If overrightarrow{x} and overrightarrow{y} ar | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given $|\overrightarrow{x}+\overrightarrow{y}|=|\overrightarrow{x}|$. Squaring both sides,we get:$|\overrightarrow{x}|^2+|\overrightarrow{y}|^2+2\

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Given $|\overrightarrow{x}+\overrightarrow{y}|=|\overrightarrow{x}|$. Squaring both sides,we get:$|\overrightarrow{x}|^2+|\overrightarrow{y}|^2+2\overrightarrow{x}\cdot\overrightarrow{y}=|\overrightarrow{x}|^2$$|\overrightarrow{y}|^2+2\overrightarrow{x}\cdot\overrightarrow{y}=0$  --- $(1)$Given that $(2\overrightarrow{x}+\lambda\overrightarrow{y})$ is perpendicular to $\overrightarrow{y}$,their dot product is zero:$(2\overrightarrow{x}+\lambda\overrightarrow{y})\cdot\overrightarrow{y}=0$$2\overrightarrow{x}\cdot\overrightarrow{y}+\lambda|\overrightarrow{y}|^2=0$  --- $(2)$From equation $(1)$,we have $2\overrightarrow{x}\cdot\overrightarrow{y}=-|\overrightarrow{y}|^2$. Substituting this into equation $(2)$:$-|\overrightarrow{y}|^2+\lambda|\overrightarrow{y}|^2=0$$(\lambda-1)|\overrightarrow{y}|^2=0$Since $\overrightarrow{y}$ is a non-zero vector,$|\overrightarrow{y}|^2 
eq 0$. Therefore,$\lambda-1=0$,which gives $\lambda=1$.

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

Similar Questions

Let $u, v$ and $w$ be three vectors such that $u+v+w=0$,$|u|=3$,$|v|=5$ and $|w|=7$. Then the angle between $u$ and $v$ is (in $^{\circ}$)

The angle $\theta$ between vectors $\vec{a} = \hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$ is . . . . . . .

The projection of the line segment joining the points $P(1, -1, 3)$ and $Q(2, -4, 11)$ on the line joining the points $A(-1, 2, 3)$ and $B(3, -2, 10)$ is

In a right trapezium $ABCD$,the diagonals are perpendicular,and the ratio of the length of bases $AD : BC = 2 : 3$. Then the ratio of the length of diagonals is

The vector $a + b$ bisects the angle between the vectors $a$ and $b$,if

Vedclass Test Series

Exam Paper Generator

Online Exam Module