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Explain Solution RD Sharma Class 12 Chapter Algebra of Vectors Exercise 22.6 Question 17 Maths.

Answers (1)

Answer:

2 \hat{\imath}-4 \hat{\jmath}+4 \hat{k}

Hint:

Unit vector  =\frac{\vec{A}}{|\vec{A}|} , new vector = 6\times  unit vector of  2 \vec{a}-\vec{b}+3 \vec{c}

Given:

\vec{a}=(\hat{\imath}+\hat{\jmath}+\hat{k}), \vec{b}=(4 \hat{\imath}-2 \hat{\jmath}+3 \hat{k}), \vec{c}=(\hat{\imath}-2 \hat{\jmath}+\hat{k})

Solution:

\begin{aligned} &\text { Let } \vec{A}=2 \vec{a}-\vec{b}+3 \vec{c} \\\\ &=2 \hat{\imath}+2 \hat{\jmath}+2 \hat{k}-4 \hat{\imath}+2 \hat{\jmath}-3 \hat{k}+3 \hat{\imath}-6 \hat{\jmath}+3 \hat{k} \\\\ &=\hat{\imath}-2 \hat{\jmath}+2 \hat{k} \end{aligned}
 

\begin{aligned} &|2 \vec{a}-\vec{b}+3 \vec{c}|=\sqrt{1^{2}+2^{2}+2^{2}} \\\\ &=\sqrt{9}=3 \end{aligned}

 

Let \vec{B}  be the new vector

\begin{aligned} &\vec{B}=6 \hat{A}=6 \frac{\vec{A}}{|\vec{A}|} \\\\ &=6 \times \frac{\vec{A}}{3}=2 \vec{A} \\\\ &=2(\hat{\imath}-2 \hat{\jmath}+2 \hat{k}) \\\\ &=2 \hat{\imath}-4 \hat{\jmath}+4 \hat{k} \end{aligned}

 

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