This question aims to find the $2$ vectors which are **orthogonal** to the given vector $U = \dfrac{-1}{4}i+\dfrac{3}{2}j$, and these two vectors should be in opposite directions.

This question is based on the concept of **orthogonal vectors.** If two vectors $A$ and $B$ have a **dot product** equal to **zero**, then the said two vectors $A$ and $B$ are said to be **orthogonal or perpendicular** to each other. It is represented as:

\[A.B=0\]

## Expert Answer

We know that for two vectors to be **orthogonal** and to be in opposite directions, their **dot product** should be equal to zero.

Let us suppose our required vector is $w$ as:

\[w= [w_1 ,w_2]\]

Given vector $u$:

\[u=\frac{-1}{4}i+\frac{3}{2}j\]

\[u.w=0\]

\[[\frac{-1}{4}+\frac{3}{2} ] . [w_1 ,w_2]=0\]

\[\frac{-1}{4}w_1+\frac{3}{2} w_2=0\]

\[\frac{-1}{4}w_1=\frac{-3}{2} w_2 \]

\[\frac{-1}{ 2}w_1=-3w_2\]

Both **negative signs will be canceled** and $2$ will be multiplied on the right-hand side, so we get:

\[w_1= 6w_2\]

as $w_1=6w_2$ so putting value of $w_1$ in vector $w$, we get:

\[[w_1, w_2]\]

\[[6w_2 , w_2]\]

Our required vector $w =[6w_2 , w_2]$ will be **orthogonal** to the given vector $u= \dfrac{-1}{4}i +\dfrac{3}{2}j$ when $w_2$ belongs to any value from the **real numbers**.

As there could be **multiple correct vectors,** let us suppose $w_2(1)=1$ and $w_2(2)=-1$.

We get vectors:

\[[6w_2 , w_2]\]

Put $w_2(1)=1$ we get the vector:

\[[6(1), 1 ]\]

\[[6, 1]\]

Now put $w_2(1)=-1$, we get the vector:

\[[6 (-1), -1]\]

\[[-6, -1]\]

So our required $2$ vectors which are **orthogonal** to given vector $u$ and opposite in direction are:

\[ [6, 1] ; [-6, -1]\]

To verify that these vectors are **orthogonal** or **perpendicular** to the given vector, we will solve for the **dot product**. If the dot product is **zero**, it means the vectors are** perpendicular**.

Given vector $u$:

\[u=\dfrac{-1}{4}i+\dfrac{3}{2}j\]

\[u.w=0\]

\[=[\dfrac{-1}{4}+\dfrac{3}{2}].[6 , 1]\]

\[=[\dfrac{-6}{4}+\dfrac{3}{2}]\]

\[=[\dfrac{-3}{2}+\dfrac{3}{2}]\]

\[=0\]

Given vector $u$:

\[u=\dfrac{-1}{4}i+\dfrac{3}{2}j\]

Vector $w$ is given as:

\[w=[-6,-1]\]

\[u.w=0\]

\[=[\frac{-1}{4}+\frac{3}{2}] . [-6,-1]\]

\[=[\frac{+6}{4}+\frac{-3}{2}]\]

\[=[\frac{3}{2}+\frac{-3}{2}]\]

\[=0\]

This verifies that both vectors are **opposite** to each other and **perpendicular** to the given vector $u$.

## Numerical Results

Our required $2$ vectors which are **orthogonal** or **perpendicular** to given vector $u=\dfrac{-1}{4}i+\dfrac{3}{2}j$ and **opposite in direction** are $[6,1]$ and $[-6,-1]$.

## Example

Find **two vectors** which are **opposite** to each other and **perpendicular** to given vector $A=\dfrac{1}{2}i-\dfrac{2}{9}j$.

let our required vector be $B=[b_1 ,b_2]$.

Given vector $A$:

\[A=\dfrac{1}{2}i-\dfrac{2}{9}j\]

\[A.B=0\]

\[[\dfrac{1}{2}-\dfrac{2}{9} ] . [b_1 ,b_2]=0\]

\[[\dfrac{1}{2}b_1- \dfrac{2}{9}b_2]=0\]

\[\dfrac{1}{2}b_1=\dfrac{2}{9} b_2\]

So $2$ will be multiplied on the right-hand side and we get equation in terms of $b_1$ as:

\[b_1=\dfrac{2 \times 2}{9}b_2\]

\[b_1=\dfrac{4}{9}b_2\]

as $b_1=\dfrac{4}{9} b_2$ so putting value of $b_1$ in vector $B$.

\[[b_1,b_2]\]

\[[\dfrac{4}{9}b_2,b_2]\]

Our required vector $B =[\dfrac{4}{9} b_2 , b_2]$ will be **orthogonal** to the given vector $A=\dfrac{1}{2}i-\dfrac{2}{9}j $ when $b_2$ belongs to any value from the **real numbers**.

As there could be multiple correct vectors, let us suppose $b_2(1)=9$ and $b_2(2)=-9$.

We get vectors as:

\[[\dfrac{4}{9} b_2 ,b_2]\]

Put $b_2(1)=9$ we get the vector as:

\[[\dfrac{4}{9} \times 9,9]\]

\[[4, 9]\]

Now put $b_2(1)=-9$ we get the vector as:

\[[\dfrac{4}{9} \times -9,-9]\]

\[[-4,-9]\]

so:

\[ B=[4i+9j], \hspace{0.4in} B=[-4i-9j] \]

Our required $2$ vectors which are **orthogonal** or **perpendicular** to given vector $A=\dfrac{1}{2}i-\dfrac{2}{9}j$ and **opposite in direction** are $[4,9]$ and $[-4,-9]$.