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  • Provide solution for RD Sharma maths class 12 chapter 23 Scaler and Dot Product exercise Multiple choice question 10

Provide solution for RD Sharma maths class 12 chapter 23 Scaler and Dot Product exercise Multiple choice question 10

Answers (1)

Answer:

Option (c) \sqrt{593}

Hint:

Use the formula |\vec{a}+\vec{b}|^{2}=|\vec{a}|^{2}+|\vec{b}|^{2}+2 \vec{a} \cdot \vec{b}

Given:

|\vec{a}|=2 \sqrt{2},|\vec{b}|=3   and angle between  \vec{a} \text { and } \vec{b} \text { is } \frac{\pi}{4}

\overrightarrow{A D}=\overrightarrow{B C}=\vec{a}-3 \vec{b} \text { and } \overrightarrow{A B}=\overrightarrow{D C}=5 \vec{a}+2 \vec{b}

Solution:

Here |\vec{a}|=2 \sqrt{2},|\vec{b}|=3

First we have to find both diagonals

Length of the two diagonal will be

\begin{aligned} &d_{1}=|(5 \vec{a}+2 \vec{b})+(\vec{a}-3 \vec{b})| \\\\ &d_{2}=|(5 \vec{a}+2 \vec{b})-(\vec{a}-3 \vec{b})| \\\\ &\Rightarrow d_{1}=|6 \vec{a}-\vec{b}| \text { and } d_{2}=|4 \vec{a}+5 \vec{b}| \end{aligned}

\begin{aligned} &d_{1}=\sqrt{36 a^{2}+b^{2}-2 \times 6 \times|\vec{a}| \cdot|\vec{b}| \cdot \cos \frac{\pi}{4}} \\\\ &=\sqrt{36 \times 8+9-12 \times 2 \sqrt{2} \times 3 \times \frac{1}{\sqrt{2}}} \end{aligned}

\begin{aligned} &=\sqrt{238+9-72} \\\\ &=\sqrt{225}=15 \end{aligned}

And again

\begin{aligned} &d_{2}=\sqrt{|4 \vec{a}|^{2}+|5 \vec{b}|^{2}+2 \times|4 \vec{a}| \times|5 \vec{b}| \cos \frac{\pi}{4}} \\\\ &=\sqrt{16 \times 8+25 \times 9+40 \times 2 \sqrt{2} \times 3 \times \frac{1}{\sqrt{2}}} \end{aligned}

\begin{aligned} &=\sqrt{128+225+240} \\\\ &=\sqrt{593}>15 \\\\ &\therefore d_{2}>d_{1} \end{aligned}

Hence length of the longest diagonal =\sqrt{593}

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