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  • Need solution for RD Sharma maths class 12 chapter 23 Scaler and Dot Product exercise Multiple choice question 3

Need solution for RD Sharma maths class 12 chapter 23 Scaler and Dot Product exercise Multiple choice question 3

Answers (1)

Answer:

Option (d) \frac{\pi }{3}

Hint:

You must know the formula of  \vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta

Given:

\vec{a}+\vec{b}+\vec{c}=0,|\vec{a}|=3,|\vec{b}|=5,|\vec{c}|=7

Solution:

\begin{aligned} &\vec{a}+\vec{b}+\vec{c}=0 \\ &\vec{a}+\vec{b}=-\vec{c} \end{aligned}

Squaring on both sides,

\begin{aligned} &(\vec{a}+\vec{b})^{2}=(-\vec{c})^{2} \\\\ &(\vec{a})^{2}+2 \vec{a} \cdot \vec{b}+(\vec{b})^{2}=\vec{c}^{2} \quad\left[\because(a+b)^{2}=a^{2}+b^{2}+2 a b\right] \end{aligned}

Now, |\vec{a}|=\vec{a}=3,|\vec{b}|=\vec{b}=5,|\vec{c}|=\vec{c}=7

(3)^{2}+2(3) \cdot(5) \cos \theta+(5)^{2}=(7)^{2} \quad[\because \vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta]

\begin{aligned} &9+30 \cos \theta+25=49 \\\\ &30 \cos \theta=49-34 \\\\ &\cos \theta=\frac{15}{30} \quad\left[\because \cos \frac{\pi}{3}=\frac{1}{2}\right] \end{aligned}

\begin{aligned} &\cos \theta=\frac{1}{2} \\\\ &\theta=\frac{\pi}{3} \end{aligned}

Hence, option (d) is correct

 

 

 

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