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  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 23 Scalar or Dot Products Exercise 23.1 Question 5 Sub Question 5 Maths Textbook Solution.

Provide Solution For R.D.Sharma Maths Class 12 Chapter 23 Scalar or Dot Products  Exercise 23.1 Question 5 Sub Question 5 Maths Textbook Solution.

Answers (1)

Answer:   \cos ^{-1}\left ( \frac{-\sqrt{2}}{3} \right )

Hint: you must know the rules of solving and finding angle b/w two vectors

Given: \vec{a}=\hat{i}+\hat{2j}-\hat{k}  and \vec{b}=\hat{i}-\hat{j}+\hat{k}

Solution:

\vec{a}=\hat{i}+\hat{2j}-\hat{k}

\vec{b}=\hat{i}-\hat{j}+\hat{k}

We\: \: know \quad \vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta

Now, (\vec{a} \cdot \vec{b})=(\hat{\imath}+2 \hat{\jmath}-\hat{k})(\hat{\imath}-\hat{\jmath}+\hat{k})

==(1-2-1)

=-2

Magnitude of |\vec{a}|=\sqrt{(1)^{2}+(2)^{2}+(-1)^{2}}

=\sqrt{1+4+1} \quad=\sqrt{6}

Magnitude of |\vec{b}|=\sqrt{(1)^{2}+(1)^{2}+(1)^{2}}

=\sqrt{1+1+1} \quad=\sqrt{3}

Put in 1

\begin{aligned} &\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta \\\\ &(-2)=(\sqrt{6})(\sqrt{3}) \cos \theta \\\\ &-2=\sqrt{18} \cos \theta \\\\ &\frac{-2}{\sqrt{3 \times 3 \times 2}}=\cos \theta \end{aligned}

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

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