What is the most recommended solution guide from the previous batch of students to their juniors?

The RD Sharma books are the most recommended reference guides by the previous batch of students to their juniors.

Where can I find the correct answers for the questions in class 12, Scalar Triple Product VSA section?

The right answers for the questions asked in the VSA part of the Scalar Triple Product can be found in the RD Sharma Class 12th Chapter 25 VSA book.

How many questions are asked in the Very Short Answer part of chapter 25 in class 12 mathematics?

The number of questions asked in the class 12 chapter 25 VSA part is eleven. The solved sums can be found in the RD Sharma Class 12th Chapter 25 VSA reference guide.

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When you access the RD Sharma books online via the Career 360 website, you need not pay even a single rupee to view the solution books.

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

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

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

Updated on 24 Jan 2022, 09:16 PM IST

The hectic schedule for the class 12 students in completing their homework and preparing for the tests and exams has left them with no time to do other chores. Even most students do not find the time to attend separate tuition apart from the school timings. With such little time, it is essential to make use of each minute in the right manner. Students who find mathematics and concepts like the Scalar Triple Product must start using the RD Sharma Class 12th Chapter 25 VSA solution book.

RD Sharma Class 12 Solutions Chapter25VSA Scalar Triple Product - Other Exercise

Scalar Triple Product Excercise: 25 VSA

Scale Triple Product Exercise Very Short Answer Question, question 1.

Answer:
24
Hint:
Use scalar triple product formula.
Given:
$\left [2 \hat{i} \; \; 3 \hat{j}\; \; 4 \hat{k} \right ]$
Solution:
$\begin{aligned} &{\left[\begin{array}{lll} 2 \hat{i} & 3 \hat{j} & 4 \hat{k} \end{array}\right]} \\ &=(2 \hat{i}) \cdot(3 \hat{j}) \times(4 \hat{k})\} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\because[\overrightarrow{\boldsymbol{a}} \overrightarrow{\boldsymbol{b}} \overrightarrow{\boldsymbol{c}}]=\overrightarrow{\boldsymbol{a}} \cdot(\overrightarrow{\boldsymbol{b}} \times \overrightarrow{\boldsymbol{c}})] \\ &=(2 \hat{i}) \cdot 12(\hat{j} \times \hat{k}) \\ &=(2 \hat{i}) \cdot 12(\hat{i}) \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;\; \; \; \; \; \; \; \; \; \quad[\because \hat{\boldsymbol{j}} \times \widehat{\boldsymbol{k}}=\hat{\boldsymbol{i}}] \\ &=(24)(i\cdot i) \\ &=24.1 \\ &=24 \end{aligned}$

Scale Triple Product Exercise Very Short Answer Question, question 2.

