Question:
Let $\vec{a}, \vec{b}, \vec{c}$ be the position vectors of the vertices of a triangle $ABC$. Through the vertices, lines are drawn parallel to the sides to form the triangle $A'B'C'$. Then the centroid of $\Delta A'B'C'$ is
Let $\vec{a}, \vec{b}, \vec{c}$ be the position vectors of the vertices of a triangle $ABC$. Through the vertices, lines are drawn parallel to the sides to form the triangle $A'B'C'$. Then the centroid of $\Delta A'B'C'$ is
Show Hint
Centroid of triangle from position vectors is always the average of the three vectors.
Updated On: May 18, 2025
- $\dfrac{\vec{a} + \vec{b} + \vec{c}}{9}$
- $\dfrac{\vec{a} + \vec{b} + \vec{c}}{6}$
- $\dfrac{\vec{a} + \vec{b} + \vec{c}}{3}$
- $\dfrac{2(\vec{a} + \vec{b} + \vec{c})}{3}$
Hide Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
The triangle $A'B'C'$ is formed by drawing lines through vertices parallel to the triangle’s sides. This creates a congruent triangle.
Since translations don’t change centroids, the centroid of $A'B'C'$ is same as for $ABC$.
Centroid of a triangle formed by position vectors $\vec{a}, \vec{b}, \vec{c}$ is $\dfrac{\vec{a} + \vec{b} + \vec{c}}{3}$.
Since translations don’t change centroids, the centroid of $A'B'C'$ is same as for $ABC$.
Centroid of a triangle formed by position vectors $\vec{a}, \vec{b}, \vec{c}$ is $\dfrac{\vec{a} + \vec{b} + \vec{c}}{3}$.
Was this answer helpful?
0
0
Top Questions on Vectors
- The diagonals of a parallelogram are given by \( \mathbf{a} = 2 \hat{i} - \hat{j} + \hat{k} \) and \( \mathbf{b} = \hat{i} + 3 \hat{j} - \hat{k}\) . Find the area of the parallelogram.
- If \( \mathbf{a} \) and \( \mathbf{b} \) are position vectors of two points \( P \) and \( Q \) respectively, then find the position vector of a point \( R \) in \( QP \) produced such that \[ QR = \frac{3}{2} QP. \]
- If \( \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} = 0 \), \( |\overrightarrow{a}| = \sqrt{37} \), \( |\overrightarrow{b}| = 3 \), and \( |\overrightarrow{c}| = 4 \), then the angle between \( \overrightarrow{b} \) and \( \overrightarrow{c} \) is:
- If $\mathbf{a}$ and $\mathbf{b}$ are position vectors of point A and point B, respectively, find the position vector of point C on $\overrightarrow{BA}$ such that $BC = 3BA$.
If vector \( \mathbf{a} = 3 \hat{i} + 2 \hat{j} - \hat{k} \) \text{ and } \( \mathbf{b} = \hat{i} - \hat{j} + \hat{k} \), then which of the following is correct?
View More Questions
Questions Asked in AP EAPCET exam
- In a container of volume 16.62 m$^3$ at 0°C temperature, 2 moles of oxygen, 5 moles of nitrogen and 3 moles of hydrogen are present, then the pressure in the container is (Universal gas constant = 8.31 J/mol K)
- AP EAPCET - 2025
- Ideal gas equation
- If \[ A = \begin{bmatrix} x & 2 & 1 \\ -2 & y & 0 \\ 2 & 0 & -1 \end{bmatrix}, \] where \( x \) and \( y \) are non-zero real numbers, trace of \( A = 0 \), and determinant of \( A = -6 \), then the minor of the element 1 of \( A \) is:}
- AP EAPCET - 2025
- Complex numbers
- Two objects of masses 5 kg and 10 kg are placed 2 meters apart. What is the gravitational force between them?
(Use \(G = 6.67 \times 10^{-11}\, \mathrm{Nm^2/kg^2}\))- AP EAPCET - 2025
- Gravitation
- The number of solutions of the equation $4 \cos 2\theta \cos 3\theta = \sec \theta$ in the interval $[0, 2\pi]$ is
- AP EAPCET - 2025
- Trigonometric Identities
- The area (in sq. units) of the triangle formed by the tangent and normal to the ellipse \( 9x^2 + 4y^2 = 72 \) at the point (2, 3) with the X-axis is
- AP EAPCET - 2025
- Coordinate Geometry
View More Questions