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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 23 Scalar or Dot Products Exercise 23.1 Question 28 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 23 Scalar or Dot  Products  Exercise 23.1 Question 28 Maths Textbook Solution.

Answers (1)

Answer:  \frac{\pi }{3},\frac{1}{\sqrt2}\hat{i},\frac{1}{2}\hat{j},\frac{1}{2}\hat{k}

Hint:  You must know the rules of solving vectors.

Given:   A unit vector \vec{a} makes angle \frac{\pi }{4} and \frac{\pi }{3}  with \hat{i} and \hat{j} respectively and an acute angle \theta and \hat{k}. Find the angle \theta and components of \vec{a}

Solution: Let

\vec{a}=a_{1}^{2}+a_{2}^{2}+a_{3}^{2}=1 \Rightarrow(1)                                                            [because\: \: \vec{a}\: \: is \: \: a \: \: unit\: \: vector ]

Now,

\begin{aligned} &\vec{a} \cdot \hat{\imath}=a_{1} \\ &\Rightarrow|\vec{a}||\hat{\imath}| \cos \frac{\pi}{4}=a_{1}\left[\text { because the angle between } \vec{a} \text { and } \hat{\imath} \text { is } \frac{\pi}{4}\right] \\ &\Rightarrow a_{1}=(1)(1)\left(\frac{1}{\sqrt{2}}\right) \\ &\Rightarrow a_{1}=\frac{1}{\sqrt{2}} \end{aligned}

Again,

\begin{aligned} &\Rightarrow|\vec{a}||\hat{\imath}| \cos \frac{\pi}{3}=a_{2} \\ &\Rightarrow(1)(1)\left(\frac{1}{2}\right)=a_{2} \\ &\Rightarrow \frac{1}{2}=a_{2} \end{aligned}

Now from (1)

\left(\frac{1}{\sqrt{2}}\right)^{2}+\left(\frac{1}{2}\right)^{2}+a_{2}^{2}=1\left[\right. because the angle between \vec{a} and \hat{\jmath} is \left.\frac{\pi}{3}\right]

\begin{aligned} &\frac{1}{2}+\frac{1}{4}+a_{3}^{2}=1 \\ &\frac{3}{4}+a_{3}^{2}=1 \\ &a_{3}^{2}=1-\frac{3}{4} \end{aligned}

a_{3}=\frac{1}{2}

NOw,

\begin{aligned} &\vec{a} \cdot \hat{k}=a_{3} \\ &\Rightarrow|\vec{a}||\hat{k}| \cos \theta=\frac{1}{2}[\text { because the angle between } \vec{a} \text { and } \hat{k} \text { is } \theta] \\ \end{aligned}

\begin{aligned} &\Rightarrow(1)(1) \cos \theta=\frac{1}{2} \\\\ &\theta=\cos ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3} \end{aligned}

And,

\vec{a}=\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{2}\hat{j}+\frac{1}{2}\hat{k}

Posted by

infoexpert21

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