Practicing Success
If $|\vec{a}+\vec{b}|=15, |\vec{a}-\vec{b}|=10, |\vec{a}|=\frac{11}{2}$ then value of $|\vec{b}|$ is/are : |
$\frac{23}{2}$ $±\frac{23}{2}$ $±23$ $\frac{23}{\sqrt{2}}$ |
$\frac{23}{2}$ |
The correct answer is Option (1) → $\frac{23}{2}$ $|(\vec{a}+\vec{b}).(\vec{a}-\vec{b})|=|a^2-b^2|$ $(\vec{a}+\vec{b}).(\vec{a}-\vec{b})=|a^2-b^2|^2$ $⇒|\vec{a}|^2+|\vec{b}|^2+2\vec{a}.\vec{b}=|a^2+b^2|^2$ ...(1) similarly $|\vec{a}|^2+|\vec{b}|^2-2\vec{a}.\vec{b}=|a^2-b^2|^2$ ...(2) eq. (1) + eq. (2) $⇒2(|\vec{a}|^2+|\vec{b}|^2)=|a^2+b^2|^2+|a^2-b^2|^2$ $2(\frac{121}{4}+|\vec{b}|^2)=225+100$ $⇒|\vec{b}|^2=\frac{529}{4}⇒|\vec{b}|=\frac{23}{2}$ |