Answer:
2
Hint:
Use scalar triple product formula.
Given:
$\left[\begin{array}{lll} \hat{i}+\hat{j} & \hat{j}+\hat{k} & \hat{k}+\hat{i} \end{array}\right]$
Solution:
$\begin{aligned} &{[\hat{i}+\hat{j} \hat{j}+\hat{k} \hat{k}+\hat{i}]} \\ &=(\hat{i}+\hat{j}) \cdot\{(\hat{j}+\hat{k}) \times(\hat{k}+\hat{i})\} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\because[\overrightarrow{\boldsymbol{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\boldsymbol{c}}]=\overrightarrow{\boldsymbol{a}} \cdot(\overrightarrow{\boldsymbol{b}} \times \overrightarrow{\boldsymbol{c}})] \\ &=(\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{j}}) \cdot\{(\hat{\boldsymbol{j}} \times \hat{\boldsymbol{k}})+(\hat{\boldsymbol{j}} \times \hat{\boldsymbol{\imath}})+(\widehat{\boldsymbol{k}} \times \widehat{\boldsymbol{k}})+(\widehat{\boldsymbol{k}} \times \hat{\boldsymbol{\imath}})\} \end{aligned}$
$\begin{aligned} &=(\hat{i}+\hat{j}) \cdot(\hat{i}+(-\hat{k})+0+\hat{j}) \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\begin{array}{l} \because \hat{\boldsymbol{j}} \times \widehat{\boldsymbol{k}}=\hat{\boldsymbol{\imath}}, \hat{\boldsymbol{j}} \times \hat{\boldsymbol{\imath}}=-\widehat{\boldsymbol{k}} \\ \hat{\boldsymbol{k}} \times \widehat{\boldsymbol{k}}=\mathbf{0}, \widehat{\boldsymbol{k}} \times \hat{\boldsymbol{\imath}}=\hat{\boldsymbol{j}} \end{array}\right] \\ &=(\hat{i}+\hat{j}) \cdot(\hat{i}-\hat{k}+\hat{j}) \\ &=(\hat{i} \hat{i})-(\hat{i} \hat{k})+(\hat{i} \cdot \hat{j})+(\hat{j} \hat{i})-(\hat{j} \hat{k})+(\hat{j} \cdot \hat{j}) \end{aligned}$
$\begin{aligned} &=1-0+0+0-0+1 \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\because \hat{\imath} . \hat{\boldsymbol{\imath}}=\hat{\boldsymbol{\jmath}} \cdot \hat{\boldsymbol{j}}=\mathbf{1}, \hat{\boldsymbol{\imath}} . \widehat{\boldsymbol{k}}=\hat{\boldsymbol{\imath}} . \hat{\boldsymbol{\jmath}}=\hat{\boldsymbol{\jmath}} \cdot \hat{\boldsymbol{\imath}}=\hat{\boldsymbol{\jmath}} \cdot \widehat{\boldsymbol{k}}=\mathbf{0}] \\ &=2 \end{aligned}$

Scale Triple Product Exercise Very Short Answer Question, question 3.

Answer:
0
Hint:
Use scalar triple product formula.
Given:
$\left[\begin{array}{lll} \hat{i}-\hat{j} & \hat{j}-\hat{k} & \hat{k}-\hat{i} \end{array}\right]$
Solution:
$\left[\begin{array}{lll} \hat{i}-\hat{j} & \hat{j}-\hat{k} & \hat{k}-\hat{i} \end{array}\right]$
$\begin{aligned} &=(\hat{i}-\hat{j}) \cdot\{(\hat{j}-\hat{k}) \times(\hat{k}-\hat{i})\} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\because[\overrightarrow{\boldsymbol{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]=\overrightarrow{\boldsymbol{a}} \cdot(\overrightarrow{\boldsymbol{b}} \times \overrightarrow{\boldsymbol{c}})] \\ &=(\hat{i}-\hat{j}) \cdot((\hat{j} \times \hat{k})-(\hat{j} \times \hat{i})-(\hat{k} \times \hat{k})+(\hat{k} \times \hat{i})\} \end{aligned}$
$\begin{aligned} &=(\hat{i}-\hat{j}) \cdot\{\hat{i}-(-\hat{k})-0+\hat{j}\} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\begin{array}{l} \because \hat{\jmath} \times \widehat{\boldsymbol{k}}=\hat{\boldsymbol{k}}, \hat{\boldsymbol{j}} \times \hat{\boldsymbol{\imath}}=-\widehat{\boldsymbol{k}} \\ \widehat{\boldsymbol{k}} \times \hat{\boldsymbol{i}}=\hat{\boldsymbol{j}}, \widehat{\boldsymbol{k}} \times \widehat{\boldsymbol{k}}=\mathbf{0} \end{array}\right] \\ &=(\hat{i}-\hat{j}) \cdot(\hat{i}+\hat{j}+\hat{k}) \\ &=(\hat{i} \hat{i})+(\hat{i} \cdot \hat{j})+(\hat{i} \cdot \hat{k})-(\hat{j} \hat{i})-(\hat{j} . \hat{j})-(\hat{j} \cdot \hat{k}) \end{aligned}$
$\begin{aligned} &=1+0+0-0-1-0\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; &\left[\begin{array}{l} \because \hat{\imath} . \hat{\jmath}=\hat{\imath} . \hat{\boldsymbol{k}}=\hat{\jmath} \cdot \hat{\boldsymbol{i}}=\hat{\boldsymbol{j}} \cdot \hat{\boldsymbol{k}}=0 \\ \hat{\boldsymbol{\imath}} \cdot \hat{\boldsymbol{i}}=\hat{\boldsymbol{\jmath}} \cdot \hat{\boldsymbol{j}}=\mathbf{1} \end{array}\right] \end{aligned}$
$=1-1=0$

