Question

Which of the following pairs of vectors are parallel?
\[\begin{align}
  & \text{A}\text{. }\vec{A}=\hat{i}-2\hat{j};\vec{B}=\hat{i}-5\hat{j} \\
 & \text{B}\text{. }\vec{A}=\hat{i}-10\hat{j};\vec{B}=2\hat{i}-5\hat{j} \\
 & \text{C}\text{. }\vec{A}=\hat{i}-5\hat{j};\vec{B}=\hat{i}-10\hat{j} \\
 & \text{D}\text{. }\vec{A}=\hat{i}-5\hat{j};\vec{B}=2\hat{i}-10\hat{j} \\
\end{align}\]

Answer 1

Hint: We are given four pairs of vectors and are asked to find which of the pairs are parallel to each other. To find which of them are parallel, we know that when two vectors are parallel to each other their cross product will be zero. Thus by finding the cross product in all the four cases we will get the solution.
Formula used:
\[\vec{A}\times \vec{B}=\left| A \right|\left| B \right|\sin \theta \]

Complete answer:
In the question we are given four pairs of vectors and we are asked to find which of them are parallel.
We know that when two vectors are parallel to each other their cross product will be zero, i.e. if we have two vectors ‘$\vec{A}$’ and ‘$\vec{B}$’, when they are parallel to each other the angle ‘$\text{ }\!\!\theta\!\!\text{ }$’ between them will be zero, thus we will get,
$\vec{A}\times \vec{B}=\left| A \right|\left| B \right|\sin 0$
Since $\sin 0=0$,
$\Rightarrow \vec{A}\times \vec{B}=\left| A \right|\left| B \right|\times 0$
$\Rightarrow \vec{A}\times \vec{B}=0$
Let us consider the first case given in the question.
The two vectors given are,
\[\vec{A}=\hat{i}-2\hat{j}\text{ and }\vec{B}=\hat{i}-5\hat{j}\]
By taking the cross product of this we will get,
$\vec{A}\times \vec{B}=\left( \hat{i}-2\hat{j} \right)\times \left( \hat{i}-5\hat{j} \right)$
\[\Rightarrow \vec{A}\times \vec{B}=\left| \begin{matrix}
   i & j & k \\
   1 & -2 & 0 \\
   1 & -5 & 0 \\
\end{matrix} \right|\]
\[\Rightarrow \vec{A}\times \vec{B}=\hat{i}\left( \left( -2\times 0 \right)-\left( 0\times 5 \right) \right)-\hat{j}\left( \left( 1\times 0 \right)-\left( 1\times 0 \right) \right)+\hat{k}\left( \left( 1\times -5 \right)-\left( -2\times 1 \right) \right)\]
\[\Rightarrow \vec{A}\times \vec{B}=\hat{i}\left( 0 \right)-\hat{j}\left( 0 \right)+\hat{k}\left( -5+2 \right)\]
\[\Rightarrow \vec{A}\times \vec{B}=\hat{k}\left( -3 \right)\]
$\therefore \vec{A}\times \vec{B}=-3\hat{k}$
Therefore we can see that the cross product here is not zero.
Hence we can say that \[\vec{A}=\hat{i}-2\hat{j}\text{ and }\vec{B}=\hat{i}-5\hat{j}\] are not parallel.
Now let us consider the second case.
The vectors given are,
\[\vec{A}=\hat{i}-10\hat{j}\text{ and }\vec{B}=2\hat{i}-5\hat{j}\]
By taking cross product, we will get
$\vec{A}\times \vec{B}=\left( \hat{i}-10\hat{j} \right)\times \left( 2\hat{i}-5\hat{j} \right)$
\[\Rightarrow \vec{A}\times \vec{B}=\left| \begin{matrix}
   i & j & k \\
   1 & -10 & 0 \\
   2 & -5 & 0 \\
\end{matrix} \right|\]
\[\Rightarrow \vec{A}\times \vec{B}=\hat{i}\left( \left( -10\times 0 \right)-\left( 0\times 5 \right) \right)-\hat{j}\left( \left( 1\times 0 \right)-\left( 0\times 2 \right) \right)+\hat{k}\left( \left( 1\times -5 \right)-\left( 2\times -10 \right) \right)\]
\[\Rightarrow \vec{A}\times \vec{B}=\hat{i}\left( 0 \right)-\hat{j}\left( 0 \right)+\hat{k}\left( -5-\left( -20 \right) \right)\]
\[\Rightarrow \vec{A}\times \vec{B}=\hat{k}\left( -5+20 \right)\]
$\therefore \vec{A}\times \vec{B}=15\hat{k}$
Here also we can see that the cross product is not zero.
Therefore we can say that \[\vec{A}=\hat{i}-10\hat{j}\text{ and }\vec{B}=2\hat{i}-5\hat{j}\]are also not parallel.
In the third case the two vectors given are,
\[\vec{A}=\hat{i}-5\hat{j}\text{ and }\vec{B}=\hat{i}-10\hat{j}\]
By taking the cross product of these two vectors, we will get
$\vec{A}\times \vec{B}=\left( \hat{i}-5\hat{j} \right)\times \left( \hat{i}-10\hat{j} \right)$
\[\Rightarrow \vec{A}\times \vec{B}=\left| \begin{matrix}
   i & j & k \\
   1 & -5 & 0 \\
   1 & -10 & 0 \\
\end{matrix} \right|\]
\[\Rightarrow \vec{A}\times \vec{B}=\hat{i}\left( \left( -5\times 0 \right)-\left( 0\times -10 \right) \right)-\hat{j}\left( \left( 1\times 0 \right)-\left( 0\times 1 \right) \right)+\hat{k}\left( \left( 1\times -10 \right)-\left( -5\times 1 \right) \right)\]
\[\Rightarrow \vec{A}\times \vec{B}=\hat{i}\left( 0 \right)-\hat{j}\left( 0 \right)+\hat{k}\left( -10-\left( -5 \right) \right)\]
\[\Rightarrow \vec{A}\times \vec{B}=\hat{k}\left( -10+5 \right)\]
$\therefore \vec{A}\times \vec{B}=-5\hat{k}$
Therefore the cross product of the two vectors \[\vec{A}=\hat{i}-5\hat{j}\text{ and }\vec{B}=\hat{i}-10\hat{j}\] is also not equal to zero. Hence they are also not parallel.
In the fourth case we are given the two vectors as \[\vec{A}=\hat{i}-5\hat{j}\text{ and }\vec{B}=2\hat{i}-10\hat{j}\].
By taking its cross product we will get,
$\vec{A}\times \vec{B}=\left( \hat{i}-5\hat{j} \right)\times \left( 2\hat{i}-10\hat{j} \right)$
\[\Rightarrow \vec{A}\times \vec{B}=\left| \begin{matrix}
   i & j & k \\
   1 & -5 & 0 \\
   2 & -10 & 0 \\
\end{matrix} \right|\]
\[\Rightarrow \vec{A}\times \vec{B}=\hat{i}\left( \left( -5\times 0 \right)-\left( 0\times -10 \right) \right)-\hat{j}\left( \left( 1\times 0 \right)-\left( 0\times 2 \right) \right)+\hat{k}\left( \left( 1\times -10 \right)-\left( -5\times 2 \right) \right)\]
\[\Rightarrow \vec{A}\times \vec{B}=\hat{i}\left( 0 \right)-\hat{j}\left( 0 \right)+\hat{k}\left( -10-\left( -10 \right) \right)\]
\[\Rightarrow \vec{A}\times \vec{B}=\hat{k}\left( -10+10 \right)\]
\[\Rightarrow \vec{A}\times \vec{B}=\hat{k}\left( 0 \right)\]
$\therefore \vec{A}\times \vec{B}=0$
Therefore we can see that the cross product of \[\vec{A}=\hat{i}-5\hat{j}\text{ and }\vec{B}=2\hat{i}-10\hat{j}\] is zero.
We know that when the cross product of two vectors is zero, then those vectors are parallel to each other.
Therefore we can say that \[\vec{A}=\hat{i}-5\hat{j}\text{ and }\vec{B}=2\hat{i}-10\hat{j}\]are parallel to each other.

Hence the correct answer is option D.

Note:
Any quantity that has both direction and magnitude is called a vector quantity. If the quantity has only magnitude and no direction, then it is a scalar quantity.
Here we find the cross product to see if the vectors are parallel to each other. We can find the same by checking if we can write one of the vectors as a product of the other vector, i.e. in the forth case we have the vectors \[\vec{A}=\hat{i}-5\hat{j}\text{ and }\vec{B}=2\hat{i}-10\hat{j}\]
Here we can write the B vector as,
\[\vec{B}=2\left( \hat{i}-5\hat{j} \right)\], i.e. B vector is a multiple of vector A. Therefore they are parallel to each other.