Answer
Verified
449.7k+ views
Hint: Use the concept of cross product or vector product of two vectors and find their product then take the modulus of resultant vector to find the value of \[\left| {\overrightarrow a \times \overrightarrow b } \right|\].
Complete step by step solution: Given two vectors,
\[\overrightarrow a = \widehat i + 3\widehat j - 2\widehat k\] and \[\overrightarrow b = 2\widehat i + \widehat j - 3\widehat k\]
To find the cross product and their magnitude
We know that cross product of \[\overrightarrow a = {a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k\] and \[\overrightarrow b = {b_1}\widehat i + {b_2}\widehat j + {b_3}\widehat k\] is equal to
\[\overrightarrow a \times \overrightarrow b = \left( {\begin{array}{*{20}{c}}
{\widehat i}&{\widehat j}&{\widehat k} \\
{{a_1}}&{{a_2}}&{{a_3}} \\
{{b_1}}&{{b_2}}&{{b_3}}
\end{array}} \right)\]
On opening the determinant we get,
\[\overrightarrow a \times \overrightarrow b = \widehat i({a_2}{b_3} - {b_2}{a_3}) - \widehat j({a_1}{b_3} - {b_1}{a_3}) + \widehat k({a_1}{b_2} - {b_1}{a_2})\]
Now put the value of vector a and vector b in above formula
We get,
\[\overrightarrow a \times \overrightarrow b = \left( {\begin{array}{*{20}{c}}
{\widehat i}&{\widehat j}&{\widehat k} \\
1&3&{ - 2} \\
2&1&{ - 3}
\end{array}} \right)\]
\[\overrightarrow a \times \overrightarrow b = \widehat i[3 \bullet \left( { - 3} \right) - 1 \bullet \left( { - 2} \right)] - \widehat j[1 \bullet \left( { - 3} \right) - 2 \bullet \left( { - 2} \right)] + \widehat k[1 \bullet 1 - 2 \bullet 3]\]
On simplification,
\[\overrightarrow a \times \overrightarrow b = - 7i - \widehat j - 5\widehat k\]
Now we have to find the modulus or magnitude of $\overrightarrow a \times \overrightarrow b $
We know that \[\left| {\overrightarrow a } \right| = \sqrt {{a^2}_1 + {a^2}_2 + {a^2}_3} \] of a vector \[\overrightarrow a = {a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k\]
So modulus of $\overrightarrow a \times \overrightarrow b $
On putting the value
\[\left| {\overrightarrow a \times \overrightarrow b } \right| = \sqrt {{{\left( { - 7} \right)}^2} + {{\left( { - 1} \right)}^2} + {{\left( { - 5} \right)}^2}} \]
On simplifying the above we get
\[\left| {\overrightarrow a \times \overrightarrow b } \right| = \sqrt {49 + 1 + 25} \]
\[\left| {\overrightarrow a \times \overrightarrow b } \right| = \sqrt {75} \]
Or \[\left| {\overrightarrow a \times \overrightarrow b } \right| = 5\sqrt 3 \]
Hence \[\overrightarrow a \times \overrightarrow b = - 7i - \widehat j - 5\widehat k\] and \[\left| {\overrightarrow a \times \overrightarrow b } \right| = 5\sqrt 3 \]
Additional information: Cross product or vector product of two vectors is always a vector quantity which has direction as well as the magnitude, while the dot product or scalar product of two vectors is always a scalar quantity which has no direction only magnitude. The modulus of any vector gives its magnitude.
Note: Magnitude or modulus of any vector can not be negative in any condition because it is always a distance from origin.
Complete step by step solution: Given two vectors,
\[\overrightarrow a = \widehat i + 3\widehat j - 2\widehat k\] and \[\overrightarrow b = 2\widehat i + \widehat j - 3\widehat k\]
To find the cross product and their magnitude
We know that cross product of \[\overrightarrow a = {a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k\] and \[\overrightarrow b = {b_1}\widehat i + {b_2}\widehat j + {b_3}\widehat k\] is equal to
\[\overrightarrow a \times \overrightarrow b = \left( {\begin{array}{*{20}{c}}
{\widehat i}&{\widehat j}&{\widehat k} \\
{{a_1}}&{{a_2}}&{{a_3}} \\
{{b_1}}&{{b_2}}&{{b_3}}
\end{array}} \right)\]
On opening the determinant we get,
\[\overrightarrow a \times \overrightarrow b = \widehat i({a_2}{b_3} - {b_2}{a_3}) - \widehat j({a_1}{b_3} - {b_1}{a_3}) + \widehat k({a_1}{b_2} - {b_1}{a_2})\]
Now put the value of vector a and vector b in above formula
We get,
\[\overrightarrow a \times \overrightarrow b = \left( {\begin{array}{*{20}{c}}
{\widehat i}&{\widehat j}&{\widehat k} \\
1&3&{ - 2} \\
2&1&{ - 3}
\end{array}} \right)\]
\[\overrightarrow a \times \overrightarrow b = \widehat i[3 \bullet \left( { - 3} \right) - 1 \bullet \left( { - 2} \right)] - \widehat j[1 \bullet \left( { - 3} \right) - 2 \bullet \left( { - 2} \right)] + \widehat k[1 \bullet 1 - 2 \bullet 3]\]
On simplification,
\[\overrightarrow a \times \overrightarrow b = - 7i - \widehat j - 5\widehat k\]
Now we have to find the modulus or magnitude of $\overrightarrow a \times \overrightarrow b $
We know that \[\left| {\overrightarrow a } \right| = \sqrt {{a^2}_1 + {a^2}_2 + {a^2}_3} \] of a vector \[\overrightarrow a = {a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k\]
So modulus of $\overrightarrow a \times \overrightarrow b $
On putting the value
\[\left| {\overrightarrow a \times \overrightarrow b } \right| = \sqrt {{{\left( { - 7} \right)}^2} + {{\left( { - 1} \right)}^2} + {{\left( { - 5} \right)}^2}} \]
On simplifying the above we get
\[\left| {\overrightarrow a \times \overrightarrow b } \right| = \sqrt {49 + 1 + 25} \]
\[\left| {\overrightarrow a \times \overrightarrow b } \right| = \sqrt {75} \]
Or \[\left| {\overrightarrow a \times \overrightarrow b } \right| = 5\sqrt 3 \]
Hence \[\overrightarrow a \times \overrightarrow b = - 7i - \widehat j - 5\widehat k\] and \[\left| {\overrightarrow a \times \overrightarrow b } \right| = 5\sqrt 3 \]
Additional information: Cross product or vector product of two vectors is always a vector quantity which has direction as well as the magnitude, while the dot product or scalar product of two vectors is always a scalar quantity which has no direction only magnitude. The modulus of any vector gives its magnitude.
Note: Magnitude or modulus of any vector can not be negative in any condition because it is always a distance from origin.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Which are the Top 10 Largest Countries of the World?
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Difference Between Plant Cell and Animal Cell
Give 10 examples for herbs , shrubs , climbers , creepers
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you graph the function fx 4x class 9 maths CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE