Mathematics - Vector Question with Solution | TestHub

$\vec{a}$ is in direction of $\vec{x}\times (\vec{y}\times \vec{z})$ i.e. $(\vec{x}\cdot \vec{z})\vec{y}-(\vec{x}\cdot \vec{y})\vec{z}$ $\Rightarrow \overrightarrow{\mathrm{a}}={\lambda }_1\left[2\times \frac{1}{2}(\overrightarrow{\mathrm{y}}-\overrightarrow{\mathrm{z}})\right]$

$\overrightarrow{\mathrm{a}}={\lambda }_1(\overrightarrow{\mathrm{y}}-\overrightarrow{\mathrm{z}})$.............(1)

Now $\vec{a}\cdot \vec{y}={\lambda }_1(\vec{y}\cdot \vec{y}-\vec{y}\cdot \vec{z})$.

$={\lambda }_1(2-1)\Rightarrow {\lambda }_1=\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{y}}$..............(2)

From (1) and (2), $\vec{a}=\vec{a}\cdot y(\vec{y}-\vec{z})$ Similarly, $\overrightarrow{\mathrm{b}}=(\overrightarrow{\mathrm{b}}\cdot \overrightarrow{\mathrm{z}})(\overrightarrow{\mathrm{z}}-\overrightarrow{\mathrm{x}})$ Now, $\vec{a}\cdot \vec{b}=(\vec{a}\cdot \vec{y})(\vec{b}\cdot \vec{z})[(\vec{y}-\vec{z})\cdot (\vec{z}-\vec{x})]$ $=(\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{y}})(\overrightarrow{\mathrm{b}}\cdot \overrightarrow{\mathrm{z}})[1-1-2+1]=-(\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{y}})(\overrightarrow{\mathrm{b}}\cdot \overrightarrow{\mathrm{z}})$.