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  • Provide Solution For R. D. Sharma Maths Class 12 Chapter 23 Scalar or Dot Products Exercise 23.1 Question 6 Maths Textbook Solution.

Provide Solution For R. D. Sharma Maths Class 12 Chapter 23 Scalar or Dot Products  Exercise 23.1 Question 6 Maths Textbook Solution.

Answers (1)

Answer: \frac{\pi }{3},\frac{2\pi }{3},\frac{\pi }{4}

Hint: you must know the rules of finding angle

Given:  \vec{a}=\hat{i}-\hat{j}+\sqrt{2}\hat{k} makes with co-ordinate ones.

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

\begin{aligned} &\therefore|\vec{a}|=\sqrt{(1)^{2}+(-1)^{2}+(\sqrt{2})^{2}} \\ &=\sqrt{1+1+2} \\ &=\sqrt{2}=2 \end{aligned}

\vec{b}=\hat{i}(Because it is a unit vector along x-axis)

\begin{aligned} &|\vec{b}|=\sqrt{(1)^{2}+0+0}=1 \\ &\vec{a} \cdot \vec{b}=(\hat{\imath}-\hat{\jmath}+\sqrt{2} \hat{k})(\hat{\imath}) \quad\left|\begin{array}{l} \hat{\imath} \cdot \hat{\imath}=1 \\ \hat{\imath} \cdot \hat{\jmath}=0 \end{array}\right| \\ &\Rightarrow 1 \\ &\text { Now, }(\vec{a} \cdot \vec{b})=|\vec{a}||\vec{b}| \cos \theta \end{aligned}

\begin{aligned} &1=(2)(1) \cos \theta \\ &\frac{1}{2}=\cos \theta \\ &\theta=\frac{\pi}{3} \\ \end{aligned}

Now, Let  \theta _{2} be the angle between \vec{a} and y-axis.

\begin{aligned} &|\vec{a}|=\sqrt{(1)^{2}+(-1)^{2}+(\sqrt{2})^{2}}=\sqrt{4} \quad=2 \\ &\vec{b}=\hat{j} \\ \end{aligned}(become it is a unit vector along y-axis)

\mid \vec{b}|=\sqrt{(1)^{2}+0+0} \quad=\sqrt{1} \quad \Rightarrow 1

\begin{aligned} &\vec{a} \cdot \vec{b}=(\hat{\imath}-\hat{\jmath}+\sqrt{2} \hat{k})(\hat{j}) \\ &\Rightarrow(-1) \quad\left|\begin{array}{l} \hat{i} \cdot \hat{\imath}=1 \\ \hat{\imath} \cdot \hat{\jmath}=0 \end{array}\right| \end{aligned}

Now, (\vec{a} \cdot \vec{b})=|\vec{a}||\vec{b}| \cos \theta

\begin{aligned} &-1=(2)(1) \cos \theta \\ &-1=2 \cos \theta_{2} \\ &\cos \theta_{2}=\frac{-1}{2} \\ &\theta_{2}=\frac{2 \pi}{3} \end{aligned}

Now, Let \theta _{3} be the angle between \vec{a}and y-axis.

\begin{aligned} &|\vec{a}|=\sqrt{(1)^{2}+(-1)^{2}+(\sqrt{2})^{2}}=\sqrt{4} \quad=2 \\ &\vec{b}=\hat{k} \\ \end{aligned}(become\: \: \hat{k} is \: \: unit\: \: vector \: \: along \: \: z-axis)

\begin{aligned} &|\vec{b}|=\sqrt{0+0+(1)^{2}}=1 \\ &\vec{a} \cdot \vec{b}=(\hat{\imath}-\hat{\jmath}+\sqrt{2} \hat{k})(\hat{k}) \quad\left|\begin{array}{l} \hat{k} \cdot \hat{k}=1 \\ \hat{k} \cdot \hat{\imath}=0 \end{array}\right| \\ &=(\sqrt{2}) \end{aligned}

Now, (\vec{a} \cdot \vec{b})=|\vec{a}||\vec{b}| \cos \theta

\begin{aligned} &\sqrt{2}=(2)(1) \cos \theta \\ &\frac{\sqrt{2}}{2}=\cos \theta \\ &\frac{2}{\sqrt{2}}=\cos \theta \\ &\theta=\frac{\pi}{4} \end{aligned}

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