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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 2 Maths Textbook Solution.

Answers (1)

Answer: $\cos ^{-1}\left ( \frac{-34}{63} \right )$

Hint: you must know the properly of findings angle b/w two vectors

Given: $\vec{a}=3\hat{i}-2\hat{j}+6\hat{k}$ and $\vec{b}=4\hat{i}-\hat{j}+8\hat{k}$

Solution: $\vec{a}=3\hat{i}-2\hat{j}+6\hat{k}$ and $\vec{b}=4\hat{i}-\hat{j}+8\hat{k}$

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

Now, $\left ( \vec{a}.\vec{b} \right )$ $=\left ( 3\hat{i}-2\hat{j}+6\hat{k} \right )$ $\left ( 4\hat{i}-\hat{j}+8\hat{k} \right )$

$=12+2-48$ $\Rightarrow 14-48=-34$

Magnitude of $\mid \vec{a}\mid =\sqrt{\left ( 3 \right )^{2}+\left ( -2 \right )^{2}+\left ( -6 \right )^{2}}$

$=\sqrt{9+4+36}$ $=\sqrt{49}$

$\Rightarrow 7$

Magnitude of $\mid \vec{b}\mid =\sqrt{\left ( 4 \right )^{2}-\left ( -1 \right )^{2}+\left ( -8 \right )^{2}}$

$=\sqrt{16+1+64}$

$=\sqrt{81}$

$\Rightarrow 9$

Put values in 1

$\begin{aligned} &\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta \\\\ &-34=(7)(9) \cos \theta \\\\ &-34=(63) \cos \theta \\\\ \end{aligned}$

$\begin{aligned} &\frac{-34}{63}=\cos \theta \\\\ &\theta=\cos ^{-1}\left(\frac{-34}{63}\right) \end{aligned}$

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