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
Verified531k+ views
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.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success
Master Class 12 Physics: Engaging Questions & Answers for Success
Master Class 12 English: Engaging Questions & Answers for Success
Master Class 12 Social Science: Engaging Questions & Answers for Success
Master Class 12 Maths: Engaging Questions & Answers for Success
Master Class 12 Business Studies: Engaging Questions & Answers for Success
Trending doubts
Which are the Top 10 Largest Countries of the World?
What are the major means of transport Explain each class 12 social science CBSE
Draw a labelled sketch of the human eye class 12 physics CBSE
Why cannot DNA pass through cell membranes class 12 biology CBSE
Differentiate between insitu conservation and exsitu class 12 biology CBSE
Draw a neat and well labeled diagram of TS of ovary class 12 biology CBSE