Question:
Let $a, b$ and $c$ be three unit vectors such that $|\vec{a}-\vec{b}|^{2}+|\vec{a}-\vec{c}|^{2}=8$. Then $|\vec{a}+2 \vec{b}|^{2}+|\vec{a}+2 \vec{c}|^{2}$ is equal to__________.
Solution:
$|\vec{a}|=|\vec{b}|=|\vec{c}|=1$
$|\vec{a}-\vec{b}|^{2}+|\vec{a}-\vec{c}|^{2}=8$
$\Rightarrow \vec{a} \cdot \vec{b}+\vec{a} \cdot \vec{c}=-2$
Now, $|\vec{a}+2 \vec{b}|^{2}+|\vec{a}+2 \vec{c}|^{2}$
$=2|\vec{a}|^{2}+4|\vec{b}|^{2}+4|\vec{c}|^{2}+4(\vec{a} \cdot \vec{b}+\vec{a} \cdot \vec{c})=2$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.