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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 23 Scalar or Dot Products Exercise 23.1 Question 27 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 23 Scalar or Dot  Products  Exercise 23.1 Question 27 Maths Textbook Solution.

Answers (1)

Answer:  Proved

Hint:  You must know the rules of solving vectors.

Given:  If \vec{a}=5\hat{i}-\hat{j}-3\hat{k} and \vec{b}=\hat{i}+3\hat{j}-5\hat{k} then show that the vectors \vec{a}+\vec{b} and \vec{a}-\vec{b} are orthogonal.

Solution: \vec{a}=5\hat{i}-\hat{j}-3\hat{k} and \vec{b}=\hat{i}+3\hat{j}-5\hat{k}

\begin{aligned} \therefore \vec{a}+\vec{b}=5 \hat{\imath}-\hat{\jmath}-3 \hat{k}+\hat{\imath}+3 \hat{\jmath}-5 \hat{k} \\ \end{aligned}

                                                                                \begin{aligned} \vec{a}+\vec{b}=6 \hat{i}+2 \hat{\jmath}-8 \hat{k} \end{aligned}

And

\begin{aligned} &\vec{a}-\vec{b}=5 \hat{\imath}-\hat{\jmath}-3 \hat{k}-\hat{\imath}+3 \hat{j}-5 \hat{k} \\\\ &\vec{a}-\vec{b}=4 \hat{\imath}-4 \hat{\jmath}+2 \hat{k} \\ \end{aligned}

Now,

\begin{aligned} &(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=(6 \hat{\imath}+2 \hat{\jmath}-8 \hat{k}) \cdot(4 \hat{\imath}-4 \hat{\jmath}+2 \hat{k}) \\\\ &(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=24-8-16 \end{aligned}

\left ( \vec{a}.\vec{b} \right ) . \left ( \vec{a}-\vec{b} \right )=0

So,\vec{a}+\vec{b}is orthogonal of \vec{a}-\vec{b}

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