# Solve this following

Question:

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

Solution:

$|\overrightarrow{\mathrm{x}}+\overrightarrow{\mathrm{y}}|=|\overrightarrow{\mathrm{x}}|$

$\sqrt{|\overrightarrow{\mathrm{x}}|^{2}+|\overrightarrow{\mathrm{y}}|^{2}+2 \overrightarrow{\mathrm{x}} \cdot \overrightarrow{\mathrm{y}}}=|\overrightarrow{\mathrm{x}}|$

$|\overrightarrow{\mathrm{y}}|^{2}+2 \overrightarrow{\mathrm{x}} \cdot \overrightarrow{\mathrm{y}}=0$ ...........(1)

Now $(2 \overrightarrow{\mathrm{x}}+\lambda \overrightarrow{\mathrm{y}}) \cdot \overrightarrow{\mathrm{y}}=0$

$2 \overrightarrow{\mathrm{x}} \cdot \overrightarrow{\mathrm{y}}+\lambda|\overrightarrow{\mathrm{y}}|^{2}=0$

from (1)

$-|\overrightarrow{\mathrm{y}}|^{2}+\lambda|\overrightarrow{\mathrm{y}}|^{2}=0$

$(\lambda-1)|\vec{y}|^{2}=0$

given $|\vec{y}| \neq 0 \quad \Rightarrow \lambda=1$