RD Sharma Class 12 Exercise FBQ Scalar and dot product Solutions Maths

RD Sharma Class 12 Exercise FBQ Scalar and dot product Solutions Maths - Download PDF Free Online

Edited By Lovekush kumar saini | Updated on Jan 27, 2022 05:02 PM IST

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RD Sharma Class 12 Solutions Chapter23FBQ Scaler and dot product - Other Exercise
Scalar or Dot Products Excercise: FBQ
RD Sharma Chapter-wise Solutions

RD Sharma Class 12 Solutions Chapter23FBQ Scaler and dot product - Other Exercise

Scalar or Dot Products Excercise: FBQ

Scalar or dot product exercise fill in the blanks question 1

Answer:

$\sqrt{3}$
Hint:
Let $|\vec {a}|=|\vec {b}|=1$
Given:
If $\vec {a}$ and $\vec {b}$ are two unit vectors such that $\left |\vec {a+b} \right |$ is unit vector, thena-b $\left |\vec {a} -\vec {b} \right |$
Solution:
As $\left |\vec {a+b} \right |$ is unit vector, $|\vec {a}|=|\vec {b}|=1$
$\begin{aligned} &|\vec{a}+\vec{b}|=1 \\ &|\vec{a}+\vec{b}|^{2}=1 \\ &|\vec{a}|^{2}+|\vec{b}|^{2}+2 a b=1 \\ \end{aligned}$
$\begin{aligned} &2+2(a \cdot b)=1 \\ &-2(a \cdot b)=1 \\ &|\vec{a}-\vec{b}|^{2}=|\vec{a}|^{2}+|\vec{b}|^{2}-2 a b \\ \end{aligned}$
$\begin{aligned} &=1+1+1 \\ &|\vec{a}-\vec{b}|^{2}=3 \\ &|\vec{a}-\vec{b}|=\sqrt{3} \end{aligned}$

Scalar or dot product exercise fill in the blanks question 2

Answer:

$\pm \frac{3}{4}$ Hint:
$\vec{a}+\lambda\vec{b}$ is perpendicular to $\vec{a}-\lambda\vec{b}$
Given:
$\left | \vec{a} \right |=3,\left | \vec{b} \right |=4$ and $\vec{a}+\lambda\vec{b}$ is perpendicular to $\vec{a}-\lambda\vec{b}$ , then $\lambda =$
Solution:
$\vec{a}+\lambda\vec{b}$ is perpendicular to $\vec{a}-\lambda\vec{b}$
$\begin{aligned} &(a+\lambda b)(a-\lambda b)=0 \\ &a^{2}-\lambda \times(a \cdot b)+\lambda \times(b \cdot a)-\lambda^{2} \times|b|^{2}=0 \\ \end{aligned}$
$\begin{aligned} &a^{2}-\lambda^{2} \times b^{2}=0 \\ &\lambda=\pm \frac{|a|}{|b|}=\pm \frac{3}{4}\end{aligned}$

Scalar or dot product exercise fill in the blanks question 3

Answer:

-5
Hint:
Use $\vec{a}.m\vec{b}=120$
Given:
If $\hat{a}=2\hat{i}-7\hat{j}+\hat{k}$ , $\hat{b}=\hat{i}+3\hat{j}-5\hat{k}$ and $a.mb=120$ then m=
Solution:
$\begin{aligned} &a \cdot b=2(1)+(-7)(3)+1(-5) \\ &=2-21-5 \\ &=-24 \\ \end{aligned}$
$\begin{aligned} &a \cdot m b=m(a \cdot b) \\ &m(-24)=120 \\ &m=\frac{-120}{24} \\ &m=-5\end{aligned}$

Scalar or dot product exercise fill in the blanks question 4

Answer:

