Find the unit vector in plane which makes $45{}^\circ $ with $\hat{i}+\hat{j}$ and $60{}^\circ $ with $3\hat{i}+4\hat{j}$
Answer
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Hint:To answer this question, we use the basic concepts for a unit vector which is given as $\hat{a}=\dfrac{{\vec{a}}}{\left| {\vec{a}} \right|}.$
Here, $\hat{a}$ represents the unit vector, $\vec{a}$ represents the vector and $\left| {\vec{a}} \right|$ represents the magnitude of the vector. We will also use the formula for the dot product of two vectors given as $\vec{a}.\vec{b}=\left| {\vec{a}} \right|\left| {\vec{b}} \right|\cos \theta .$ Here $\vec{a}$ and $\vec{b}$ represent two vectors and $\theta $ is the angle between the two vectors.
Complete step-by-step solution:
We are required to find the unit vector in the plane which makes $45{}^\circ $ with $\hat{i}+\hat{j}$ and $60{}^\circ $ with $3\hat{i}+4\hat{j}$ . In order to solve this, let us assume the vector is given by r,
$\Rightarrow \hat{r}=x\hat{i}+y\hat{j}\ldots \ldots \left( 1 \right)$
Magnitude of this vector is equal to,
$\Rightarrow \left| {\vec{r}} \right|=\sqrt{{{\left( x\hat{i} \right)}^{2}}+{{\left( y\hat{j} \right)}^{2}}}$
We also know that ${{\hat{i}}^{2}}={{\hat{j}}^{2}}=1.$ Therefore,
$\Rightarrow \left| {\vec{r}} \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}$
This is a unit vector hence this is equal to 1.
$\Rightarrow \left| {\vec{r}} \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}=1$
Now we take the dot product of this vector given by equation 1 with the vector $\hat{i}+\hat{j}$ .For this, the formula is given by $\vec{a}.\vec{b}=\left| {\vec{a}} \right|\left| {\vec{b}} \right|\cos \theta .$ Here $\vec{a}$ and $\vec{b}$ represent two vectors and $\theta $ is the angle between the two vectors. For the vector $\hat{i}+\hat{j}$ , the angle with the unit vector is given as $45{}^\circ $ . Therefore, substituting in the above equation,
$\Rightarrow \left( x\hat{i}+y\hat{j} \right).\left( \hat{i}+\hat{j} \right)=\left| {\vec{r}} \right|\left| \hat{i}+\hat{j} \right|\cos 45$
Dot product of two vectors is given by simply multiplying the corresponding terms for $\hat{i}$ ,$\hat{j}$ , and $\hat{k}$ and taking the sum of all the terms.
$\Rightarrow \left( x\hat{i}.\hat{i}+y\hat{j}.\hat{j} \right)=\left| {\vec{r}} \right|\left| \hat{i}+\hat{j} \right|\cos 45$
We know that $\hat{i}.\hat{i}=\hat{j}.\hat{j}=\hat{k}.\hat{k}=1.$
$\Rightarrow x+y=\left| {\vec{r}} \right|\left| \hat{i}+\hat{j} \right|\cos 45$
We also know that the magnitude of $\hat{i}+\hat{j}$ is given by,
$\Rightarrow \sqrt{{{{\hat{i}}}^{2}}+{{{\hat{j}}}^{2}}}=\sqrt{1+1}=\sqrt{2}$
We know the value of $\cos 45{}^\circ $ is $\dfrac{1}{\sqrt{2}}.$ Substituting these,
$\Rightarrow x+y=\sqrt{{{x}^{2}}+{{y}^{2}}}.\sqrt{2}.\dfrac{1}{\sqrt{2}}$
Cancelling the terms,
$\Rightarrow x+y=\sqrt{{{x}^{2}}+{{y}^{2}}}$
Since we know that $\Rightarrow \left| {\vec{r}} \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}=1,$
$\Rightarrow x+y=1\ldots \ldots \left( 2 \right)$
We apply the same formula for the second vector given as $3\hat{i}+4\hat{j}$ . For this case, the angle is $60{}^\circ $ .
