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  • Provide Solution For R.D. Sharma Maths Class 12 Chapter 22 Algebra of Vectors Exercise 22.9 Question 5 Maths Textbook Solution.

Provide Solution For  R.D. Sharma Maths Class 12 Chapter 22 Algebra of Vectors  Exercise 22.9 Question 5 Maths Textbook Solution.

Answers (1)

Answer:4\left ( \sqrt{2\hat{i}}+\hat{j}+\hat{k} \right )

Given: A vector \vec{\mu } is inclined at equal acute angles to x-axis at 450and y-axis at 600.If \mid \vec{\mu }\mid =8 units,

 find \vec{\mu }

Hint: Use \mu=|\vec{\mu}|(l \hat{i}+m \hat{j}+n \hat{k})

Explanation: Here  \vec{\mu } makes an angle 450 with OX and 600 with OY.

So, I=\cos 45^{\circ}=\frac{1}{\sqrt{2}} \text { and } m=\cos 60^{\circ}=\frac{1}{2}

Now we know if l , m , n be the direction cosines then l^{2}+m^{2}+n^{2}=1

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

Therefore \begin{aligned} &\vec{\mu}=|\vec{u}|(l \hat{i}+m \hat{j}+n \hat{k}) \\ \end{aligned}

\begin{aligned} &=8\left(\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{2} \hat{j} \pm \frac{1}{2} \hat{k}\right) \\ &=4(\sqrt{2} \hat{i}+\hat{j} \pm \hat{k})[\because|\vec{\mu}|=8] \\ &\vec{\mu}=8\left(\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{2} \hat{j} \pm \frac{1}{2} \hat{k}\right) \end{aligned}

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