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Find two vectors in opposite directions that are orthogonal to the vector u. U=(-1/4)i +(3/2)j

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 $[-4,-9]$.

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