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Please solve RD Sharma class 12 chapter 22 Algebra of Vectors exercise Fill in the blanks question 9 maths textbook solution

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Answer:- 3 \sqrt{2} \hat{i}+3 \hat{j}+3 \hat{k}

Hint:-To solve this equation, we use standard formula l^{2}+n^{2}+m^{2}=1 . .

Given:-A vector of magnitude 6; making angle \frac{\pi }{4}  with x-axis; \frac{\pi }{3}  with y axis and an acute angle with z axis

Solution:-we have |\vec{a}|=6

\begin{aligned} &\text { So } l=\cos \alpha=\frac{\sqrt{2}}{2} \\\\ &m=\cos \beta=\frac{1}{2} \\\\ &n=\cos \gamma=\cos \theta \end{aligned} 

 

 

We know that l^{2}+n^{2}+m^{2}=1

 

 

\begin{aligned} &\left(\frac{\sqrt{2}}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}+n^{2}=1 \\\\ &\frac{2}{4}-\frac{1}{4}+n^{2}=1 \\\\ &\frac{3}{4}+n^{2}=1 \end{aligned}

\begin{aligned} &n^{2}=1-\frac{3}{4}=\frac{4-3}{4}=\frac{1}{4} \\\\ &n=\pm \frac{1}{2} \\\\ &\hat{a}=l \hat{i}+m \hat{j}+n \hat{k}=\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{2} \hat{j}+\frac{1}{2} \hat{k} \end{aligned}

\begin{aligned} &\hat{a}=|\hat{a}|(\hat{i}+m \hat{j}+n \hat{k}) \\\\ &=6\left(\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{2} \hat{j}+\frac{1}{2} \hat{k}\right) \\\\ &=3 \sqrt{2} \hat{i}+3 \hat{j}+3 \hat{k} \end{aligned}

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