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If $|\vec{a}|=7,|\vec{b}|=11$ and $|\vec{a}+\vec{b}|=10 \sqrt{3}$, then $|\vec{a}-\vec{b}|$ is equal
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Verified Answer
The correct answer is:
$2 \sqrt{10}$
$|\overrightarrow{\mathrm{a}}|=7,|\overrightarrow{\mathrm{b}}|=11$ and $|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|=10 \sqrt{3}$
Now $|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|^{2}=|\overrightarrow{\mathrm{a}}|^{2}+|\overrightarrow{\mathrm{b}}|^{2}+2|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}| \cos \theta$
$\begin{array}{ll}\therefore & (10 \sqrt{3})^{2}=49+121+2 \times 7 \times 7 \cos \theta \\ \therefore & 300=170+154 \cos \theta \\ & \frac{300-170}{154}=\cos \theta \\ \therefore & \frac{65}{77}=\cos 0 \\ & \quad \text { Now, }|\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|^{2}=|\overrightarrow{\mathrm{a}}|^{2}+|\overrightarrow{\mathrm{b}}|^{2}-2|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}| \cos \theta \\ & =(7)^{2}+(11)^{2}-2 \times 7 \times 11 \times \frac{65}{77} \\ & =49+121-2 \times 65 \\ & \equiv 170-130 \equiv 40 \\ \therefore & |\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|=\sqrt{40}=2 \sqrt{10}\end{array}$
$\therefore \quad$ Option (d) is correct.
Now $|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|^{2}=|\overrightarrow{\mathrm{a}}|^{2}+|\overrightarrow{\mathrm{b}}|^{2}+2|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}| \cos \theta$
$\begin{array}{ll}\therefore & (10 \sqrt{3})^{2}=49+121+2 \times 7 \times 7 \cos \theta \\ \therefore & 300=170+154 \cos \theta \\ & \frac{300-170}{154}=\cos \theta \\ \therefore & \frac{65}{77}=\cos 0 \\ & \quad \text { Now, }|\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|^{2}=|\overrightarrow{\mathrm{a}}|^{2}+|\overrightarrow{\mathrm{b}}|^{2}-2|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}| \cos \theta \\ & =(7)^{2}+(11)^{2}-2 \times 7 \times 11 \times \frac{65}{77} \\ & =49+121-2 \times 65 \\ & \equiv 170-130 \equiv 40 \\ \therefore & |\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|=\sqrt{40}=2 \sqrt{10}\end{array}$
$\therefore \quad$ Option (d) is correct.
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