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Explain Solution R.D. Sharma Class 12 Chapter 22 Algebra of Vectors  Exercise Very Short Answer Type Question 30 Maths Textbook Solution.

Answers (1)

Answer: \frac{1}{3}\left ( \hat{i}+2\hat{j} +2\hat{k}\right )

Hint: You must know the rules of vector functions

Given: \vec{a}=\hat{i}+2\hat{j}-3\hat{k}and  \vec{b}=2\hat{i}+4\hat{j}+9\hat{k}find vector parallel to \vec{a}+\vec{b}

Solution:

\vec{a}=\hat{i}+2\hat{j}-3\hat{k}

\vec{b}=2\hat{i}+4\hat{j}+9\hat{k}

Now,\begin{aligned} &\vec{a}+\vec{b}=\hat{i}+2 \hat{j}-3 \hat{k}+2 \hat{i}+4 \hat{j}+9 \hat{k} \\ \end{aligned}

\begin{aligned} &\vec{a}+\vec{b}=3 \hat{i}+6 \hat{j}+6 \hat{k} \\ \end{aligned}

=3(\hat{i}+2 \hat{j}+2 \hat{k}) \\

|\vec{a}+\vec{b}|=\sqrt{(3)^{2}+(6)^{2}+(6)^{2}} \\

=\sqrt{9+36+36} \\

=\sqrt{81} \\

=9 \\

Vector parallel to \vec{a}+\vec{b}

\frac{\vec{a}+\vec{b}}{|\vec{a}+\vec{b}|}=\frac{3(\hat{i}+2 \hat{j}+2 \hat{k})}{9}=\frac{1}{3}(\hat{i}+2 \hat{j}+2 \hat{k})

 

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