Scalar Triple Product Exercise Very Short Answer Question, question 5.

Answer:

Hint:
Use scalar triple product formula.
Given:
Given vectors are

.
Solution:
Let
$\left.\begin{array}{c} \vec{\alpha}=\hat{\imath}+\hat{\jmath} \\ \vec{\beta}=\hat{\imath}+2 \hat{\jmath} \\ \vec{\gamma}=\hat{\imath}+\hat{\jmath}+\pi \hat{k} \end{array}\right\}$ Eq.(i)
We know that the volume of a parallel piped whose three adjacent edge are
$\vec{\alpha} \; \; \vec{\beta}\: \text{and} \: \vec{\gamma}$ is equal to modulus of scalar triple product i.e. $\left |\left [\vec{\alpha} \; \; \vec{\beta}\: \: \vec{\gamma} \right ] \right |$
So, first we will find $\left[\begin{array}{lll} \vec{\alpha} & \vec{\beta} & \vec{\gamma} \end{array}\right] \text { then }\left|\left[\begin{array}{lll} \vec{\alpha} & \vec{\beta} & \vec{\gamma} \end{array}\right]\right|$
$\begin{aligned} &\therefore\left[\begin{array}{lll} \vec{\alpha} & \vec{\beta} & \vec{\gamma} \end{array}\right]=\vec{\alpha} \cdot(\vec{\beta} \times \vec{\gamma}) \\ &=(\hat{\imath}+\hat{\jmath}) \cdot\{(\hat{\imath}+2 \hat{\jmath}) \times(\hat{\imath}+\hat{\jmath}+\pi \hat{k})\} \end{aligned}$ From (1)
$\begin{aligned} &=(\hat{\imath}+\hat{\jmath}) \cdot\left|\begin{array}{lll} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 1 & 2 & 0 \\ 1 & 1 & \pi \end{array}\right| \\ &=(\hat{\imath}+\hat{\jmath}) \cdot\{\hat{\imath}(2 \pi-0)-\hat{\jmath}(\pi-0)+\hat{k}(1-2)\} \\ &=(\hat{\imath}+\hat{\jmath}) \cdot\{2 \pi \hat{\imath}-\pi \hat{\jmath}-\hat{k}\} \end{aligned}$
$\begin{aligned} &=2 \pi(\hat{\imath} . \hat{\imath})-\pi(\hat{\imath} . \hat{\jmath})-(\hat{\imath} . \hat{k})+2 \pi(\hat{\jmath} \cdot \hat{\imath})-\pi(\hat{\jmath} \cdot \hat{\jmath})-(\hat{\jmath} \cdot \hat{k})\\ &=2 \pi-\pi=\pi \end{aligned}$
Now, $\left |\left [\vec{\alpha} \; \; \vec{\beta}\: \: \vec{\gamma} \right ] \right |=\left |\pi \right |=\pi$ cubic units

Scalar Triple Product Exercise Very Short Answer Question, question 7.

