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

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

Answers (1)

Answer:     \vec{r}= \pm 2\left ( \hat{i}+\hat{j}+\hat{k} \right )

Given: A vector \vec{r} is inclined at equal angles to the three axis. If the magnitude of \vec{r} is 2\sqrt{3},find \vec{r}

Hint: Use \cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1

Explanation: Let\alpha ,\beta ,\gamma be the angles inclined to the three axis.

Since \vec{r} is inclined equal angle to axis

\begin{aligned} &\therefore \alpha=\beta=\gamma \\ &\Rightarrow 3 \cos ^{2} \alpha=1 \\ &\therefore \cos \alpha=\pm \frac{1}{\sqrt{3}} \\ \end{aligned}

So,

\begin{aligned} &\vec{r}=\pm \frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k}) \\ &\therefore \vec{r}=|\vec{r}| \hat{r} \\ &=2 \sqrt{3} \times\left(\pm \frac{1}{\sqrt{3}}\right)(\hat{i}+\hat{j}+\hat{k}) \\ &=\pm 2(\hat{i}+\hat{j}+\hat{k}) \\ &\therefore \vec{r}=\pm 2(\hat{i}+\hat{j}+\hat{k}) \end{aligned}

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