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  • Please Solve RD Sharma Class 12 Chapter 22 Algebra of Vectors Exercise 22.6 Question 3 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter 22 Algebra of Vectors Exercise 22.6 Question 3 Maths Textbook Solution.

Answers (1)

Answer:

The unit vector will be  \frac{1}{\sqrt{21}}(4 \hat{\imath}+2 \hat{\jmath}-\hat{k})

Hint:

Use vector magnitude formula.

Given:

\vec{a}=\hat{\imath}-\hat{\jmath}+3 \hat{k}, \vec{b}=2 \hat{\imath}+\hat{\jmath}-2 \hat{k} \text { and } \overrightarrow{\mathrm{c}}=\hat{\imath}+2 \hat{\jmath}-2 \hat{k}

Solution:

To find the resultant vector we add all the vectors by vector addition.

\vec{a}=\hat{\imath}-\hat{\jmath}+3 \hat{k}, \vec{b}=2 \hat{i}+\hat{\jmath}-2 \hat{k} \text { and } \overrightarrow{\mathrm{c}}=\hat{\imath}+2 \hat{\jmath}-2 \hat{k}

So, resultant vector is

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

So, the unit vector   \vec{P}=\frac{\vec{p}}{|\vec{P}|}

Magnitude of vector

\begin{aligned} &|\vec{P}|=\sqrt{4^{2}+2^{2}+(-1)^{2}} \\ &\quad=\sqrt{16+4+1}=\sqrt{21} \\ &\therefore \vec{P}=\frac{4 \hat{\imath}+2 \hat{\jmath}-\hat{k}}{\sqrt{21}}=\frac{1}{\sqrt{21}}(4 \hat{\imath}+2 \hat{j}-\hat{k}) \end{aligned}

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