Question

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

Answer 1

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

= - \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 $$