$A, B, C, D$ are four points in a plane with position vectors $\overline{a}, \overline{b}, \overline{c}, \overline{d}$ respectively such that $(\overline{a}-\overline{d}) \cdot(\overline{b}-\overline{c})=(\overline{b}-\overline{d}) \cdot(\overline{c}-\overline{a})=0$. Then the point $D$ is the $\dots$ of $\triangle ABC$.
- Acentroid
- Bcircumcentre
- Cincentre
- Dorthocentre
Explore More
Similar Questions
Let $\vec{c}$ be the projection vector of $\vec{b}=\lambda \hat{i}+4 \hat{k}, \lambda>0$,on the vector $\vec{a}=\hat{i}+2 \hat{j}+2 \hat{k}$. If $|\vec{a}+\vec{c}|=7$,then the area of the parallelogram formed by the vectors $\vec{b}$ and $\vec{c}$ is . . . . . . .
If $\overrightarrow{a}, \overrightarrow{b}$ and $\overrightarrow{c}$ are unit vectors such that $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}$,then the value of $3 \overrightarrow{a} \cdot \overrightarrow{b}+2 \overrightarrow{b} \cdot \overrightarrow{c}+\overrightarrow{c} \cdot \overrightarrow{a}$ is
MediumKCET 2010
View SolutionIf $a$ makes an acute angle with $b$,$r \cdot a = 0$ and $r \times b = c \times b$,then $r=$
If $a$,$b$,and $c$ are unit vectors,then $|a - b|^2 + |b - c|^2 + |c - a|^2$ does not exceed
DifficultIIT 2001
View SolutionIf $a$ and $b$ are unit vectors and $a - b$ is also a unit vector,then the angle between $a$ and $b$ is
Easy
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo