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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 4 Maths Textbook Solution.

Answers (1)

Answer: $\cos ^{-1}\left ( \frac{-3}{\sqrt{84}} \right )$

Hint: You must know the rules of finding angle between two vectors.

Given: $\vec{a}=2\hat{i}-3\hat{j}+\vec{k}$ and $\vec{b}=\hat{i}+\hat{j}-2\vec{k}$

Solution:

$\vec{a}=2\hat{i}-3\hat{j}+\vec{k}$

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

we Know $\vec{a}.\vec{b}=\mid \vec{a}\mid \mid \vec{b}\mid \cos \theta$ ....(i)

Now, $\left ( \vec{a}.\vec{b} \right )=$ $\left ( 2\hat{i}-3\hat{j}+\vec{k} \right )$ $\left ( \hat{i}+\hat{j}-2\vec{k} \right )$ $\left|\begin{array}{l} \hat{\imath} \cdot \hat{\imath}=1 \\ \hat{\imath} \cdot \hat{\jmath}=0 \end{array}\right|$

$\left ( 2-3-2 \right )\Rightarrow -3$

Magnitude of $\mid \vec{a}\mid =\sqrt{\left ( 2 \right )^{2}+\left ( -3 \right )^{2}+\left ( -2 \right )^{2}}$

= $\sqrt{4+9+4}$ $=\sqrt{8+9}=\sqrt{17}$

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

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

Put in (1)

$\vec{a}.\vec{b}=\mid \vec{a}\mid \mid \vec{b}\mid \cos \theta$

$-3=\left ( \sqrt{17} \right )\left ( \sqrt{6} \right )\cos \theta$

$-3=\sqrt{84}\cos \theta$

$\cos \theta =\frac{-3}{\sqrt{84}}$

$\theta =\cos ^{-1}\left ( \frac{-3}{\sqrt{84}} \right )$

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