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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 3 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 3 Maths Textbook Solution.

Answers (1)

Answer:  \frac{\pi }{2}

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

Given:     \vec{a}=2\hat{i}-\hat{j}+2\hat{k} and \vec{b}=4\hat{i}+4\hat{j}-2\hat{k}

Solution:

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

\vec{b}=4\hat{i}+4\hat{j}-2\hat{k}

We know   \vec{a}.\vec{b}=\mid \vec{a}\mid \mid \vec{b}\mid \cos \theta

Now,\left ( \vec{a}.\vec{b} \right )=\left ( 2\hat{i}-\hat{j}+2\hat{k} \right ).\left( 4\hat{i}+4\hat{j}-2\hat{k} \right )                                \left|\begin{array}{l} \hat{l} \cdot \hat{l}=1 \\ \hat{\imath} \cdot \hat{j}=0 \end{array}\right|

\left ( 8-4-4 \right )\Rightarrow 0

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

=\sqrt{4+1+4}                                            =\sqrt{9}=3

Magnitude of\mid \vec{b}\mid =\sqrt{\left ( 4 \right )^{2}+\left ( 4 \right )^{2}+\left ( -2 \right )^{2}}

=\sqrt{16+16+4}                            =\sqrt{36}=6

Put in 1

\begin{aligned} &\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta \\\\ &0=(3)(6) \cos \theta \\\\ &0=\cos \theta \\\\ &\theta=\frac{\pi}{2} \end{aligned}

 

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