$\pi$
Hint:
Angle between $\vec{a}$ and $\vec{c}$ .
Given:
The non-zero vector $\vec{a},\vec{b},\vec{c}$ are related by $\vec{a}=8b$ and $\vec{c}=-7b$ then angle between $\vec{a}$ and $\vec{c}$ is
Solution:
We have,
$\vec{a}=8b$ and $\vec{c}=-7b$
Let,
$\begin{aligned} &\vec{b}=b_{1} \hat{\imath}+b_{2} \hat{\jmath}+b_{3} \hat{k} \\ \end{aligned}$
Then,
$\begin{aligned} &\vec{a}=8\left(b_{1} \hat{\imath}+b_{2} \hat{\jmath}+b_{3} \hat{k}\right) \\ \end{aligned}$
And $\begin{aligned} &\vec{c}=-7\left(b_{1} \hat{\imath}+b_{2} \hat{\jmath}+b_{3} \hat{k}\right) \\ \end{aligned}$
Consider angle between $\vec{a}$ and $\vec{c}$ is $\theta$ .
Now, we know the rule of dot product.
$\begin{aligned} &\vec{a} . \vec{c}=|\vec{a} \| \vec{c}| \cos \theta \\ &\therefore \cos \theta=\frac{\vec{a} \cdot \vec{c}}{|\vec{a}||\vec{c}|} \\ \end{aligned}$
$\begin{aligned} &\cos \theta=\frac{\left[\left(8\left(b_{1} \hat{\imath}+b_{2} \hat{\jmath}+b_{3} \hat{k}\right) \cdot\left(-7\left(b_{1} \hat{\imath}+b_{2} \hat{\jmath}+b_{3} \hat{k}\right)\right)\right)\right]}{|-7| \times|8| \cdot \sqrt{b_{1}^{2}+b_{2}^{2}+b_{3}^{2}} \cdot \sqrt{b_{1}^{2}+b_{2}^{2}+b_{3}^{2}}} \\ \end{aligned}$
$\begin{aligned} &\cos \theta=\frac{-56\left(b_{1}^{2}+b_{2}^{2}+b_{3}^{2}\right)}{|-56|\left(b_{1}^{2}+b_{2}^{2}+b_{3}^{2}\right)} \quad\left[\begin{array}{l} \imath{\imath} . \hat{\imath}=1, \hat{\jmath} \cdot \hat{\jmath}=1, \\ \hat{k} \cdot \hat{k}=1 \end{array}\right] \\ \end{aligned}$
$\begin{aligned} &\cos \theta=\frac{-56}{56} \\ &\cos \theta=-1 \end{aligned}$
Therefore,
$\begin{aligned} \theta=\pi &&& \quad[\because \cos \pi=-1] \end{aligned}$
Hence, angle between $\vec{a}$ and $\vec{c}$ is $\pi$ .

Scalar or dot product exercise fill in the blanks question 5

Answer:

$\sqrt{2}$
Hint:
a is perpendicular to b
Given:
If a and b are mutually perpendicular unit vector, then $\left |\vec{a}+\vec{b} \right |$
Solution:
$\left |\vec{a} \right |=\left |\vec{b} \right |=1$ , $\vec{a} .\vec{b} =0$ …as they are perpendicular … (i)
Now,
$\begin{aligned} &|\vec{a}+\vec{b}|^{2}=|\vec{a}|^{2}+|\vec{b}|^{2}+2 \vec{a} \cdot \vec{b} \\ \end{aligned}$
$\begin{aligned} &=1+1+2(0)=2 &&&&&& [Use (i)]\\ \end{aligned}$
$\begin{aligned} &|\vec{a}+\vec{b}|=\sqrt{2} &&&&&& \left[\because \sqrt{1^{2}+1^{2}}=\sqrt{2}\right] \end{aligned}$

Scalar or dot product exercise Fill in the blanks question 6

Answer:

0
Hint:
Find $\alpha$
Given:
If the angle between vector such that $\hat{i}+\hat{k}$ and $\hat{i}- \hat{j}+\alpha \hat{k}$ is $\frac{\pi}{3}$ .
Find $\alpha$
Solution:
$\begin{aligned} &\cos \theta=\frac{\vec{a} \vec{b}}{|\vec{a}| \cdot|\vec{b}|^{\prime}} \theta=\frac{\pi}{3}\\ \end{aligned}$
$\begin{aligned} &\cos \theta=\frac{(\hat{\imath}+\hat{k}) \cdot(\hat{\imath}-\hat{\jmath}+\alpha \hat{k})}{\sqrt{1^{2}+1^{2}} \cdot \sqrt{1^{2}+1^{2}+\alpha^{2}}}\\ \end{aligned}$
$\begin{aligned} &\cos \frac{\pi}{3}=\frac{(\hat{\imath}+\hat{k}) \cdot(\hat{\imath}-\hat{\jmath}+\alpha \hat{k})}{\sqrt{2} \cdot \sqrt{2+\alpha^{2}}}\\ \end{aligned}$
$\begin{aligned} &\frac{1}{2}=\frac{(1+\alpha)}{\sqrt{2} \cdot \sqrt{2+\alpha^{2}}}\\ \end{aligned}$
$\begin{aligned} &\frac{1}{4}=\frac{(1+\alpha)^{2}}{2\left(2+\alpha^{2}\right)} \\ \end{aligned}$ ..............…squaring the equation
$\begin{aligned} &2(1+\alpha)^{2}=2+\alpha^{2}\\ &2\left(1+\alpha^{2}+2 \alpha\right)=2+\alpha^{2}\\ &2+2 \alpha^{2}+4 \alpha=2+\alpha^{2}\\ \end{aligned}$
$\begin{aligned} &\alpha^{2}+4 \alpha=0\\ &\alpha(\alpha+4)=0\\ &\alpha=0, \alpha=-4\\ \end{aligned}$
Hence $\begin{aligned} & \alpha=0\end{aligned}$

