RD Sharma Class 12 Exercise 25 FBQ Scalar Triple Product Solutions Maths - Download PDF Free Online

RD Sharma Class 12 Exercise 25 FBQ Scalar Triple Product Solutions Maths - Download PDF Free Online

Updated on 24 Jan 2022, 09:16 PM IST

The Class 12 RD Sharma chapter 25 exercise FBQ solution deals with the chapter of Scalar triple product which is one of the interesting chapters of the RD Sharma solution and most students find solving it easily. The RD Sharma class 12th exercise FBQ should be used by every student to make sure they have the complete knowledge and understanding about the chapters and its each concept as no other study material resource will provide this brief explanation of the chapter as in comparison of the RD Sharma solutions.

RD Sharma Class 12 Solutions Chapter25FBQ Scalar Triple Product - Other Exercise

Scalar Triple Product Excercise: 25 FBQ

Scalar Triple Product Exercise Fill In The Blanks Question 1

Answer:$-1$
Hint :- Just simplify using dot product.
Solution: $\begin{aligned} &\lceil\mathrm{i} \mathrm{k} \mathrm{j}]+\lceil\mathrm{k} \mathrm{j} \mathrm{i}]+\lceil\mathrm{j} \mathrm{k} \mathrm{i}]\\ \end{aligned}$
$\begin{aligned} &=(i \times k) \cdot j+(k \times j) \cdot i+\lceil j \times k\rceil . i\\ &=-j . j+(-i) \cdot i+i\cdot i\\ &=-1-1+1\\ &=-1 \end{aligned}$

Scalar Triple Product Exercise Fill In The Blanks Question 3

Answer: 1
HINT : Simplify the numerator
Given:$\frac{(\vec{b} \times \vec{c}) \cdot(\vec{a}+\vec{b}+\vec{c})}{[\vec{a} \vec{b} \vec{c}]}$
Solution:$\frac{(\vec{b} \times \vec{c}) \cdot(\vec{a}+\vec{b}+\vec{c})}{[\vec{a} \vec{b} \vec{c}]}$
$=\frac{(\vec{b} \times \vec{c}) \cdot \vec{a}+(\vec{b} \times \vec{c}) \cdot \vec{b}+(\vec{b} \times \vec{c}) \cdot \vec{c}}{|\vec{a} \vec{b} \vec{c}|}$
$\begin{aligned} &\frac{[\vec{a} \vec{b} \vec{c}]+0+0}{\lfloor\vec{a} \vec{b} \vec{c}]} =1 \end{aligned}$

Scalar Triple Product Exercise Fill In The Blanks Question 4

Answer: 0
HINT :- Simplify the expression
Given: $\vec{a} \cdot\{(\vec{b}+\vec{c}) \times(\vec{a}+\vec{b}+\vec{c})\}$
Solution:$\vec{a} \cdot\{(\vec{b}+\vec{c}) \times(\vec{a}+\vec{b}+\vec{c})\}$
$\begin{aligned} &=\vec{a} \cdot\{(\vec{b} \times \vec{a})+(\vec{b} \times \vec{b})+(\vec{b} \times \vec{c})+(\vec{c} \times \vec{a})+(\vec{c} \times \vec{b})+(\vec{c}\times \vec{c})\} \\\\ &=\vec{a} \cdot\left\{(\vec{b} \times \vec{a})+0+(\vec{b} \times \vec{c})+\left(\vec{c} \times \vec{a}\right)+(\vec{c} \times \vec{b})+0\right\} \end{aligned}$
$\begin{aligned} &=\vec{a} \cdot(\vec{b} \times \vec{a})+\vec{a} \cdot(\vec{b} \times \vec{c})+\vec{a} \cdot(\vec{c} \times \vec{a})+\vec{a} \cdot(\vec{c} \times \vec{b}) \\ \end{aligned}$
$\begin{aligned} &=0+[ \vec{a}\hspace{0.2cm} \vec{b}\hspace{0.2cm} \vec{c}]+0+[ \vec{a} \hspace{0.2cm}\hspace{0.2cm}\vec{c} \hspace{0.2cm}\vec{b}] \\\\ &=[ \vec{a} \hspace{0.2cm}\vec{b}\hspace{0.2cm} \vec{c}]+[\vec{a}\hspace{0.2cm} \vec{c}\hspace{0.2cm} \vec{b}] \\\\ &=(\therefore [ \vec{a}\hspace{0.2cm} \vec{b}\hspace{0.2cm} \vec{c}]=-[ \vec{a}\hspace{0.2cm} \vec{c}\hspace{0.2cm} \vec{b}] \end{aligned}$
$=0$

Scalar Triple Product Exercise Fill In The Blanks Question 5

Answer: 0
HINT :- Just simplify using scalar triple product
Given: $[ \overrightarrow{a} -\overrightarrow{b}\overrightarrow{b}-\overrightarrow{c}\overrightarrow{c}-\overrightarrow{a}]$

