$A, B, C, D$ 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
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If $\vec{a}, \vec{b}$ and $\vec{c}$ are unit vectors and $\vec{a}+\vec{b}+\vec{c}=\vec{0}$,then the value of $\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}$ is . . . . . . . (in $/2$)
The vector projection of $\overline{PQ}$ on $\overline{AB}$,where $P \equiv (-2, 1, 3)$,$Q \equiv (3, 2, 5)$,$A \equiv (4, -3, 5)$ and $B \equiv (7, -5, -1)$ is
Let $\vec{a}=6 \hat{i}+9 \hat{j}+12 \hat{k}$,$\vec{b}=\alpha \hat{i}+11 \hat{j}-2 \hat{k}$ and $\vec{c}$ be vectors such that $\vec{a} \times \vec{c}=\vec{a} \times \vec{b}$. If $\vec{a} \cdot \vec{c}=-12$ and $\vec{c} \cdot (\hat{i}-2 \hat{j}+\hat{k})=5$,then $\vec{c} \cdot (\hat{i}+\hat{j}+\hat{k})$ is equal to $.............$.
DifficultJEE MAIN 2023
View SolutionIf $\overline{a}+\overline{b}+\overline{c}=\overline{0}$ with $|\overline{a}|=3, |\overline{b}|=5$ and $|\overline{c}|=7$,then the angle between $\overline{a}$ and $\overline{b}$ is:
Let $a$,$b$,and $c$ be $3$ non-zero vectors such that no $2$ of these are collinear. If vector $a + 2b$ is collinear with $c$ and $b + 3c$ is collinear with $a$,then $a + 2b + 6c$ equals:
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