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

Edited By Lovekush kumar saini | Updated on Jan 27, 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.

Scalar or Dot Products Excercise: FBQ

Scalar or dot product exercise fill in the blanks question 1

$\sqrt{3}$
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

$\pm \frac{3}{4}$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

-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

$\pi$
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

$\sqrt{2}$
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

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

$\frac{2\pi}{3}$
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 betweenaandbis
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

$\frac{\pi}{3}$
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

$\vec{a}.\vec{b}$
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

$\frac{\pi}{4}$
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

$\frac{\pi}{3}$
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

$\sqrt{a^{2}+b^{2}+c^{2}}$
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

$\pi$
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

$\left | \vec{r} \right |^{2}$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

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

$\vec{r}$
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

$\frac{\pi}{4}$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

$\frac{3}{5\sqrt{2}}$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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