Hence α=0

Scalar or dot product exercise Fill in the blanks question 7

Answer:

$\frac{2\pi}{3}$
Hint:
Angle between $\vec{a}$ andn $\vec{b}$
Given:
If $\vec{a},\vec{b},\vec{c}$ are unit vector such that $\vec{a}+\vec{b}-\vec{c}=0$ , then angle betweena

andb

is
Solution:
$\left |\vec{a} \right |=\left |\vec{b} \right |=\left |\vec{c} \right |=1$
We have,
$\vec{a}+\vec{b}-\vec{c}=0$
$\vec{a}+\vec{b}=\vec{c}$
on squaring both sides,
$\begin{aligned} &(\vec{a}+\vec{b})^{2}=(\vec{c})^{2} \\ &|\vec{a}|^{2}+|\vec{b}|^{2}+2 \cdot a \cdot b \cos \theta=(\vec{c})^{2} \\ \end{aligned}$
$\begin{aligned} &1+1+2 a \cdot b \cos \theta=1 \\ &2 a \cdot b \cos \theta=1-2 \\ \end{aligned}$
$\begin{aligned} &\cos \theta(1)(1)=\frac{-1}{2} \\ &\cos \theta=\frac{-1}{2} \\ &\theta=\frac{2 \pi}{3} \end{aligned}$

Scalar or dot product exercise Fill in the blanks question 8

Answer:

$\frac{\pi}{3}$
Hint:
Angle between $\vec{a}$ and $\vec{b}$
Given:
If $\vec{a}, \vec{b}$ are unit vector such that, $\left | \vec{a}+ \vec{b} \right |=\sqrt{3}$ , then angle between $\vec{a}$ and $\vec{b}$ is
Solution:
$\begin{aligned} &|\vec{a}+\vec{b}|=\sqrt{3},|\vec{a}|=|\vec{b}|=|\vec{c}|=1 \\ &|\vec{a}+\vec{b}|^{2}=(\sqrt{3})^{2} \\ \end{aligned}$
$\begin{aligned} &(\vec{a}+\vec{b})^{2}=(\sqrt{3})^{2} \\ &|\vec{a}|^{2}+|\vec{b}|^{2}+2(1)(1) \cos \theta=3 \\ \end{aligned}$
$\begin{aligned} &1+1+2 a \cdot b \cos \theta=3 \\ &2+2 \cos \theta=3 \\ &\cos \theta=1 \\ \end{aligned}$
$\begin{aligned} &\cos \theta=\frac{1}{2} \\ &\cos \theta=\cos 60^{\circ} \\ &\theta=\frac{\pi}{3} \end{aligned}$

Scalar or dot product exercise fill in the blanks question 9

Answer:

$\vec{a}.\vec{b}$
Hint:
Use projection $\vec{a}$ on $\vec{b}$
Given:
If $\vec{a},\vec{b}$ are two non-zero vector, then projection of $\vec{a}$ on $\vec{b}$ is
Solution:
Projection of $\vec{a}$ on $\vec{b}$ = $\frac{1}{|b|}\vec{a}.\vec{b}$
$\left |\vec{a} \right |=\left |\vec{b} \right |=1\\ \left |\vec{b} \right |=1$
Hence Projection of $\vec{a}$ on $\vec{b}$ = $\vec{a}.\vec{b}$

