Question

What will be the value of the expression \[\left( \widehat{k}\times \widehat{j} \right).\widehat{i}+\widehat{j}.\widehat{k}\] ?

Answer 1

Hint: To solve this question we need to know that,
Cross-product of two vectors $\overrightarrow{a}\,and\,\overrightarrow{b}$ is given by , \[\overrightarrow{\mathbf{a}}\times \overrightarrow{\mathbf{b}}~=|\overrightarrow{\mathbf{a}}||\overrightarrow{\mathbf{b}}|\sin \left( \theta \right)\widehat{n}\]
And dot-product of two vectors $\overrightarrow{a}\,and\,\overrightarrow{b}$ is given by , \[\overrightarrow{\mathbf{a}}.\overrightarrow{\mathbf{b}}~=|\overrightarrow{\mathbf{a}}||\overrightarrow{\mathbf{b}}|\cos \left( \theta \right)\]
Where $\theta $ is the angle between the two vectors. And $\widehat{n}$ is the unit vector in the perpendicular direction of both $\overrightarrow{a}\,and\,\overrightarrow{b}$.

Complete step by step answer:
We know that the unit vectors $\widehat{i},\widehat{j}\,and\,\widehat{k}$ are mutually perpendicular vectors to each other, so the angle between any two of them is ${{90}^{\circ }}$.
And we need to know that the,
Cross-product of two vectors $\overrightarrow{a}\,and\,\overrightarrow{b}$ is given by , \[\overrightarrow{\mathbf{a}}\times \overrightarrow{\mathbf{b}}~=|\overrightarrow{\mathbf{a}}||\overrightarrow{\mathbf{b}}|\sin \left( \theta \right)\widehat{n}\], and
dot-product of two vectors $\overrightarrow{a}\,and\,\overrightarrow{b}$ is given by , \[\overrightarrow{\mathbf{a}}.\overrightarrow{\mathbf{b}}~=|\overrightarrow{\mathbf{a}}||\overrightarrow{\mathbf{b}}|\cos \left( \theta \right)\],
Now, we need to find value of the expression \[\left( \widehat{k}\times \widehat{j} \right).\widehat{i}+\widehat{j}.\widehat{k}\] where $\widehat{i},\widehat{j}\,and\,\widehat{k}$ are mutually perpendicular unit vectors along the x, y and z axis respectively.
Now,
Applying the cross product and dot product to the given vectors, we get
\[\left( \widehat{k}\times \widehat{j} \right).\widehat{i}+\widehat{j}.\widehat{k}=\left( |\widehat{k}||\widehat{j}|\sin \left( {{90}^{\circ }} \right)\left( -\widehat{i} \right) \right).\widehat{i}+\widehat{j}.\widehat{k}\]
Here \[\widehat{n}\] we get as \[-\widehat{i}\] by using the right hand rule on the cross-product \[\left( \widehat{k}\times \widehat{j} \right)\] i.e. when we keep our palm in the direction of $\widehat{k}$ and rotate it in the direction of $\widehat{j}$ then we get the direction of our thumb in the direction of \[-\widehat{i}\].
Also we know that \[|\widehat{i|}=\,|\widehat{j|}=\,|\widehat{k|}=1\] i.e. the magnitude of the unit vector is always 1. So we get
\[\begin{align}
  & \left( \widehat{k}\times \widehat{j} \right).\widehat{i}+\widehat{j}.\widehat{k}=\left( -\widehat{i} \right).\widehat{i}+\widehat{j}.\widehat{k} \\
 & \left( \widehat{k}\times \widehat{j} \right).\widehat{i}+\widehat{j}.\widehat{k}=\left( -\widehat{i} \right).\widehat{i}+|\widehat{j}||\widehat{k}|\cos {{90}^{\circ }} \\
 & \left( \widehat{k}\times \widehat{j} \right).\widehat{i}+\widehat{j}.\widehat{k}=\left( -\widehat{i} \right).\widehat{i}+1\times 1\times 0 \\
 & \left( \widehat{k}\times \widehat{j} \right).\widehat{i}+\widehat{j}.\widehat{k}=-|\widehat{i}|.|\widehat{i}|\cos {{0}^{\circ }}+1\times 1\times 0 \\
 & \left( \widehat{k}\times \widehat{j} \right).\widehat{i}+\widehat{j}.\widehat{k}=-1\times 1\times \cos {{0}^{\circ }}+0 \\
\end{align}\]
As we know $\cos {{0}^{\circ }}=1$ so we get,
\[\left( \widehat{k}\times \widehat{j} \right).\widehat{i}+\widehat{j}.\widehat{k}=-1+0\]
\[\left( \widehat{k}\times \widehat{j} \right).\widehat{i}+\widehat{j}.\widehat{k}=-1\]

Hence the value of the given expression is -1.

Note: You need to always remember the formula for the cross-product and the dot-product mentioned in the solution as they both are very important and useful formulas for vectors and are frequently used in the vector related problems. And you need to be careful while finding out the direction after the cross product as there will be two perpendicular directions to the plane so use the right hand thumb rule for finding the right direction.