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  • Provide Solution For R.D. Sharma Maths Class 12 Chapter 23 Scalar or Dot Products Exercise 23.1 Question 12 Maths Textbook Solution.

Provide Solution For  R.D. Sharma Maths Class 12 Chapter 23 Scalar or Dot Products  Exercise 23.1 Question 12 Maths Textbook Solution.

Answers (1)

Answer: proved

Hint: you must know the rules of solving vectors.

Given: show that the vector \hat{i}+\hat{j}+\hat{k} is equally inclined to coordinate

Solution: Let \theta, be the angle b/w \ \ \ \vec{a} and x-axis,

\begin{aligned} &|\vec{a}|=\sqrt{(1)^{2}+(1)^{2}+(1)^{2}}=\sqrt{1+1+1}=\sqrt{3} \\\\ &\vec{b}=\hat{i}(\text { Because } \hat{i} \text { is the unit vector along } \mathrm{x}-\mathrm{axis}) \\\\ &|\vec{b}|=\sqrt{(1)^{2}+0+0}=\sqrt{1} \quad \Rightarrow 1 \\\\ &\vec{a} \cdot \vec{b}=1+0+0=1 \end{aligned}

\begin{aligned} &|\vec{b}|=\sqrt{(1)^{2}+0+0}=\sqrt{1} \quad \Rightarrow 1 \\\\ &\vec{a} \cdot \vec{b}=1+0+0=1 \\\\ &\cos \theta_{1}=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}=\frac{1}{(\sqrt{3})(1)} \quad \Rightarrow \frac{1}{\sqrt{3}} \\\\ &\therefore \theta_{1}=\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right) \end{aligned}1

\\\text{Let} \ \theta_{2} \text{ be the angle between } \vec{a} \: and \: \mathrm{y}-axis

\begin{aligned} &|\vec{a}|=\sqrt{(1)^{2}+(1)^{2}+(1)^{2}=\sqrt{3}} \\\\ &\vec{b}=\hat{J} \\\\ &|\vec{b}|=\sqrt{0+(1)^{2}+0} \quad=\sqrt{1} \Rightarrow 1 \\\\ &(\vec{a} \cdot \vec{b})=0+1+0=1 \\\\ &\Rightarrow 1 \end{aligned}

we Know \vec{a} \cdot \vec{b}=\cos \theta|\vec{a}||\vec{b}|

\begin{aligned} &1=\sqrt{3} \cos \theta_{2} \\ &\frac{1}{\sqrt{3}}=\cos \theta_{2} \\ &\therefore \theta_{2}=\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right) \end{aligned}

Again ,  Let \theta _{3} be the angle between \vec{a} and z-axis

\begin{aligned} &|\vec{a}|=\sqrt{(1)^{2}+(1)^{2}+(1)^{2}=\sqrt{3}} \\\\ &\vec{b}=\hat{k} \quad \text { (Because } \hat{k} \text { is unit vector along } z-\text { axis }) \\\\ &|\vec{b}|=\sqrt{0+0+(1)^{2} \quad=\sqrt{1}}=1 \\\\ &\vec{a} \cdot \vec{b}=(\hat{\imath}-\hat{\jmath}+\hat{k}) \cdot(\hat{k}) \\\\ &\Rightarrow(0+0+1) \quad \Rightarrow 1 \end{aligned}

We Know \vec{a} \cdot \vec{b}=\cos \theta|\vec{a}||\vec{b}|

1=\sqrt{3} \cos \theta

\frac{1}{\sqrt{3}}=\cos \theta

\therefore \theta=\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right)

From 1, 2, 3 the given vector is equally inclined to the co-ordinate ones.

 

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