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VECTOR ALGEBRA — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSVECTOR ALGEBRA3 Marks
Question
Show that $\big|\vec{a}|\ \vec{b}+\big|\vec{b}\big|\ \vec{a}$ is perpendicular to $\big|\vec{a}|\ \vec{b}-\big|\vec{b}\big|\ \vec{a},$ for any two nonzero vectors $\vec{a}\ \text{and}\ \vec{b}.$ 
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Answer

$\text{Let}\ \vec{c}=\big|\vec{a}\big|\ \vec{b}+\big|\vec{b}\big|\ \vec{a}=l\vec{b}+m\vec{a},$ $\text{where}\ l=\big|\vec{a}\big|\ \text{and}\ m=\big|\vec{b}\big|$ $\text{Let}\ \ \ \ \ \ \vec{d}=\big|\vec{a}\big|\ \vec{b}-\big|\vec{b}\big|\ \vec{a}=l\vec{b}-m\vec{a}$ $\text{Now}\ \ \ \ \ \ \big|\vec{c}\big|.\Big|\vec{d}\Big|=\big(l\vec{b}+m\vec{a}\big)\big(l\vec{b}-m\vec{a}\big)$ $=l^2\vec{b}.\vec{b}-lm\ \vec{b}.\vec{a}+lm\vec{a}.\vec{b}-m^2\vec{a}.\vec{a}$$=l^2\Big|\vec{b}\Big|^2-lm\vec{a}.\vec{b}+lm\vec{a}.\vec{b}-m^2\big|\vec{a}|^2$
$=l^2\Big|\vec{b}\Big|^2-m^2\big|\vec{a}|^2$
$\text{Putting},\ l=\big|\vec{a}\big|\ \text{and}\ m=\Big|\vec{b}\Big|,$$=\big|\vec{a}\big|^2\big|\vec{b}\big|^2-\big|\vec{b}\big|^2\cdot|\vec{a}|^2$
$\Rightarrow\ \ \ \ \ \ \big|\vec{c}\big|.\Big|\vec{d}\Big|=0$

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