RD Sharma Class 12 Exercise 25 FBQ Scalar Triple Product Solutions Maths

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

Edited By Lovekush kumar saini | Updated on Jan 24, 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.

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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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Frequently Asked Questions (FAQs)

1. What is a scalar triple product?

The scalar triple product is simply defined as the dot product of any one of the given vectors with the cross product of the other two vectors. This is also called mixed product, triple scalar product, and box product.

2. What is the formula of the triple vector product?

The formula for triple vector product is:

A x (b x c) = (a . c) b – (b . c)a

3. What is the scalar product of two vectors?

It is said to be an algebraic operation that takes two equal-length sequences of numbers and returns a single number when the dot product and the scalar product is mixed.

4. What does the vector triple product represent?

The scalar triple product formula represents the parallelepiped volume whose three coterminous edges represent the three vectors a, b and c.

5. What is a scalar triple product used for?

The value of the vector triple product can be found by the product of a vector with the cross product of the other two vectors. Vector triple product arises in the discussion of rotational motion, and generally, it is used to simplify the different vector formulas.