Scalar or dot product exercise Fill in the blanks question 10

Answer:

$\frac{\pi}{4}$
Hint:
Angle between $\vec{a}$ and $\vec{b}$
Given:
Let $\vec{a}$ , $\vec{b}$ are unit vector such that, $\vec{a}-\sqrt{2}\vec{b}$ is also a unit vector, then angle between $\vec{a}$ and $\vec{b}$
Solution:
$\begin{aligned} &|\vec{a}|=|\vec{b}|=1 \\ &|\vec{a}-\sqrt{2} \vec{b}|=1 \\ &|\vec{a}-\sqrt{2} \vec{b}|^{2}=1^{2} \\ \end{aligned}$ ….squaring on both sides
$\begin{aligned} &|\vec{a}|^{2}+|\sqrt{2} \vec{b}|^{2}-2|\vec{a}| \cdot|\sqrt{2} \vec{b}| \cos \theta=1 \\ &1+2(1)^{2}-2(1)(\sqrt{2}(1)) \cos \theta=1 \end{aligned}$
$\begin{aligned} &1+2-2 \sqrt{2} \cos \theta=1 \\ &3-2 \sqrt{2} \cos \theta=1 \\ \end{aligned}$
$\begin{aligned} &3-1=2 \sqrt{2} \cos \theta \\ &2=2 \sqrt{2} \cos \theta \\ \end{aligned}$
$\begin{aligned} &\cos \theta=\frac{2}{2 \sqrt{2}} \\ &\cos \theta=\frac{1}{\sqrt{2}} \\ &\theta=\frac{\pi}{4} \end{aligned}$

Scalar or dot product exercise Fill in the blanks question 11

Answer:

$\frac{\pi}{3}$
Hint:
Angle between $\vec{a}$ and $\vec{b}$
Given:
$\left |\vec{a} \right |=1$ , $\left |\vec{b} \right |=3$ and $\left |a-b \right |=\sqrt{7}$ . Find Angle between a and b.
Solution:
$\left |\vec{a} \right |=1$ , $\left |\vec{b} \right |=3$ and $\left |a-b \right |=\sqrt{7}$ … (i)
$\left |a-b \right |=\sqrt{7}$
Squaring on both side we get
$\begin{aligned} &|a-b|^{2}=(\sqrt{7})^{2} \\ &|\vec{a}|^{2}+|\vec{b}|^{2}-2|\vec{a}| \cdot|\vec{b}| \cos \theta=(\sqrt{7})^{2} \end{aligned}$
From (i)
$\begin{aligned} &(1)^{2}+(3)^{2}-2(1)(3) \cos \theta=7 \\ &1+9-6 \cos \theta=7 \\ &10-7=6 \cos \theta \\ \end{aligned}$
$\begin{aligned} &\frac{3}{6}=\cos \theta \\ &\cos \theta=\frac{1}{2} \\ &\theta=\frac{\pi}{3} \end{aligned}$

Scalar or dot product exercise fill in the blanks question 12

Answer:

$\sqrt{a^{2}+b^{2}+c^{2}}$
Hint:
Use $a.b=0$
Given:
If a,b,c are mutually perpendicular vector, then $\left | \vec{a}+\vec{b}+\vec{c} \right |$
Solution:
$\begin{aligned} &|\vec{a}|=1,|\vec{b}|=1,|\vec{c}|=1 \\ &\vec{a} \cdot \vec{b}=0, \vec{b} \cdot \vec{c}=0, \vec{c} \cdot \vec{a}=0 \\ &\vec{b} \cdot \vec{a}=0, \vec{c} \cdot \vec{b}=0, \vec{a} \cdot \vec{c}=0 \\ \end{aligned}$
$\begin{aligned} &|\vec{a}+\vec{b}+\vec{c}|^{2}=(\vec{a}+\vec{b}+\vec{c}) \cdot(\vec{a}+\vec{b}+\vec{c}) \\ \end{aligned}$
$\begin{aligned} &=\vec{a} \cdot \vec{a}+\vec{a} \cdot \vec{b}+\vec{c} \cdot \vec{c}+\vec{b} \cdot \vec{a}+\vec{b} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}+\vec{c} \cdot \vec{b}+\vec{c} \cdot \vec{c} \\ \end{aligned}$
$\begin{aligned} &=|\vec{a}|^{2}+0+0+0+|\vec{b}|^{2}+0+0+0+|\vec{c}|^{2} \\ &=|\vec{a}|^{2}+|\vec{b}|^{2}+|\vec{c}|^{2} \\ \end{aligned}$
$\begin{aligned} &|\vec{a}+\vec{b}+\vec{c}|^{2}=|\vec{a}|^{2}+|\vec{b}|^{2}+|\vec{c}|^{2} \\ &\vec{a}+\vec{b}+\vec{c}=\sqrt{a^{2}+b^{2}+c^{2}} \end{aligned}$

