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  • Provide solution for RD Sharma Maths Class 12 Chapter 25 Scalar Triple Product Exercise Very Short Answer Question, question 7.

Provide solution for RD Sharma Maths Class 12 Chapter 25 Scalar Triple Product Exercise Very Short Answer Question, question 7.

Answers (1)

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}

                        

 

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