$\Rightarrow \left( x\hat{i}+y\hat{j} \right).\left( 3\hat{i}+4\hat{j} \right)=\left| {\vec{r}} \right|\left| 3\hat{i}+4\hat{j} \right|\cos 60$
Taking the dot product and substituting the value of $\cos 60{}^\circ $ as $\dfrac{1}{2},$
$\Rightarrow \left( 3x\hat{i}.\hat{i}+4y\hat{j}.\hat{j} \right)=\left| {\vec{r}} \right|\left| 3\hat{i}+4\hat{j} \right|.\dfrac{1}{2}$
We also know that the magnitude of $3\hat{i}+4\hat{j}$ is given by,
$\Rightarrow \sqrt{{{\left( 3\hat{i} \right)}^{2}}+{{\left( 4\hat{j} \right)}^{2}}}=\sqrt{9+16}=\sqrt{25}$
Root of 25 is 5,
$\Rightarrow \sqrt{{{\left( 3\hat{i} \right)}^{2}}+{{\left( 4\hat{j} \right)}^{2}}}=5$
Substituting this,
$\Rightarrow 3x+4y=\sqrt{{{x}^{2}}+{{y}^{2}}}.5.\dfrac{1}{2}$
Multiplying the terms on the right-hand side,
$\Rightarrow 3x+4y=\dfrac{5}{2}\sqrt{{{x}^{2}}+{{y}^{2}}}$
Since we know that $\Rightarrow \left| {\vec{r}} \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}=1,$
$\Rightarrow 3x+4y=\dfrac{5}{2}\ldots \ldots \left( 3 \right)$
To solve the two equations, we multiply equation 2 by 3 on both sides.
$\Rightarrow 3x+3y=3\ldots \ldots \left( 4 \right)$
We subtract the above two equations 3 and 4,
$\begin{align}
& \Rightarrow 3x+4y=\dfrac{5}{2} \\
& \underline{\text{ -}3x-3y=-3\text{ }} \\
& \text{ }0+y=-\dfrac{1}{2}\text{ } \\
\end{align}$
Hence, we get y value as $-\dfrac{1}{2}.$ Substituting this in equation 2,
$\Rightarrow x-\dfrac{1}{2}=1$
Adding $\dfrac{1}{2}$ on both sides,
$\Rightarrow x=\dfrac{3}{2}$
Substituting these in equation 1,
$\Rightarrow \hat{r}=\dfrac{3}{2}\hat{i}-\dfrac{1}{2}\hat{j}$
Hence, the vector $\hat{r}$ which makes $45{}^\circ $ with $\hat{i}+\hat{j}$ and $60{}^\circ $ with $3\hat{i}+4\hat{j}$ is $\dfrac{\left( 3\hat{i}-\hat{j} \right)}{2}.$ This above vector is not a unit vector since its magnitude does not add up to 1.
Note:Students need to know the concepts of dot product of vectors and basic trigonometric values for standard angles to solve this question. We can verify whether the answer is a unit vector or not by finding the magnitude of that. If it is equal to 1, it is a unit vector; otherwise it is not.
Here, $\hat{a}$ represents the unit vector, $\vec{a}$ represents the vector and $\left| {\vec{a}} \right|$ represents the magnitude of the vector. We will also use the formula for the dot product of two vectors given as $\vec{a}.\vec{b}=\left| {\vec{a}} \right|\left| {\vec{b}} \right|\cos \theta .$ Here $\vec{a}$ and $\vec{b}$ represent two vectors and $\theta $ is the angle between the two vectors.
Complete step-by-step solution:
We are required to find the unit vector in the plane which makes $45{}^\circ $ with $\hat{i}+\hat{j}$ and $60{}^\circ $ with $3\hat{i}+4\hat{j}$ . In order to solve this, let us assume the vector is given by r,
$\Rightarrow \hat{r}=x\hat{i}+y\hat{j}\ldots \ldots \left( 1 \right)$
Magnitude of this vector is equal to,
$\Rightarrow \left| {\vec{r}} \right|=\sqrt{{{\left( x\hat{i} \right)}^{2}}+{{\left( y\hat{j} \right)}^{2}}}$
We also know that ${{\hat{i}}^{2}}={{\hat{j}}^{2}}=1.$ Therefore,
$\Rightarrow \left| {\vec{r}} \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}$
This is a unit vector hence this is equal to 1.
$\Rightarrow \left| {\vec{r}} \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}=1$
Now we take the dot product of this vector given by equation 1 with the vector $\hat{i}+\hat{j}$ .For this, the formula is given by $\vec{a}.\vec{b}=\left| {\vec{a}} \right|\left| {\vec{b}} \right|\cos \theta .$ Here $\vec{a}$ and $\vec{b}$ represent two vectors and $\theta $ is the angle between the two vectors. For the vector $\hat{i}+\hat{j}$ , the angle with the unit vector is given as $45{}^\circ $ . Therefore, substituting in the above equation,
$\Rightarrow \left( x\hat{i}+y\hat{j} \right).\left( \hat{i}+\hat{j} \right)=\left| {\vec{r}} \right|\left| \hat{i}+\hat{j} \right|\cos 45$
Dot product of two vectors is given by simply multiplying the corresponding terms for $\hat{i}$ ,$\hat{j}$ , and $\hat{k}$ and taking the sum of all the terms.