Scalar or dot product exercise Fill in the blanks question 13

Answer:

$\pi$
Hint:
Angle between $\vec{a}$ and $\vec{b}$
Given:
If $\vec{a}$ , $\vec{b}$ are two non-zero vector $\vec{a}.\vec{b}=-\left |\vec{a} \right | . \left |\vec{b} \right |$ , then angle between $\vec{a}$ and $\vec{b}$ is
Solution:
$\vec{a}.\vec{b}=-\left |\vec{a} \right | . \left |\vec{b} \right |$ … (i)
And we know that $\vec{a}.\vec{b}=-\left |\vec{a} \right | . \left |\vec{b} \right |.\cos\theta$
From (i) we get
$-\vec{a}.\vec{b}=-\left |\vec{a} \right | . \left |\vec{b} \right |.\cos\theta\\ \cos \theta =-1\\ \theta =\pi$

Scalar or dot product exercise fill in the blanks question 14

Answer:

$\left | \vec{r} \right |^{2}$ Hint:
Squaring both sides
Given:
For any vector $\left ( r.\hat{i} \right )^{2}+ \left ( r.\hat{j} \right )^{2}+\left ( r.\hat{k} \right )^{2}=$ ___________
Solution:
$\begin{aligned} &|r|=x \hat{\imath}+y \hat{\jmath}+z \hat{k} \\ &|\vec{r}|=\sqrt{x^{2}+y^{2}+z^{2}} \\ &|\vec{r}|^{2}=x^{2}+y^{2}+z^{2} \\ \end{aligned}$
$\begin{aligned} (r \cdot \hat{\imath})^{2}+(r \cdot \hat{\jmath})^{2}+(r \cdot \hat{k})^{2}=((x \hat{\imath}+y \hat{\jmath}+z \hat{k}) \cdot \hat{\imath})^{2}+((x \hat{\imath}+y \hat{\jmath}+z \hat{k}) \cdot \hat{\jmath})^{2}+((x \hat{\imath}+y \hat{\jmath}+z \hat{k}) \cdot \hat{k})^{2} \\ \end{aligned}$
$\begin{aligned} =x^{2}+y^{2}+z^{2} \\ (r \cdot \hat{\imath})^{2}+(r \cdot \hat{\jmath})^{2}+(r \cdot \hat{k})^{2}=x^{2}+y^{2}+z^{2}=|\vec{r}|^{2} \end{aligned}$

Scalar or dot product exercise fill in the blanks question 15

Answer:

4
Hint:
$\left |a \right |=?$
Given:
If a,b non-zero vector of same magnitude. Such that angle between $\vec{a}$ and $\vec{b}$ are $\frac{2\pi}{3}$ and $a.b=-8$ then $\left | a \right |$
Solution:
Let
$\left | a \right |=m\\ \vec{a}.\vec{b}=-8\\ \left |\vec{a} \right |.\left |\vec{b} \right |.\cos \theta =-8\\$
$m^{2}\left ( -\frac{1}{2} \right ) =-8\\$ ,since a and b have same magnitude
$m^{2}=16$
m=4,m cannot be negative

Scalar or dot product exercise fill in the blanks question 16

Answer:

$\vec{r}$
Hint:
$\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$
Given:
For any non-vector $\vec{r}$ , the expression $\vec{r}=(r.i)\hat{i}+(r.j)\hat{j}+(r.k)\hat{k}$ equals to
Solution:
$\begin{aligned} &\vec{r}=x \hat{\imath}+y \hat{\jmath}+z \hat{k} \\ \end{aligned}$
$\begin{aligned} &r \cdot \hat{\imath}=x \\ &r \cdot \hat{\jmath}=y \\ &r \cdot \hat{k}=z \\ \end{aligned}$
$\begin{aligned} &(r . i) \hat{\imath}+(r . j) \hat{\jmath}+(r \cdot k) \hat{k}=x \hat{\imath}+y \hat{\jmath}+z \hat{k} \\ &x \hat{\imath}+y \hat{\jmath}+z \hat{k}=\vec{r} \end{aligned}$