Answer:
2
Hint:
Use scalar triple product formula.
Given:
Vectors are $\left(\sec ^{2} A\right) \hat{\imath}+\hat{\jmath}+\hat{k}, \hat{\imath}+\left(\sec ^{2} B\right) \hat{\jmath}+\hat{k} \text { and } \hat{\imath}+\hat{\jmath}+\left(\sec ^{2} C\right) \hat{k}$ are co planar.
Solution:
Let
$\left.\begin{array}{l} \vec{\alpha}=\left(\sec ^{2} A\right) \hat{\imath}+\hat{\jmath}+\hat{k} \\ \vec{\beta}=\hat{\imath}+\left(\sec ^{2} B\right) \hat{\jmath}+\hat{k} \\ \vec{\gamma}=\hat{\imath}+\hat{\jmath}+\left(\sec ^{2} C\right) \hat{k} \end{array}\right\}$ Eq.(i)
We know if three vectors are co planar then their scalar triple product is zero.
So if α,β and γ are co planar then the scalar triple product must be zero.
$\therefore\left[\begin{array}{lll} \vec{\alpha} & \vec{\beta} & \vec{\gamma} \end{array}\right]=0$
$\Rightarrow \vec{\alpha} \cdot(\vec{\beta} \times \vec{\gamma})=0 \quad[\because[\vec{a} \quad \vec{b} \quad \vec{c}]=\vec{a} \cdot(\vec{b} \times \vec{c})]$
$\Rightarrow\left\{\left(\sec ^{2} A\right) \hat{\imath}+\hat{\jmath}+\hat{k}\right\} \cdot\left\{\hat{\imath}+\left(\sec ^{2} B\right) \hat{\jmath}+\hat{k}\right\} \times\left\{\hat{\imath}+\hat{\jmath}+\left(\sec ^{2} C\right) \hat{k}\right\}=0$
$\Rightarrow\left\{\left(\sec ^{2} A\right) \hat{\imath}+\hat{\jmath}+\hat{k}\right\} \cdot\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 1 & \sec ^{2} B & 1 \\ 1 & 1 & \sec ^{2} C \end{array}\right|=0$
$\Rightarrow\left\{\left(\sec ^{2} A\right) \hat{i}+\hat{j}+\hat{k}\right\} \cdot\left\{\hat{i}\left(\sec ^{2} B \cdot \sec ^{2} C-1\right)-\hat{j}\left(\sec ^{2} C-1\right)+\hat{k}\left(1-\sec ^{2} B\right)\right\}=0$
$\begin{aligned} &\Rightarrow \sec ^{2} A\left(\sec ^{2} B \cdot \sec ^{2} C-1\right)(\hat{\imath} . \hat{\imath})+\left(\sec ^{2} B \cdot \sec ^{2} C-1\right)(\hat{j} \cdot \hat{\imath})\\ &+\left(\sec ^{2} B \cdot \sec ^{2} C-1\right)(\hat{k} \cdot \hat{\imath})\\ &-\left(\sec ^{2} C-1\right) \sec ^{2} A(\hat{\imath} \cdot \hat{j})-\left(\sec ^{2} C-1\right)(\hat{j} \cdot \hat{\jmath})-\left(\sec ^{2} C-1\right)(\hat{k} \cdot \hat{j})\\ &+\left(1-\sec ^{2} B\right) \sec ^{2} A(\hat{\imath} . \hat{k})+\left(1-\sec ^{2} B\right)(\hat{\jmath} . \hat{k})+\left(1-\sec ^{2} B\right)(\hat{k} \cdot \hat{k})=0 \end{aligned}$
$\begin{aligned} &\Rightarrow \sec ^{2} A\left(\sec ^{2} B \cdot \sec ^{2} C-1\right) \cdot 1+0+0-0-\left(\sec ^{2} C-1\right) \cdot 1-0+0+0 \\ &+\left(1-\sec ^{2} B\right) \cdot 1=0 \end{aligned}$ $\left[\begin{array}{l} \because \hat{\imath} \cdot \hat{\boldsymbol{\imath}}=\hat{\boldsymbol{j}} \cdot \hat{\boldsymbol{j}}=\widehat{\boldsymbol{k}} \cdot \widehat{\boldsymbol{k}}=\mathbf{1} \\ \hat{\boldsymbol{\imath}} \cdot \hat{\boldsymbol{j}}=\hat{\boldsymbol{j}} \cdot \hat{\boldsymbol{\imath}}=\mathbf{0}, \hat{\boldsymbol{\imath}} . \widehat{\boldsymbol{k}}=\widehat{\boldsymbol{k}} \cdot \hat{\boldsymbol{\imath}}=\mathbf{0} \\ \hat{\boldsymbol{j}} \cdot \widehat{\boldsymbol{k}}=\widehat{\boldsymbol{k}} \cdot \hat{\boldsymbol{j}}=\mathbf{0} \end{array}\right]$
