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  • Explain Solution R.D. Sharma Class 12 Chapter 23 Scalar or Dot Products Exercise 23.1 Question 48 Maths Textbook Solution.

Explain Solution R.D. Sharma Class 12 Chapter 23 Scalar or Dot Products Exercise 23.1 Question 48 Maths Textbook Solution.

Answers (1)

Answer:  \frac{-11}{2}

Hint: You must know the rules of solving vectors.

Given: If \vec{a} and \vec{b} are two non-collinear unit vectors such that \mid \vec{a}+\vec{b}\mid =\sqrt{3}  .Find \left ( 2\vec{a}-5\vec{b} \right )\cdot \left ( 3\vec{a}+\vec{b} \right )

Solution: We have,

\mid \vec{a}+\vec{b}\mid =\sqrt{3}

Squaring both sides

\begin{aligned} &\Rightarrow|\vec{a}+\vec{b}|^{2}=3 \\ &\Rightarrow|\vec{a}|^{2}+|\vec{b}|^{2}+2 \vec{a} \cdot \vec{b}=3 \\ &\Rightarrow 1+1+2 \vec{a} \cdot \vec{b}=3 \end{aligned}                                                            [because\: \vec{a} \: and\: \vec{b} \: are\: unit \: vector ]

\begin{aligned} &\Rightarrow 2+2 \vec{a} \cdot \vec{b}=3 \\ &\Rightarrow 2 \vec{a} \cdot \vec{b}=3-2 \\ &\Rightarrow \vec{a} \cdot \vec{b}=\frac{1}{2} \end{aligned}

Now,

\left ( 2\vec{a}-5\vec{b} \right )\cdot \left ( 3\vec{a}+\vec{b} \right )

\begin{aligned} &\Rightarrow 6|\vec{a}|^{2}+2 \vec{a} \cdot \vec{b}-15 \vec{b} \cdot \vec{a}-5|\vec{b}|^{2} \\ &\Rightarrow 6|\vec{a}|^{2}+2 \vec{a} \cdot \vec{b}-15 \vec{b} \cdot \vec{a}-5|\vec{b}|^{2} \quad[\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{a}] \\ &\Rightarrow 6|\vec{a}|^{2}-13 \vec{a} \cdot \vec{b}-5|\vec{b}|^{2} \\ &\Rightarrow 6(1)-13\left(\frac{1}{2}\right)-5(1) \\ &\Rightarrow 1-\frac{13}{2} \\ &\Rightarrow \frac{-11}{2} \end{aligned}

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