Scalar or dot product exercise fill in the blanks question 17

Answer:

$\frac{\pi}{4}$ Hint:
Find magnitude
Given:
The vector $\vec{a}=3\hat{i}-2\hat{j}+2\hat{k}$ and $\vec{b}=-\hat{i}-2\hat{k}$ are adjacent side of parallelogram
The acute angle between diagonal is
Solution:
Let $\vec{AB}=3\hat{i}-2\hat{j}+2\hat{k}$ & $\vec{BC}=-\hat{i}-2\hat{k}$
AC & BD are the diagonals of the parallelogram,
$\vec{AC}=\vec{AB}+\vec{BC}\\ =(3-1)\hat{i}-2\hat{j}+(2-2)\hat{k}\\ =2\hat{i}-2\hat{j}$
$\begin{aligned} &\overrightarrow{B D}=\overrightarrow{A B}-\overrightarrow{B C} \\ &=(3-(-1)) \hat{\imath}-2 \hat{\jmath}+4 \hat{k} \\ &\overrightarrow{B D}=4 \vec{\imath}-2 \vec{\jmath}+4 \vec{k} \\ \end{aligned}$
$\begin{aligned} &\overrightarrow{A C} \cdot \overrightarrow{B D}=|\overrightarrow{A C}| \cdot|\overrightarrow{B D}| \cos \theta \\ &12=\sqrt{4+4} \cdot \sqrt{16+4+16} \cos \theta \\ \end{aligned}$
$\begin{aligned} &12=\sqrt{8} \cdot \sqrt{36} \cos \theta \\ &12=2 \sqrt{2} \cdot 6 \cos \theta \\ \end{aligned}$
$\begin{aligned} &\frac{1}{\sqrt{2}}=\cos \theta \\ &\frac{\pi}{4}=\theta \end{aligned}$

Scalar or dot product exercise fill in the blanks question 18

Answer:

$\frac{3}{5\sqrt{2}}$ Hint:
$\vec{a}$ and $\vec{b}$ projection
Given:
$\vec{a}=2\hat{i}-\hat{j}+2\hat{k}$ and $\vec{b}=5\hat{i}-3\hat{j}-4\hat{k}$
Solution:
$\vec{a}=2\hat{i}-\hat{j}+2\hat{k} \\ \vec{b}=5\hat{i}-3\hat{j}-4\hat{k}$
We know
Projection of vector $\vec{a}$ and $\vec{b}$ $=\frac{1}{|\vec{b}|} \cdot(\vec{a} \cdot \vec{b})$
$\begin{aligned} (\vec{a} \cdot \vec{b}) &=(2 \times 5)+(-1 \times-3)+(2 \times-4) \\ &=10+3-8 \\ &=5 \\ \end{aligned}$
$\begin{aligned} |\vec{b}| &=\sqrt{25+9+16} \\ &=\sqrt{25+25} \\ &=\sqrt{50} \end{aligned}$
Projection of vector $\vec{a}$ on $\vec{b}$ $=\frac{1}{\sqrt{2}.5} \times 5\\$
$=\frac{1}{\sqrt{2}}$
Projection of $\vec{b}$ and $\vec{a}$ $=\frac{1}{|\vec{a}|} \cdot(\vec{b} \cdot \vec{a})$
$\begin{aligned} \vec{b} \cdot \vec{a}=&(5 \times 2)+(-3 \times-1)+(-4 \times 2) \\ &=10+3-8 \\ &=5 \\ \end{aligned}$
$\begin{aligned} |\vec{a}|=& \sqrt{2^{2}+1^{2}+2^{2}} \\ =& \sqrt{4+1+4} \\ &=\sqrt{9}\end{aligned}$
$\begin{aligned} &=3 \\ &\frac{\text { Projection of } \vec{a} \text { and } \vec{b}}{\text { Projection of } \vec{b} \text { and } \vec{a}}=\frac{\frac{1}{\sqrt{2}}}{\frac{5}{3}} \\ &\qquad=\frac{3}{5 \sqrt{2}} \end{aligned}$

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