Solution:$[ \overrightarrow{a} -\overrightarrow{b}\overrightarrow{b}-\overrightarrow{c}\overrightarrow{c}-\overrightarrow{a}]$
$=\left ( \overrightarrow{a}-\vec{b} \right ).\left \{ \left ( \overrightarrow{b}- \overrightarrow{c} \right ) \times \left ( \overrightarrow{c}-\overrightarrow{a} \right )\right \}$
$=\left ( \overrightarrow{a}-\vec{b} \right ).\left \{ \left ( \overrightarrow{b}\times \overrightarrow{c} \right ) - \left ( \overrightarrow{b}\times \overrightarrow{a} \right )-\left ( \overrightarrow{c}\times\overrightarrow{c} \right )+\left ( \overrightarrow{c}\times \overrightarrow{a} \right )\right \}$
$=\left ( \overrightarrow{a}-\vec{b} \right ).\left \{ \left ( \overrightarrow{b}\times \overrightarrow{c} \right ) - \left ( \overrightarrow{b}\times \overrightarrow{a} \right )+\left ( \overrightarrow{c}\times \overrightarrow{a} \right )\right \}$
$=\overrightarrow{a}.\left ( \overrightarrow{b}\times \vec{c} \right )-\overrightarrow{a}.\left ( \vec{b}\times \vec{a} \right )+\vec{a}.\left ( \vec{c}\times \vec{a} \right )$
$=-\overrightarrow{b}.\left ( \overrightarrow{b}\times \vec{c} \right )+\overrightarrow{b}.\left ( \vec{b}\times \vec{a} \right )-\vec{b}.\left ( \vec{c}\times \vec{a} \right )$

$=0$

Scalar Triple Product Exercise Fill In The Blanks Question 6

Answer: $\left \lceil \overrightarrow{a} \right \rceil^{2}\left \lceil \overrightarrow{b} \right \rceil^{2}$
HINT :- Solve for the given vectors & simplify it.
Given: $\left \lfloor (\vec{a}\: \: \vec{b} \: \: \vec{a}) \times \vec{b} \right \rfloor+(\vec{a} \cdot \vec{b})^{2}$
Solution:$\left \lfloor (\vec{a}\: \: \vec{b} \: \: \vec{a}) \times \vec{b} \right \rfloor+(\vec{a} \cdot \vec{b})^{2}$
$\begin{aligned} &=[(\vec{a} \times \vec{b}) \cdot(\vec{a} \times \vec{b})]+(\vec{a} \cdot \vec{b})^{2} \\\\ &=(\vec{a} \times \vec{b})^{2}+(\vec{a} \cdot \vec{b})^{2} \\\\ &=\left[\vec { a } | ^ { 2 } \left[\left.\vec{b}\right|^{2} \operatorname{Sin}^{2} \theta+\left[\vec { a } | ^ { 2 } \left[\left.\vec{b}\right|^{2} \cos ^{2} \theta\right.\right.\right.\right. \end{aligned}$
$\begin{aligned} &=\left[\left.\vec{a}\right|^{2}|\vec{b}|^{2}\left(\operatorname{Sin}^{2} \theta+\cos ^{2} \theta\right)\right. \\\\ &=\left[\left.\vec{a}\right|^{2}|\vec{b}|^{2}\right. \end{aligned}$

Scalar Triple Product Exercise Fill In The Blanks Question 7

Answer: $\pm 12$
HINT :- Simplify the given value.
Given: Volume of parallelopiped =6 cubic units
Solution:
We need to find,
$\left \lceil \overrightarrow{a}+\overrightarrow{b}\overrightarrow{b}+\overrightarrow{c}\overrightarrow{c}+\overrightarrow{a} \right \rceil$$=\left ( \overrightarrow{a}+\overrightarrow{b} \right ).\left \lceil \left ( \overrightarrow{b}+\overrightarrow{c} \right )\times \left ( \overrightarrow{c}+\overrightarrow{a} \right )\right \rceil$

(using scalar triple product)

$=(\vec{a}+\vec{b}) \cdot[(\vec{b} \times \vec{c})+(\vec{b} \times \vec{a})+(\vec{c} \times \vec{c})+(\vec{c} \times \vec{a})]$

$=(\vec{a}+\vec{b}) \cdot[(\vec{b} \times \vec{c})+(\vec{b} \times \vec{a})+0+(\vec{c} \times \vec{a})]$

$=\vec{a} \cdot(\vec{b} \times \vec{c})+\vec{a} \cdot(\vec{b} \times \vec{a})+\vec{a} \cdot\left(\vec{c} \times\overrightarrow{a}\right)$

$+\vec{b} \cdot(\vec{b} \times \vec{c})+\vec{b} \cdot(\vec{b} \times \vec{a})+\vec{b} \cdot\left(\vec{c} \times\overrightarrow{a}\right)$

we Know $\overrightarrow{a}.\left ( \overrightarrow{b}\times \overrightarrow{a} \right )=0$

