Let O be the origin, vec{OP} = vec{a} and vec | vedclass

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

Question

(B) Given that $O$ is the origin, $\vec{OP} = \vec{a}$ and $\vec{OQ} = \vec{b}$.Since $\vec{OP} = 5\vec{OR}$, we have $\vec{OR} = \frac{1}{5}\vec{a}$.

Vedclass · Accepted Answer

$\frac{1}{5}(\vec{b}-4\vec{a})$

Vedclass · Answer

(B) Given that $O$ is the origin, $\vec{OP} = \vec{a}$ and $\vec{OQ} = \vec{b}$.Since $\vec{OP} = 5\vec{OR}$, we have $\vec{OR} = \frac{1}{5}\vec{a}$.Given $\vec{OQ} = 5\vec{RM}$, we have $\vec{RM} = \frac{1}{5}\vec{b}$.We know that $\vec{RM} = \vec{OM} - \vec{OR}$, so $\vec{OM} = \vec{OR} + \vec{RM}$.Substituting the values, $\vec{OM} = \frac{1}{5}\vec{a} + \frac{1}{5}\vec{b}$.Now, $\vec{PM} = \vec{OM} - \vec{OP} = (\frac{1}{5}\vec{a} + \frac{1}{5}\vec{b}) - \vec{a}$.$\vec{PM} = \frac{1}{5}\vec{b} + (\frac{1}{5} - 1)\vec{a} = \frac{1}{5}\vec{b} - \frac{4}{5}\vec{a} = \frac{1}{5}(\vec{b} - 4\vec{a})$.

Let $O$ be the origin, $\vec{OP} = \vec{a}$ and $\vec{OQ} = \vec{b}$. If $R$ is the point on $\vec{OP}$ such that $\vec{OP} = 5\vec{OR}$, and $M$ is the point such that $\vec{OQ} = 5\vec{RM}$, then $\vec{PM}$ is equal to:

Similar Questions

If the orthocentre of the triangle whose vertices are $2 \hat{i}+3 \hat{j}+5 \hat{k}$,$5 \hat{i}+2 \hat{j}+3 \hat{k}$,and $3 \hat{i}+5 \hat{j}+2 \hat{k}$ is $x \hat{i}+y \hat{j}+z \hat{k}$,then:

The triad $(x, y, z)$ of real numbers such that $(3 \hat{i}-\hat{j}+2 \hat{k})=(2 \hat{i}+3 \hat{j}-\hat{k}) x+(\hat{i}-2 \hat{j}+2 \hat{k}) y+(-2 \hat{i}+\hat{j}-2 \hat{k}) z$ is

Let $H$ be the orthocentre of an acute angled $\triangle ABC$ and $O$ be its circumcenter. Then,$\vec{HA} + \vec{HB} + \vec{HC}$

If $A(a)$, $B(b)$, and $C(c)$ are vertices of $\Delta ABC$. Point $D$ divides segment $BC$ internally in the ratio $2 : 1$. Point $E$ divides segment $AD$ internally in the ratio $1 : 2$, then the position vector of $E$ is

Three non-zero non-collinear vectors $\vec{a}, \vec{b}$ and $\vec{c}$ are such that $\vec{a}+3\vec{b}$ is collinear with $\vec{c}$, and $3\vec{b}+2\vec{c}$ is collinear with $\vec{a}$. Then $\vec{a}+3\vec{b}+2\vec{c}$ equals to

Vedclass Test Series

Exam Paper Generator

Online Exam Module