$\begin{aligned} &\Rightarrow \sec ^{2} A \sec ^{2} B \sec ^{2} C-\sec ^{2} A-\sec ^{2} C+1+1-\sec ^{2} B=0 \\ &\Rightarrow \sec ^{2} A \sec ^{2} B \sec ^{2} C-\sec ^{2} A-\sec ^{2} B-\sec ^{2} C+2=0 \\ &\Rightarrow\left(1+\tan ^{2} A\right)\left(1+\tan ^{2} B\right)\left(1+\tan ^{2} C\right)-\left(1+\tan ^{2} A\right) \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\begin{array}{l} \because \sec ^{2} \theta-\tan ^{2} \theta=1 \\ \Rightarrow \sec ^{2} \theta=1+\tan ^{2} \theta \end{array}\right] \end{aligned}$
$\begin{aligned} &\Rightarrow\left(1+\tan ^{2} B+\tan ^{2} A+\tan ^{2} A \tan ^{2} B\right)\left(1+\tan ^{2} C\right)-1-\tan ^{2} A-1-\tan ^{2} B \\ &-1-\tan ^{2} C+2=0 \\ &\Rightarrow 1+\tan ^{2} B+\tan ^{2} A+\tan ^{2} A \tan ^{2} B+\tan ^{2} C+\tan ^{2} B \tan ^{2} C+\tan ^{2} A \tan ^{2} C \\ &+\tan ^{2} A \tan ^{2} B \tan ^{2} C-\tan ^{2} A-\tan ^{2} B-\tan ^{2} C-3+2=0 \\ &\Rightarrow 1+\tan ^{2} A \tan ^{2} B+\tan ^{2} B \tan ^{2} C+\tan ^{2} A \tan ^{2} C+\tan ^{2} A \tan ^{2} B \tan ^{2} C-1=0 \\ &\Rightarrow \tan ^{2} A \tan ^{2} B+\tan ^{2} B \tan ^{2} C+\tan ^{2} A \tan ^{2} C+\tan ^{2} A \tan ^{2} B \tan ^{2} C=0 \\ &\Rightarrow \tan ^{2} A \tan ^{2} B+\tan ^{2} B \tan ^{2} C+\tan ^{2} A \tan ^{2} C=-\tan ^{2} A \tan ^{2} B \tan ^{2} C \end{aligned}$
$\begin{aligned} &\Rightarrow \frac{\tan ^{2} A \tan ^{2} B+\tan ^{2} B \tan ^{2} C+\tan ^{2} A \tan ^{2} C}{\tan ^{2} A \tan ^{2} B \tan ^{2} C}=-1 \\ &\Rightarrow \frac{\tan ^{2} A \tan ^{2} B}{\tan ^{2} A \tan ^{2} B \tan ^{2} C}+\frac{\tan ^{2} B \tan ^{2} C}{\tan ^{2} A \tan ^{2} B \tan ^{2} C}+\frac{\tan ^{2} A \tan ^{2} C}{\tan ^{2} A \tan ^{2} B \tan ^{2} C}=-1 \\ &\Rightarrow \frac{1}{\tan ^{2} C}+\frac{1}{\tan ^{2} A}+\frac{1}{\tan ^{2} B}=-1 \\ &\Rightarrow\left(\frac{1}{\tan C}\right)^{2}+\left(\frac{1}{\tan A}\right)^{2}+\left(\frac{1}{\tan B}\right)^{2}=-1 \end{aligned}$
$\begin{aligned} &\Rightarrow(\cot C)^{2}+(\cot A)^{2}+(\cot B)^{2}=-1\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \quad\left[\because \frac{1}{\tan \theta}=\cot \theta\right] \\ &\Rightarrow \cot ^{2} C+\cot ^{2} A+\cot ^{2} B=-1 \\ &\Rightarrow\left(\cos e c^{2} C-1\right)+\left(\cos e c^{2} A-1\right)+\left(\operatorname{cosec}^{2} B-1\right)=-1 \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\begin{array}{l} \because \cos e c^{2} \theta-\cot ^{2} \theta=1 \\ \Rightarrow \cot ^{2} \theta=\cos e \boldsymbol{c}^{2} \boldsymbol{\theta}-1 \end{array}\right] \\ &\Rightarrow \operatorname{cosec}^{2} C+\operatorname{cosec}^{2} A+\operatorname{cosec}^{2} B-3=-1 \\ &\Rightarrow \operatorname{cosec}^{2} A+\operatorname{cosec}^{2} B+\operatorname{cosec}^{2} C=-1+3=2 \\ &\therefore \operatorname{cosec}^{2} A+\operatorname{cosec}^{2} B+\operatorname{cosec}^{2} C=2 \end{aligned}$