$=\left \lceil \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right \rceil+0+0+0+0+\left \lceil \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right \rceil$

$=2\left \lceil \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right \rceil$

Now, value of volume of parallelopiped =6

$=\pm \left \lceil \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right \rceil=6$

Substituting values we get

$=\pm 2\left ( 6 \right )$

$=\pm 12$

Scalar Triple Product Exercise Fill In The Blanks Question 8

Answer:
HINT : Use formula of $\left \lceil \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right \rceil$ & then simplify it.
Given: Volume of parallel piped = 8 Sq. units.
$\therefore \left \lceil \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right \rceil=8$ Sq.Units
Solution:

Now, For values
$\left \lceil \overrightarrow{3a}\hspace{0.2cm}\overrightarrow{4b}\hspace{0.2cm}\overrightarrow{5c} \right \rceil=\pm 3\times 4\times 5 \left \lceil \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right \rceil$
$=\pm 480$

Scalar Triple Product Exercise Fill In The Blanks Question 9

Answer: 0
HINT :- Use formula of volume of parallelepiped $=\left [ \overrightarrow{a}\: \overrightarrow{b}\: \overrightarrow{c} \right ]$
Given: Adjacent edges
$\overrightarrow{a}-\overrightarrow{b}\overrightarrow{b}-\overrightarrow{c}$ & $\overrightarrow{c}-\overrightarrow{a}$,where $\overrightarrow{a},\overrightarrow{b}$&$\overrightarrow{c}$are adjacent edges.
Solution:
∴ volume of parallelopied
$=[\vec{a}-\vec{b}\hspace{0.2cm}\vec{b}-\vec{c}\hspace{0.2cm} \vec{c}-\vec{a}]$
$=(\vec{a}-\vec{b}) \cdot\left[(\vec{b}-\vec{c}) \times\left(\vec{c}-\vec{a}\right)\right]$
$=(\vec{a}-\vec{b}) \cdot\left[(\vec{b}\times \vec{c}) -\left(\vec{b}\times \vec{a}\right)-\left ( \vec{c}\times \vec{c} \right )=\left ( \vec{c}\times \vec{a} \right )\right]$
$=\left[\vec{a}.(\vec{b}\times \vec{c}) -\vec{a}.\left(\vec{b}\times \vec{a}\right)+\vec{a}.\left ( \vec{c}\times \vec{a} \right )-\vec{b}.\left ( \vec{b}\times \vec{c} \right )+\vec{b}.\left ( \vec{b}\times \vec{a} \right )-\vec{b}.\left ( \vec{c\times \vec{a}} \right )\right]$

$\left ( \therefore \vec{a}.\left ( \vec{b}\times \vec{c} \right )=\vec{b}.\left ( \vec{c}\times \vec{a} \right )=\left [ \vec{a}\vec{b}\vec{c} \right ] \right )$
$=0$

Scalar Triple Product Exercise Fill In The Blank Question 10

Answer: 4
HINT :-We know that for adjacent edges , $\overrightarrow{a}\overrightarrow{b}$&$\overrightarrow{c}$
Volume of parallelopiped $=\left [ \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right ]$$=\left ( \overrightarrow{a}\times \overrightarrow{b} \right ).\overrightarrow{c}$

Given:

Volume of parallelepiped, $\left [ \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right ]=24$

And $=\left (\overrightarrow{a}\times \overrightarrow{b} \right )=6$

Solution:

Now , we have values of volume of parallelopiped

$=\left (\overrightarrow{a}\times \overrightarrow{b} \right ).\vec{c}=24$

$6\overrightarrow{c}=24$


$\therefore \overrightarrow{c}=4$

Where ,$\overrightarrow{c}$represents the height of the parallelopiped

Scalar Triple Product Exercise Fill In The Blank Question 11

Answer: 0
HINT :- When the vectors are is same plane, they are coplanar & their scalar triple product is zero.
Given:
$\overrightarrow{r}.\overrightarrow{a}=0$ $\overrightarrow{r}.\overrightarrow{b}=0$ &$\overrightarrow{r}.\overrightarrow{c}=0$, for some non- zero vector $\overrightarrow{r},$
Solution:
As $\overrightarrow{r}.\overrightarrow{a}=0$
then either $\overrightarrow{a}$ is zero or if their dot product is zero, then, $\overrightarrow{r}$ is a vector perpendicular to all.
Thus, as all lie on same planes $\vec{a} \cdot\left(\vec{b} \times \vec{c}\right)=[\vec{a}\hspace{0.2cm} \vec{b} \hspace{0.2cm}\vec{c}]$
= 0


The RD Sharma class 12 solution of Scalar Triple Product exercise FBQ is the fill in the blank type question, which consists of a total of 11 questions covering the essential concepts of the chapter mentioned below-

  • The volume of coplanar vectors

  • Volume of non-coplanar vectors

  • Scalar triple product in component form

  • The volume of the parallelepiped

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