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

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

Answers (1)

Answer:

Option (d)-1

Hint:

Use the identity (a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a \cdot b+2 b \cdot c+2 c \cdot a

Given:

|\vec{a}|=|\vec{b}|=|\vec{c}|=1 \text { and }|\vec{a}+\vec{b}+\vec{c}|=1

\vec{a} is perpendicular to \vec{b} and \vec{c} makes angle \alpha and \beta with \vec{a} and \vec{b} respectively

Solution:

\begin{aligned} &|\vec{a}+\vec{b}+\vec{c}|=1 \\\\ &(\vec{a}+\vec{b}+\vec{c})^{2}=(\vec{a})^{2}+(\vec{b})^{2}+(\vec{c})^{2}+2 \vec{a} \cdot \vec{b}+2 \vec{b} \cdot \vec{c}+2 \vec{c} \cdot \vec{a} \end{aligned}

1=1+1+1+0+2 \cos \alpha+2 \cos \beta                \begin{aligned} &{[\because|\vec{a}|=|\vec{b}|=|\vec{c}|=1]} \\ &{[\because \vec{a} \cdot \vec{b}=0]} \end{aligned}

\begin{aligned} &2 \cos \alpha+2 \cos \beta=-2 \\ &\cos \alpha+\cos \beta=-1 \end{aligned}

Hence, option (d) is correct

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