Scalar Triple Product Exercise Very Short Answer Question, question 8.

Answer:
144
Hint:
Use scalar triple product formula.
Given:
$|\vec{a}|=3 \text { and }|\vec{b}|=4 \text { and }\left[\begin{array}{lll} \overrightarrow{\boldsymbol{a}} & \vec{b} & \vec{a} \times \vec{b}]+(\vec{a} \cdot \vec{b})^{2} \end{array}\right.$
Solution:
$\left[\begin{array}{lll} \vec{a} & \vec{b} & \vec{a} \times \vec{b} \end{array}\right]+(\vec{a} \cdot \vec{b})^{2}$
$\begin{aligned} &=\{(\vec{a}) \times(\vec{b}) \cdot(\vec{a} \times \vec{b})\}+(\vec{a} \cdot \vec{b})^{2}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because\left[\begin{array}{lll} \vec{a} & \vec{b} & \vec{c}] \end{array}=(\vec{a} \times \vec{b}) \cdot \vec{c}\right]\right.\\ &=(\vec{a} \times \vec{b}) \cdot(\vec{a} \times \vec{b})+(\vec{a} \cdot \vec{b})^{2} \end{aligned}$
$\begin{aligned} &=|(\vec{a} \times \vec{b})|^{2}+(\vec{a} \cdot \vec{b})^{2} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \vec{a} \times \vec{a}=|\vec{a}|^{2}\right] \\ &=(|\vec{a}||\vec{b}| \sin \theta)^{2}+(|\vec{\alpha}||\vec{b}| \cos \theta)^{2} & {\left[\begin{array}{l} \because \vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta \\ \vec{a} \times \vec{b}=|\vec{a}||\vec{b}| \sin \theta \end{array}\right]} \end{aligned}$
$\begin{aligned} &=(|\vec{a}||\vec{b}|)^{2} \sin ^{2} \theta+(|\vec{a}||\vec{b}|)^{2} \cos ^{2} \theta \\ &=(|\vec{a}||\vec{b}|)^{2}\left(\sin ^{2} \theta+\cos ^{2} \theta\right) \\ &=(|\vec{a}||\vec{b}|)^{2} \cdot 1 \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \sin ^{2} A+\cos ^{2} A=1\right] \\ &=(3.4)^{2}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\because|\vec{a}|=3,|\vec{b}|=4] \\ &=(12)^{2}=144 \end{aligned}$

Scalar Triple Product Exercise Very Short Answer Question, question 9.

