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  • Explain solution RD sharma class 12 chapter 23 scalar or dot product exercise Fill in the blanks question 17

Explain solution RD sharma class 12 chapter 23 scalar or dot product exercise Fill in the blanks question 17

Answers (1)

\frac{\pi}{4}Hint:

               Find magnitude

Given:

The vector \vec{a}=3\hat{i}-2\hat{j}+2\hat{k}  and  \vec{b}=-\hat{i}-2\hat{k}  are adjacent side of parallelogram

The acute angle between diagonal is

Solution:

Let \vec{AB}=3\hat{i}-2\hat{j}+2\hat{k} & \vec{BC}=-\hat{i}-2\hat{k}

AC & BD are the diagonals of the parallelogram,

\vec{AC}=\vec{AB}+\vec{BC}\\ =(3-1)\hat{i}-2\hat{j}+(2-2)\hat{k}\\ =2\hat{i}-2\hat{j}

\begin{aligned} &\overrightarrow{B D}=\overrightarrow{A B}-\overrightarrow{B C} \\ &=(3-(-1)) \hat{\imath}-2 \hat{\jmath}+4 \hat{k} \\ &\overrightarrow{B D}=4 \vec{\imath}-2 \vec{\jmath}+4 \vec{k} \\ \end{aligned}

\begin{aligned} &\overrightarrow{A C} \cdot \overrightarrow{B D}=|\overrightarrow{A C}| \cdot|\overrightarrow{B D}| \cos \theta \\ &12=\sqrt{4+4} \cdot \sqrt{16+4+16} \cos \theta \\ \end{aligned}

\begin{aligned} &12=\sqrt{8} \cdot \sqrt{36} \cos \theta \\ &12=2 \sqrt{2} \cdot 6 \cos \theta \\ \end{aligned}

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

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