \[\theta = {0^ \circ }\], and then eq. (8) becomes
\[ \hat i \cdot \hat i = \cos {0^ \circ } \\
   = 1 \\ \] ………………………………… (9)
Similarly, as the angle between the unit vectors \[\hat j\] and \[\hat k\] is \[{90^ \circ }\]and applying the formulae from eq. (6) the dot product of \[\hat j\] and \[\hat k\] is given by,
\[ \hat j \cdot \hat k = \left| {\hat j} \right|\left| {\hat k} \right|\cos \theta \\
   = 1 \times 1 \times \cos {90^ \circ } \\
   = 0 \\ \] ……………………………… (10)
Now substituting the value of eq. (9) and eq. (10) in eq. (7), we will get
\[ \left( {\hat k \times \hat j} \right) \cdot \hat i + \hat j \cdot \hat k = - 1 + 0 \\
   = - 1 \\ \] …………………………… (11)
Here we got the value of the expression.

Note: Cross product of two vectors is always a vector. The dot product of two vectors is always a scalar. In alternative methods we can simply use the following formulae to evaluate the given expression.
\[ \hat i \cdot \hat i = \hat j \cdot \hat j = \hat k \cdot \hat k = 1 \\
  \hat i \times \hat i = \hat j \times \hat j = \hat k \times \hat k = 0 \\
  \hat i \cdot \hat j = \hat j \cdot \hat k = \hat k \cdot \hat i = 0 \\
  \hat i \times \hat j = \hat k,\hat j \times \hat k = \hat i,\hat k \times \hat i = \hat j \\
  \hat j \times \hat i = - \hat k,\hat k \times \hat j = - \hat i,\hat i \times \hat k = - \hat j \]