Answer:
10
Hint:
Use scalar triple product formula.
Given:
$\left[\begin{array}{llll} 3 \vec{a}+7 \vec{b} & \vec{c} & \vec{d} \end{array}\right]=\lambda\left[\begin{array}{llll} \vec{a} & \vec{c} & \vec{d} \end{array}\right]+\mu\left[\begin{array}{lll} \vec{b} & \vec{c} & \vec{d} \end{array}\right]$
Solution:
$\left[\begin{array}{llll} 3 \vec{a}+7 \vec{b} & \vec{c} & \vec{d} \end{array}\right]=\lambda\left[\begin{array}{llll} \vec{a} & \vec{c} & \vec{d} \end{array}\right]+\mu\left[\begin{array}{lll} \vec{b} & \vec{c} & \vec{d} \end{array}\right]$
L.H.S
$\begin{array}{ll} =[3 \vec{a}+7 \vec{b} \quad \vec{c} \quad \vec{d}] \\ =\{(3 \vec{a}+7 \vec{b}) \times \vec{c}\} \cdot \vec{d} &\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; {[\because[\vec{a} \quad \vec{b} \quad \vec{c}]=\vec{a} \times \vec{b} \cdot \vec{c}=\vec{a} \cdot(\vec{b} \times \vec{c})]} \\ =\{3(\vec{a} \times \vec{c})+7(\vec{b} \times \vec{c})\} \cdot \vec{d} & \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; {[\text { By distributive property }]} \end{array}$
$\begin{aligned} &=3\{(\vec{a} \times \vec{c}) \cdot \vec{d}\}+7\{(\vec{b} \times \vec{c}) \cdot \vec{d}\} \\ &=3\left[\begin{array}{llll} \vec{a} & \vec{c} & \vec{d} \end{array}\right]+7[\vec{b} \quad \vec{c} \quad \vec{d}] \end{aligned} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because(\vec{a} \times \vec{b}) \cdot \vec{c}=\left[\begin{array}{lll} \vec{a} & \vec{b} & \vec{c} \end{array}\right]\right.$
Since, L.H.S = R.H.S
i.e.$\left[\begin{array}{llll} 3 \vec{a}+7 \vec{b} & \vec{c} & \vec{d} \end{array}\right]=\lambda\left[\begin{array}{llll} \vec{a} & \vec{c} & \vec{d} \end{array}\right]+\mu\left[\begin{array}{lll} \vec{b} & \vec{c} & \vec{d} \end{array}\right]$
Comparing and equating co efficient from both sides.
We get,
$\begin{aligned} &\lambda=3 \text { and } \mu=7 \\ &\therefore \lambda+\mu=3+7=10 \end{aligned}$

Scalar Triple Product Exercise Very Short Answer Question, question 10.

Answer:
0
Hint:
Use scalar triple product formula.
Given:
a,b,c are non co planar vectors and$\frac{\vec{a} \cdot(\vec{b} \times \vec{c})}{(\vec{c} \times \vec{a}) \cdot \vec{b}}+\frac{\vec{b} \cdot(\vec{a} \times \vec{c})}{\vec{c} \cdot(\vec{a} \times \vec{b})}$
Solution:
$\frac{\vec{a} \cdot(\vec{b} \times \vec{c})}{(\vec{c} \times \vec{a}) \cdot \vec{b}}+\frac{\vec{b} \cdot(\vec{a} \times \vec{c})}{\vec{c} \cdot(\vec{a} \times \vec{b})}$
$=\frac{\vec{a} \cdot(\vec{b} \times \vec{c})}{\vec{c} \cdot(\vec{a} \times \vec{b})}+\frac{(\vec{b} \times \vec{a}) \cdot \vec{c}}{\vec{c} \cdot(\vec{a} \times \vec{b})} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\begin{array}{l} \because \text { The position of dot and cross can be interchanged } \\ \text { provided that the cyclic order of the vectors remain } \\ \text { same in scalar triple product. } \end{array}\right]$
$=\frac{\vec{a} \cdot(\vec{b} \times \vec{c})}{(\vec{a} \times \vec{b}) \cdot \vec{c}}+\left\{-\frac{(\vec{a} \times \vec{b}) \cdot \vec{c}}{(\vec{a} \times \vec{b}) \cdot \vec{c}}\right\}\; \; \; \; \; \; \; \; \quad[\because \vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{a} \text { and }(\vec{a} \times \vec{b})=-(\vec{b} \times \vec{a})]$
$\left[\because(\vec{a} \times \vec{b}) \cdot \vec{c}=\left[\begin{array}{lll} \vec{a} & \vec{b} & \vec{c} \end{array}\right]=\vec{a} \cdot(\vec{b} \times \vec{c})\right]$
$=1-1=0$