$\Rightarrow \left( x\hat{i}.\hat{i}+y\hat{j}.\hat{j} \right)=\left| {\vec{r}} \right|\left| \hat{i}+\hat{j} \right|\cos 45$
We know that $\hat{i}.\hat{i}=\hat{j}.\hat{j}=\hat{k}.\hat{k}=1.$
$\Rightarrow x+y=\left| {\vec{r}} \right|\left| \hat{i}+\hat{j} \right|\cos 45$
We also know that the magnitude of $\hat{i}+\hat{j}$ is given by,
$\Rightarrow \sqrt{{{{\hat{i}}}^{2}}+{{{\hat{j}}}^{2}}}=\sqrt{1+1}=\sqrt{2}$
We know the value of $\cos 45{}^\circ $ is $\dfrac{1}{\sqrt{2}}.$ Substituting these,
$\Rightarrow x+y=\sqrt{{{x}^{2}}+{{y}^{2}}}.\sqrt{2}.\dfrac{1}{\sqrt{2}}$
Cancelling the terms,
$\Rightarrow x+y=\sqrt{{{x}^{2}}+{{y}^{2}}}$
Since we know that $\Rightarrow \left| {\vec{r}} \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}=1,$
$\Rightarrow x+y=1\ldots \ldots \left( 2 \right)$
We apply the same formula for the second vector given as $3\hat{i}+4\hat{j}$ . For this case, the angle is $60{}^\circ $ .
$\Rightarrow \left( x\hat{i}+y\hat{j} \right).\left( 3\hat{i}+4\hat{j} \right)=\left| {\vec{r}} \right|\left| 3\hat{i}+4\hat{j} \right|\cos 60$
Taking the dot product and substituting the value of $\cos 60{}^\circ $ as $\dfrac{1}{2},$
$\Rightarrow \left( 3x\hat{i}.\hat{i}+4y\hat{j}.\hat{j} \right)=\left| {\vec{r}} \right|\left| 3\hat{i}+4\hat{j} \right|.\dfrac{1}{2}$
We also know that the magnitude of $3\hat{i}+4\hat{j}$ is given by,
$\Rightarrow \sqrt{{{\left( 3\hat{i} \right)}^{2}}+{{\left( 4\hat{j} \right)}^{2}}}=\sqrt{9+16}=\sqrt{25}$
Root of 25 is 5,
$\Rightarrow \sqrt{{{\left( 3\hat{i} \right)}^{2}}+{{\left( 4\hat{j} \right)}^{2}}}=5$
Substituting this,
$\Rightarrow 3x+4y=\sqrt{{{x}^{2}}+{{y}^{2}}}.5.\dfrac{1}{2}$
Multiplying the terms on the right-hand side,
$\Rightarrow 3x+4y=\dfrac{5}{2}\sqrt{{{x}^{2}}+{{y}^{2}}}$
Since we know that $\Rightarrow \left| {\vec{r}} \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}=1,$
$\Rightarrow 3x+4y=\dfrac{5}{2}\ldots \ldots \left( 3 \right)$
To solve the two equations, we multiply equation 2 by 3 on both sides.
$\Rightarrow 3x+3y=3\ldots \ldots \left( 4 \right)$
We subtract the above two equations 3 and 4,
$\begin{align}
& \Rightarrow 3x+4y=\dfrac{5}{2} \\
& \underline{\text{ -}3x-3y=-3\text{ }} \\
& \text{ }0+y=-\dfrac{1}{2}\text{ } \\
\end{align}$
Hence, we get y value as $-\dfrac{1}{2}.$ Substituting this in equation 2,
$\Rightarrow x-\dfrac{1}{2}=1$
Adding $\dfrac{1}{2}$ on both sides,
$\Rightarrow x=\dfrac{3}{2}$
Substituting these in equation 1,
$\Rightarrow \hat{r}=\dfrac{3}{2}\hat{i}-\dfrac{1}{2}\hat{j}$
Hence, the vector $\hat{r}$ which makes $45{}^\circ $ with $\hat{i}+\hat{j}$ and $60{}^\circ $ with $3\hat{i}+4\hat{j}$ is $\dfrac{\left( 3\hat{i}-\hat{j} \right)}{2}.$ This above vector is not a unit vector since its magnitude does not add up to 1.
Note:Students need to know the concepts of dot product of vectors and basic trigonometric values for standard angles to solve this question. We can verify whether the answer is a unit vector or not by finding the magnitude of that. If it is equal to 1, it is a unit vector; otherwise it is not.
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