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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 22 Algebra of Vectors Exercise 22.9 Question 9 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 22 Algebra of Vectors  Exercise 22.9 Question 9 Maths Textbook Solution.

Answers (1)

Answer: Direction cosines are equally inclined to axis.

Given: Show that the direction cosines of a vector equally inclined to the axis OX, OY and OZ are \frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}.

Hint: Find \mid \vec{\mu } \mid

Explanation: Let the required vector be \vec{\mu } =a\hat{i}+b\hat{j}+c\hat{k}

Direction ratios are a, b, c

Since the vector is equally inclined to axis OX, OY and OZ, thus the direction cosines are equal.

\frac{a}{\text { magnitude } \vec{\mu}}=\frac{b}{\text { magnitude } \vec{\mu}}=\frac{c}{\text { magnitude } \vec{\mu}}

Since a=b=c

\therefore The vector is \vec{\mu } =a\hat{i}+a\hat{j}+a\hat{k}

Magnitude of  \begin{aligned} &\vec{\mu}=\sqrt{(a)^{2}+(a)^{2}+(a)^{2}} \\ \end{aligned}

\begin{aligned} &\Rightarrow|\vec{\mu}|=\sqrt{3 a^{2}}=\sqrt{3} a \\ \end{aligned}

Direction cosines are \begin{aligned} &\left(\frac{a}{\sqrt{3} a}, \frac{b}{\sqrt{3} a}, \frac{c}{\sqrt{3} a}\right) \end{aligned}

=\begin{aligned} &\left(\frac{a}{\sqrt{3} a}, \frac{b}{\sqrt{3} a}, \frac{c}{\sqrt{3} a}\right) \end{aligned}

=\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}.

Hence Proved

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