Scalar Triple Product Exercise Very Short Answer Question, question 11.

Answer:
-10
Hint:
Use scalar triple product formula.
Given:
Vectors are $\vec{a}=2 \hat{\imath}+\hat{\jmath}+3 \hat{k}, \vec{b}=-\hat{\imath}+2 \hat{\jmath}+\hat{k} \text { and } \vec{c}=3 \hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}$
Solution:
We have
$\left.\begin{array}{l} \vec{a}=2 \hat{\imath}+\hat{\jmath}+3 \hat{k} \\ \vec{b}=-\hat{\imath}+2 \hat{\jmath}+\hat{k} \\ \vec{c}=3 \hat{\imath}+\hat{\jmath}+2 \hat{k} \end{array}\right\}$ Eq.(i)
$\begin{aligned} \text { Now }(\vec{b} \times \vec{c}) &=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & k \\ -1 & 2 & 1 \\ 3 & 1 & 2 \end{array}\right| \\ &=\hat{i}(4-1)-\hat{j}(-2-3)+\hat{k}(-1-6) \end{aligned}$
$=3 \hat{i}-(-5) \hat{j}+(-7) \hat{k}=3 \hat{i}+5 \hat{j}-7 \hat{k}$
$\therefore \overrightarrow{\boldsymbol{a}} \cdot(\overrightarrow{\boldsymbol{b}} \times \overrightarrow{\boldsymbol{c}})=(2 \hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}}+3 \hat{\boldsymbol{k}}) \cdot(3 \hat{\boldsymbol{\imath}}+5 \hat{\boldsymbol{\jmath}}-7 \widehat{\boldsymbol{k}})$
$=6(\hat{\imath} . \hat{\imath})+10(\hat{\imath} . \hat{\jmath})-14(\hat{\imath} . \hat{k})+3(\hat{\jmath} \cdot \hat{\imath})+5(\hat{\jmath} \cdot \hat{\jmath})-7(\hat{\jmath} \cdot \hat{k})+9(\hat{k} \cdot \hat{\imath})+15(\hat{k} \cdot \hat{\jmath})-21(\widehat{k}.\widehat{k})$
$\begin{aligned} &=6.1+0-0+0+5-0+0+0-21.1\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\begin{array}{l} \because \hat{\imath} . \hat{\imath}=\hat{\jmath} \cdot \hat{\jmath}=\hat{k} \cdot \hat{k}=1, \hat{\imath} . \hat{\jmath}=\hat{\jmath} . \hat{\imath}=0 \\ \hat{\imath} \cdot \hat{k}=\hat{k} \cdot \hat{\imath}=0, \hat{\jmath} \cdot \hat{k}=\hat{k} \cdot \hat{\jmath}=0 \end{array}\right] \\ &=6+5-21 \\ &=11-21=-10 \end{aligned}$

The 25th chapter in the class 12 mathematics book, Scalar Triple Product, consists of only one exercise, ex 25.1. The Very Short Answer section or the VSA part has 11 questions asked in the textbook. The concepts that this section revolves around are the basic ideas of vectors, coplanar vector, non-collinear vector, the product of vectors, and the application of the scalar triple product formula to find the volume of the parallelepiped. The RD Sharma Class 12 Chapter 25 VSA reference guide will direct the students to find the answers for this section.

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RD Sharma Chapter wise Solutions