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  • Please solve RD Sharma class 12 chapter 23 Scaler and Dot Product exercise Multiple choice question 1 maths textbook solution

Please solve RD Sharma class 12 chapter 23 Scaler and Dot Product exercise Multiple choice question 1 maths textbook solution

Answers (1)

best_answer

Answer:

Option (c)   \cos \theta=\frac{-4}{5}

Hint:

Apply Elimination method to find \vec{a} \text { and } \vec{b}, then find |\vec{a}| \text { and }|\vec{b}|

Given:

\vec{a} \text { and } \vec{b}satisfy the equation 2 \vec{a}+\vec{b}=\vec{p} and \vec{a}+2 \vec{b}=\vec{q}

Where, \vec{p}=\hat{i}+\hat{j}, \vec{q}=\hat{i}-\hat{j}  and \theta is the angle between \vec{a} \text { and } \vec{b}

Solution:

\begin{aligned} &2 \vec{a}+\vec{b}=\vec{p} \quad\quad\quad.....(i)\\ &\vec{a}+2 \vec{b}=\vec{q} \quad\quad\quad.....(ii)\times 2 \end{aligned}

\Rightarrow \quad \begin{aligned} &2 \vec{a}+\vec{b}=\vec{p} \\ &2 \vec{a}+4 \vec{b}=2 \vec{q} \end{aligned}

                ____________

\begin{aligned} &-3 \vec{b}=\vec{p}-2 \vec{q} \\ \Rightarrow \quad & \vec{b}=\frac{2 \vec{q}-\vec{p}}{3} \end{aligned}

Put in equation (i)

\begin{aligned} &2 \vec{a}+\frac{2 \vec{q}-\vec{p}}{3}=\vec{p} \Rightarrow 2 \vec{a}=\frac{3 \vec{p}-2 \vec{q}+\bar{p}}{3} \\\\ &\vec{a}=\frac{4 \vec{p}-2 \vec{q}}{3(2)} \Rightarrow \vec{a}=\frac{2 \vec{p}-\vec{q}}{3} \end{aligned}

Now, \vec{a}=\frac{2(\hat{i}+\hat{j})-(\hat{i}-\hat{j})}{3}=\frac{\hat{i}+3 \hat{j}}{3}                    \left[\begin{array}{l} \because \vec{p}=\hat{i}+\hat{j} \\ \because \vec{q}=\hat{i}-\hat{j} \end{array}\right]

\begin{aligned} &\vec{b}=\frac{2(\hat{i}-\hat{j})-(\hat{i}+\hat{j})}{3}=\frac{\hat{i}-3 \hat{j}}{3} \\\\ &|\vec{a}|=\frac{\sqrt{1+9}}{3}=\frac{\sqrt{10}}{3} \\\\ &|\vec{b}|=\frac{\sqrt{1+9}}{3}=\frac{\sqrt{10}}{3} \end{aligned}                    \left[\begin{array}{l} \because|\vec{a}|=\sqrt{(\hat{i})^{2}+(3 \hat{j})^{2}} \\\\ \because|\vec{b}|=\sqrt{(\hat{i})^{2}+(-3 \hat{j})^{2}} \end{array}\right]

\begin{aligned} &\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta \\\\ &\frac{1}{9}\left[\hat{i}^{2}-(3 \hat{j})^{2}\right]=\frac{\sqrt{10}}{3} \times \frac{\sqrt{10}}{3} \cos \theta \\\\ &\frac{1}{9}[1-9]=\frac{10}{9} \cos \theta \end{aligned}

\begin{aligned} &\frac{-8}{9}=\frac{10}{9} \cos \theta \\\\ &\frac{-8}{10}=\cos \theta \\\\ &\cos \theta=\frac{-4}{5} \end{aligned}

Hence, option (c) is correct

 

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