
If the three vectors A, B and C satisfy the relation $ \text{A}\cdot \text{B}=0 $ and $ \text{A}\cdot \text{C}=0 $ , then vector A is parallel to
(A) A
(B) B
(C) $ \text{A}\times \text{B} $
(D) $ \text{B}\times \text{C} $
Answer
550.2k+ views
Hint: Use the product rules of vector multiplication to find the angles between the vectors as well as their resultants. There is a difference between cross product and dot product. If the product of two vectors is scalar quantity, the product is called a scalar or dot product. if the product of two vectors is a vector quantity, the product is called a vector or cross product. If two vectors are perpendicular to each other then their scalar product is zero.
Complete step by step solution
As $ \overrightarrow{\text{A}}.\overrightarrow{\text{B}}=0 $ (Given)
i.e. $ \text{AB cos }\!\!\theta\!\!\text{ =0} $
As $ \overrightarrow{\text{A}} $ and $ \overrightarrow{\text{B}} $ are non-zero
So $ \text{cos }\!\!\theta\!\!\text{ =0} $
$ \text{ }\!\!\theta\!\!\text{ }=90{}^\circ $
$ \therefore \overrightarrow{\text{A}} $ is perpendicular to $ \overrightarrow{\text{B}} $
Similarly,
$ \overrightarrow{\text{A}}.\overrightarrow{\text{C}}=0 $ (Given)
$ \begin{align}
& \text{A C cos }\!\!\theta\!\!\text{ }=\text{0} \\
& \text{cos }\!\!\theta\!\!\text{ }=\text{0} \\
& \text{ }\!\!\theta\!\!\text{ }=90{}^\circ \\
\end{align} $
$ \therefore \overrightarrow{\text{A}} $ is perpendicular to $ \overrightarrow{\text{C}} $
Now, from the given options, we need to check whose direction is the same as $ \overrightarrow{\text{A}} $ or is parallel to it.
Option (A) $ \overrightarrow{\text{A}} $ =Here it is the same vector along which we need to find the parallel vector
(B) $ \overrightarrow{\text{C}} $ = As discussed before in $ \overrightarrow{\text{A}}.\overrightarrow{\text{C}}=0 $ , here $ \overrightarrow{\text{A}} $ is perpendicular to $ \overrightarrow{\text{C}} $
(C) $ \overrightarrow{\text{A}}\times \overrightarrow{\text{B}} $ = The vector product of two vectors is perpendicular to both $ \overrightarrow{\text{A}} $ and $ \overrightarrow{\text{B}} $
(D) $ \overrightarrow{\text{B}}\times \overrightarrow{\text{C}} $ =The vector product of $ \overrightarrow{\text{B}} $ and $ \overrightarrow{\text{C}} $ is perpendicular to both $ \overrightarrow{\text{B}} $ and $ \overrightarrow{\text{C}} $ .Also $ \overrightarrow{\text{A}} $ is perpendicular to $ \overrightarrow{\text{B}} $ and $ \overrightarrow{\text{C}} $ . Hence, $ \overrightarrow{\text{B}}\times \overrightarrow{\text{C}} $ is parallel to $ \overrightarrow{\text{A}} $ .
Therefore option (D) is correct.
Note
The cross-product of two vectors gives the direction of the resultant perpendicular to both the vectors and is given by right hand thumb rule. The cross product is used to find a vector which is perpendicular to the plane spanned by two vectors. It has many applications in physics when dealing with the rotating bodies. The dot product essentially tells us how much of the force vector is applied in the direction of the motion vector.
Complete step by step solution
As $ \overrightarrow{\text{A}}.\overrightarrow{\text{B}}=0 $ (Given)
i.e. $ \text{AB cos }\!\!\theta\!\!\text{ =0} $
As $ \overrightarrow{\text{A}} $ and $ \overrightarrow{\text{B}} $ are non-zero
So $ \text{cos }\!\!\theta\!\!\text{ =0} $
$ \text{ }\!\!\theta\!\!\text{ }=90{}^\circ $
$ \therefore \overrightarrow{\text{A}} $ is perpendicular to $ \overrightarrow{\text{B}} $
Similarly,
$ \overrightarrow{\text{A}}.\overrightarrow{\text{C}}=0 $ (Given)
$ \begin{align}
& \text{A C cos }\!\!\theta\!\!\text{ }=\text{0} \\
& \text{cos }\!\!\theta\!\!\text{ }=\text{0} \\
& \text{ }\!\!\theta\!\!\text{ }=90{}^\circ \\
\end{align} $
$ \therefore \overrightarrow{\text{A}} $ is perpendicular to $ \overrightarrow{\text{C}} $
Now, from the given options, we need to check whose direction is the same as $ \overrightarrow{\text{A}} $ or is parallel to it.
Option (A) $ \overrightarrow{\text{A}} $ =Here it is the same vector along which we need to find the parallel vector
(B) $ \overrightarrow{\text{C}} $ = As discussed before in $ \overrightarrow{\text{A}}.\overrightarrow{\text{C}}=0 $ , here $ \overrightarrow{\text{A}} $ is perpendicular to $ \overrightarrow{\text{C}} $
(C) $ \overrightarrow{\text{A}}\times \overrightarrow{\text{B}} $ = The vector product of two vectors is perpendicular to both $ \overrightarrow{\text{A}} $ and $ \overrightarrow{\text{B}} $
(D) $ \overrightarrow{\text{B}}\times \overrightarrow{\text{C}} $ =The vector product of $ \overrightarrow{\text{B}} $ and $ \overrightarrow{\text{C}} $ is perpendicular to both $ \overrightarrow{\text{B}} $ and $ \overrightarrow{\text{C}} $ .Also $ \overrightarrow{\text{A}} $ is perpendicular to $ \overrightarrow{\text{B}} $ and $ \overrightarrow{\text{C}} $ . Hence, $ \overrightarrow{\text{B}}\times \overrightarrow{\text{C}} $ is parallel to $ \overrightarrow{\text{A}} $ .
Therefore option (D) is correct.
Note
The cross-product of two vectors gives the direction of the resultant perpendicular to both the vectors and is given by right hand thumb rule. The cross product is used to find a vector which is perpendicular to the plane spanned by two vectors. It has many applications in physics when dealing with the rotating bodies. The dot product essentially tells us how much of the force vector is applied in the direction of the motion vector.
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