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

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

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

Lovekush kumar sainiUpdated on 27 Jan 2022, 05:02 PM IST

The Class 12 RD Sharma chapter 23 exercise FBQ solution books are utilized by each understudy getting ready for their class 12th board examinations. However, chapters like Scalar and spot items can be mind-boggling for the students to tackle, so they need additional resources for self-practice to expert this chapter for their examination. RD Sharma solutions Therefore, it is enthusiastically proposed that the students use the RD Sharma class 12th exercise FBQ reply for their rational help and foster their capacities to perform better in exams.

This Story also Contains

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 $\vec{a}$ $=\frac{1}{|\vec{a}|} \cdot(\vec{b} \cdot \vec{a})$
$\begin{aligned} \vec{b} \cdot \vec{a}=&(5 \times 2)+(-3 \times-1)+(-4 \times 2) \\ &=10+3-8 \\ &=5 \\ \end{aligned}$
$\begin{aligned} |\vec{a}|=& \sqrt{2^{2}+1^{2}+2^{2}} \\ =& \sqrt{4+1+4} \\ &=\sqrt{9}\end{aligned}$
$\begin{aligned} &=3 \\ &\frac{\text { Projection of } \vec{a} \text { and } \vec{b}}{\text { Projection of } \vec{b} \text { and } \vec{a}}=\frac{\frac{1}{\sqrt{2}}}{\frac{5}{3}} \\ &\qquad=\frac{3}{5 \sqrt{2}} \end{aligned}$

Therefore, the RD Sharma class 12th exercise FBQ for the chapter Increasing and a Decreasing request has accommodated the advantage of students. The solutions of RD Sharma class 12 solution chapter 23 exercise FBQ are compact answer kinds of inquiries by which students can work on their scientific thinking and time usage abilities.

The RD Sharma books help plan for tests and exams. However, the students can likewise chip away at their schoolwork and tasks with its assistance. There are just 18 inquiries gave in the RD Sharma class 12th exercise FBQ, and This chapter covers every one of the fundamental ideas the vector involves heading and size. FBQ questions are identified with Scalar as speck items, the point between vectors, the projection of vectors, and commonly opposite vectors.

Though there are various motivations to utilize the RD Sharma class 12 solution of Scalar and spot item exercise FBQ for exam planning, a portion of the advantages are recorded beneath:-

The inquiries are intended to conceal practically every one of the subjects that can be asked in the exams.
The RD Sharma class 12 chapter 23 exercise FBQ involves routinely presented requests in board exams and gives tips to settle the inquiries.
Experts in the field of maths set up the RD Sharma class 12th exercise FBQ questions.
RD Sharma class 12 solutions chapter 23 exercise FBQ handles the homework, as teachers take the help of a comparative book to give out a task. It is in like manner supportive in getting ready for exams as teachers suggest RD Sharma's class 12th exercise FBQ to design question papers.
The main part of RD Sharma's class 12 exercise FBQ is that it will in general be downloaded freed from the Career360 webpage; it doesn't cost a single penny to have this book.