
If \[\overrightarrow A = \widehat i\left( {A\,\cos \theta } \right) - \widehat j\left( {A\,\sin \theta } \right)\] be any vector. Another vector \[\overrightarrow B \], which is normal to \[\overrightarrow A \] can be expressed as:
A. \[\widehat iB\,\cos \theta - \widehat jB\,\sin \theta \]
B. \[\widehat iB\,\cos \theta + \widehat jB\,\sin \theta \]
C. \[\widehat iB\,\sin \theta - \widehat jB\,\cos \theta \]
D. \[\widehat iB\,\sin \theta + \widehat jB\,\cos \theta \]
Answer
504.9k+ views
Hint:As we know that if two vectors are normal to each other, their dot product is equal to zero. So here, we will do a dot product of \[\overrightarrow A \]with the given options, and see which of these gives the zero magnitude.
Formula used:
\[a.b = \left| a \right|\left| b \right|\cos \theta \]
Complete step by step answer:
As we know that if two vectors are normal their dot product comes out be 0 using the formula: \[a.b = \left| a \right|\left| b \right|\cos \theta \]
where, \[\left| a \right|\]is the magnitude (length) of vector \[\overrightarrow A \], \[\left| b \right|\] is the magnitude (length) of the vector \[\overrightarrow B \], \[\theta \] is the angle between \[\overrightarrow A \] and \[\overrightarrow B \] .
The formula for the dot product in terms of vector components (unit vector) would make it easier to calculate the dot product between two given vectors. The dot product is also known as the Scalar product. Since here vector is represented in the form of unit vector another formula can be used that is \[a \cdot b = {a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}\].
Now, doing dot product of \[\overrightarrow A \] with all the possible options of \[\overrightarrow B \], we can see that in option C:
\[\overrightarrow A .\overrightarrow B = A\cos \theta \times B\sin \theta + A\sin \theta \times \left( { - B\cos \theta } \right)\]
\[ \therefore AB\cos \theta \sin \theta - AB\sin \theta \cos \theta = 0\]
Therefore, option C is correct.
Note: Here vector is represented in its vector components (unit vectors). They are shown with an arrow \[\overrightarrow A \] . \[\widehat A\] denotes a unit vector. If we want to change any vector in a unit vector, divide it by the vector’s magnitude.
Formula used:
\[a.b = \left| a \right|\left| b \right|\cos \theta \]
Complete step by step answer:
As we know that if two vectors are normal their dot product comes out be 0 using the formula: \[a.b = \left| a \right|\left| b \right|\cos \theta \]
where, \[\left| a \right|\]is the magnitude (length) of vector \[\overrightarrow A \], \[\left| b \right|\] is the magnitude (length) of the vector \[\overrightarrow B \], \[\theta \] is the angle between \[\overrightarrow A \] and \[\overrightarrow B \] .
The formula for the dot product in terms of vector components (unit vector) would make it easier to calculate the dot product between two given vectors. The dot product is also known as the Scalar product. Since here vector is represented in the form of unit vector another formula can be used that is \[a \cdot b = {a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}\].
Now, doing dot product of \[\overrightarrow A \] with all the possible options of \[\overrightarrow B \], we can see that in option C:
\[\overrightarrow A .\overrightarrow B = A\cos \theta \times B\sin \theta + A\sin \theta \times \left( { - B\cos \theta } \right)\]
\[ \therefore AB\cos \theta \sin \theta - AB\sin \theta \cos \theta = 0\]
Therefore, option C is correct.
Note: Here vector is represented in its vector components (unit vectors). They are shown with an arrow \[\overrightarrow A \] . \[\widehat A\] denotes a unit vector. If we want to change any vector in a unit vector, divide it by the vector’s magnitude.
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