# Write the value of the product of vectors \[\left( {{\text{i}} \times {\text{j}}} \right).{\text{k + }}\left( {{\text{j}} \times {\text{k}}} \right).{\text{i }}\].

Last updated date: 28th Mar 2023

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Answer

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Hint: The cross product two unit vector is the vector perpendicular to both of them. And, the dot product of two unit vectors in the same direction is equal to 1.

We have , an equation \[\left( {{\text{i}} \times {\text{j}}} \right).{\text{k + }}\left( {{\text{j}} \times {\text{k}}} \right).{\text{i }}\]. The vectors i, j and k are the vectors in x-direction , y-direction and the z-direction.

There are three operations involved in these expressions.

1) Cross product of two unit vector

2) Dot product of two unit vector

3) Addition

Cross product of any two unit vector can be given by remembering small sequence

$

{\text{i}} \to {\text{j}} \to {\text{k}} \to {\text{i}} \\

{\text{i}} \times {\text{j = k}} \\

{\text{j}} \times {\text{k = i}} \\

{\text{k}} \times {\text{i = j}} \\

$

Applying it on the given equation, we have

\[\left( {{\text{i}} \times {\text{j}}} \right).{\text{k + }}\left( {{\text{j}} \times {\text{k}}} \right).{\text{i }}\]= k.k + i.i

Now, the dot product of two vectors is equal to the multiplication of length of the two vectors and cosine of the angle between them.

${\text{a}}{\text{.b = |a||b|cos}}\theta $ where $\theta $ is the angle between vectors a and b.

The length of the unit vector is 1. And if angle is 0 degree, then a.b = 1.

So,

\[\left( {{\text{i}} \times {\text{j}}} \right).{\text{k + }}\left( {{\text{j}} \times {\text{k}}} \right).{\text{i }}\]= k.k + i.i

=1 + 1

= 2

The value of \[\left( {{\text{i}} \times {\text{j}}} \right).{\text{k + }}\left( {{\text{j}} \times {\text{k}}} \right).{\text{i }}\]is 2.

Note: In these types of questions, we need to start with solving the brackets , then the whole expression. The cross product of two unit vectors in the same direction is zero. The dot product of two perpendicular unit vectors is zero.

We have , an equation \[\left( {{\text{i}} \times {\text{j}}} \right).{\text{k + }}\left( {{\text{j}} \times {\text{k}}} \right).{\text{i }}\]. The vectors i, j and k are the vectors in x-direction , y-direction and the z-direction.

There are three operations involved in these expressions.

1) Cross product of two unit vector

2) Dot product of two unit vector

3) Addition

Cross product of any two unit vector can be given by remembering small sequence

$

{\text{i}} \to {\text{j}} \to {\text{k}} \to {\text{i}} \\

{\text{i}} \times {\text{j = k}} \\

{\text{j}} \times {\text{k = i}} \\

{\text{k}} \times {\text{i = j}} \\

$

Applying it on the given equation, we have

\[\left( {{\text{i}} \times {\text{j}}} \right).{\text{k + }}\left( {{\text{j}} \times {\text{k}}} \right).{\text{i }}\]= k.k + i.i

Now, the dot product of two vectors is equal to the multiplication of length of the two vectors and cosine of the angle between them.

${\text{a}}{\text{.b = |a||b|cos}}\theta $ where $\theta $ is the angle between vectors a and b.

The length of the unit vector is 1. And if angle is 0 degree, then a.b = 1.

So,

\[\left( {{\text{i}} \times {\text{j}}} \right).{\text{k + }}\left( {{\text{j}} \times {\text{k}}} \right).{\text{i }}\]= k.k + i.i

=1 + 1

= 2

The value of \[\left( {{\text{i}} \times {\text{j}}} \right).{\text{k + }}\left( {{\text{j}} \times {\text{k}}} \right).{\text{i }}\]is 2.

Note: In these types of questions, we need to start with solving the brackets , then the whole expression. The cross product of two unit vectors in the same direction is zero. The dot product of two perpendicular unit vectors is zero.

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