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

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

Answers (1)

Answer: proved

Hint: you must know the rules of solving vectors

Given: show that the vectors

\begin{aligned} &\vec{a}=\frac{1}{7}(2 \hat{\imath}+3 \hat{\jmath}+6 \hat{k}), \quad \vec{b}=\frac{1}{7}(3 \hat{\imath}-6 \hat{\jmath}+2 \hat{k}) \\ &\vec{c}=\frac{1}{7}(6 \hat{\imath}+2 \hat{\jmath}-3 \hat{k}) \end{aligned}

Solution: we Have

\begin{aligned} \vec{a}=\frac{1}{7}((2)^{2}+(3)^{2}+(6)^{2}) \\ \end{aligned}

                                                            \begin{aligned} =\frac{1}{7} \sqrt{4+9+36} \\ \end{aligned}

\begin{aligned} =\frac{1}{7} \sqrt{49} \quad \Rightarrow \frac{7}{7} \end{aligned}

\begin{aligned} &=1 \\ \end{aligned}

\begin{aligned} &\vec{b}=\frac{1}{7} \sqrt{(3)^{2}+(-6)^{2}+(2)^{2}} \\\\ &=\frac{1}{7} \sqrt{9+36+4} \\ \end{aligned}

\begin{aligned} &=\frac{1}{7} \sqrt{49} \\\\ &=1 \\\\ &\vec{c}=\frac{1}{7} \sqrt{(6)^{2}+(2)^{2}+(-3)^{2}} \\\\ &=\frac{1}{7} \sqrt{36+4+9} \end{aligned}

=\frac{1}{7}\Rightarrow 1

And

\begin{aligned} &\vec{a} \cdot \vec{b}=\frac{1}{7}(2 \hat{\imath}+3 \hat{\jmath}+6 \hat{k}) \cdot \frac{1}{7}(3 \hat{\imath}-6 \hat{j}+2 \hat{k}) \\ &=\frac{1}{49}(6-18+12) \\ &\Rightarrow 0 \end{aligned}

\begin{aligned} &\vec{b} \cdot \vec{c}=\frac{1}{7}(3 \hat{\imath}-6 \hat{\jmath}+2 \hat{k}),(6 \hat{\imath}+2 \hat{\jmath}-3 \hat{k}) \\ &=\frac{1}{49}(18-12-6) \\ &=0 \end{aligned}

\begin{aligned} &\vec{c} \cdot \vec{a}=\frac{1}{7}(6 \hat{\imath}+2 \hat{\jmath}-3 \hat{k}) \cdot \frac{1}{7}(2 \hat{\imath}+3 \hat{\jmath}+6 \hat{k}) \\ &=\frac{1}{7}(12+6+18) \\ &\Rightarrow 0 \end{aligned}

So,\mid \vec{a}\mid =\mid \vec{b}\mid =\mid \vec{c}\mid =1

And \vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{c}=\vec{c} \cdot \vec{a}=0

Therefore, the given vectors are mutually perpendicular unit vectors.

Posted by

infoexpert21

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