Question

If $\overrightarrow{x}$and $\overrightarrow{y}$be two non-zero vectors such that $\left|\overrightarrow{x}+\overrightarrow{y}\right|=\left|\overrightarrow{x}\right|$ and $2\overrightarrow{x}+\lambda \overrightarrow{y}$is perpendicular to $\overrightarrow{y}$ ,then the value of $\lambda$ is

Answer 1

Find the value of $\lambda$.

Given that,

$\left|\overrightarrow{x}+\overrightarrow{y}\right|=\left|\overrightarrow{x}\right|$

By squaring both sides, we get

$$\begin{gathered} {\left|\overrightarrow{x}+\overrightarrow{y}\right|}^2={\left|\overrightarrow{x}\right|}^2 \\ \Rightarrow {\left|\overline{y}\right|}^2+2\overline{x}.\overline{y}=0...\left(1\right) \end{gathered}$$

Now, by taking a dot product of $2\overrightarrow{x}+\lambda \overrightarrow{y}$ with $y$, we get

$$\begin{gathered} (2\overline{x}+\lambda \overline{y}).\overline{y}=0 \\ \Rightarrow \lambda {\left(\left|\overline{y}\right|\right)}^2+2\overline{x}\overline{y}=0\dots \left(2\right) \end{gathered}$$

Now, on comparing both the equations, we get

$\lambda =1$

Hence, the required answer is 1.