Question:
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 :
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 :
Updated On: Apr 12, 2026
- $\frac{1}{5}(\vec{a} - 4\vec{b})$
- $\frac{1}{5}(\vec{b} - 4\vec{a})$
- $\frac{1}{5}(-\vec{a} + 4\vec{b})$
- $\frac{1}{5}(-\vec{b} + 4\vec{a})$
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
We represent the positions of points R and M in terms of vectors $\vec{a}$ and $\vec{b}$ using the given ratios, then find the vector $\vec{PM}$.
Step 2: Detailed Explanation:
Given $\vec{OP} = \vec{a}$ and $\vec{OQ} = \vec{b}$.
$R$ is on $\vec{OP}$ such that $\vec{OP} = 5\vec{OR} \implies \vec{OR} = \frac{1}{5}\vec{a}$.
Point $M$ satisfies $\vec{OQ} = 5\vec{RM}$.
$\vec{RM} = \vec{OM} - \vec{OR} = \vec{OM} - \frac{1}{5}\vec{a}$.
So, $\vec{b} = 5(\vec{OM} - \frac{1}{5}\vec{a}) = 5\vec{OM} - \vec{a}$.
$\vec{OM} = \frac{\vec{a} + \vec{b}}{5}$.
Now, $\vec{PM} = \vec{OM} - \vec{OP} = \frac{\vec{a} + \vec{b}}{5} - \vec{a} = \frac{\vec{a} + \vec{b} - 5\vec{a}}{5} = \frac{\vec{b} - 4\vec{a}}{5}$.
Step 3: Final Answer:
Vector $\vec{PM}$ is $\frac{1}{5}(\vec{b} - 4\vec{a})$.
We represent the positions of points R and M in terms of vectors $\vec{a}$ and $\vec{b}$ using the given ratios, then find the vector $\vec{PM}$.
Step 2: Detailed Explanation:
Given $\vec{OP} = \vec{a}$ and $\vec{OQ} = \vec{b}$.
$R$ is on $\vec{OP}$ such that $\vec{OP} = 5\vec{OR} \implies \vec{OR} = \frac{1}{5}\vec{a}$.
Point $M$ satisfies $\vec{OQ} = 5\vec{RM}$.
$\vec{RM} = \vec{OM} - \vec{OR} = \vec{OM} - \frac{1}{5}\vec{a}$.
So, $\vec{b} = 5(\vec{OM} - \frac{1}{5}\vec{a}) = 5\vec{OM} - \vec{a}$.
$\vec{OM} = \frac{\vec{a} + \vec{b}}{5}$.
Now, $\vec{PM} = \vec{OM} - \vec{OP} = \frac{\vec{a} + \vec{b}}{5} - \vec{a} = \frac{\vec{a} + \vec{b} - 5\vec{a}}{5} = \frac{\vec{b} - 4\vec{a}}{5}$.
Step 3: Final Answer:
Vector $\vec{PM}$ is $\frac{1}{5}(\vec{b} - 4\vec{a})$.
Was this answer helpful?
0
0
Top JEE Main Mathematics Questions
- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
- If , then the general value of is:
- The general solution of
- If the constant term in the binomial expansion of $\left(\frac{x^{\frac{3}{2}}}{2}-\frac{4}{x^l}\right)^9$ is $-84$ and the coefficient of $x^{-3 l}$ is $2^\alpha \beta$, where $\beta<0$ is an odd number, then $|\alpha l-\beta|$ is equal to_____
View More Questions
Top JEE Main Geometry and Vectors Questions
- Let the arc AC of a circle subtend a right angle at the center O. If the point B on the arc AC divides the arc AC such that: \[ \frac{\text{length of arc AB}}{\text{length of arc BC}} = \frac{1}{5} \] and \[ \overrightarrow{OC} = \alpha \overrightarrow{OA} + \beta \overrightarrow{OB}, \] then \( \alpha = \sqrt{2} (\sqrt{3}-1) \beta \) is equal to:
- Let P be the foot of the perpendicular from the point \( Q(10,-3,-1) \) on the line: \[ \frac{x-3}{7} = \frac{y-2}{-1} = \frac{z+1}{-2}. \] Then the area of the right-angled triangle PQR, where R is the point \( (3,-2,1) \), is:
- Let \( \frac{\overline{z} - i}{z - i} = \frac{1}{3}, \, z \in \mathbb{C} \), be the equation of a circle with center at \( C \). If the area of the triangle, whose vertices are at the points \( (0, 0), C \) and \( (\alpha, 0) \), is 11 square units, then \( \alpha^2 \) equals:
- Let the area of a triangle \( \triangle PQR \) with vertices \( P(5, 4) \), \( Q(-2, 4) \), and \( R(a, b) \) be 35 square units. If its orthocenter and centroid are \( O(2, \frac{14}{5}) \) and \( C(c, d) \) respectively, then \( c + 2d \) is equal to:
- Let the position vectors of the vertices A, B, and C of a tetrahedron ABCD be \( \hat{i} + 2\hat{j} + \hat{k} \), \( \hat{i} + 3\hat{j} - 2\hat{k} \), and \( 2\hat{i} + \hat{j} - \hat{k} \) respectively. The altitude from the vertex D to the opposite face ABC meets the median line segment through A of the triangle ABC at the point E. If the length of AD is \( \frac{\sqrt{10}}{3} \) and the volume of the tetrahedron is \( \frac{\sqrt{805}}{6\sqrt{2}} \), then the position vector of E is:
View More Questions
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
View More Questions