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
- $\dfrac{\vec{a} + \vec{b} + \vec{c}}{9}$
- $\dfrac{\vec{a} + \vec{b} + \vec{c}}{6}$
- $\dfrac{\vec{a} + \vec{b} + \vec{c}}{3}$
- $\dfrac{2(\vec{a} + \vec{b} + \vec{c})}{3}$
The Correct Option is C
Solution and Explanation
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 $\dfrac{\vec{a} + \vec{b} + \vec{c}}{3}$.
Top Questions on Vectors
- Consider the vectors $$ \vec{x} = \hat{i} + 2\hat{j} + 3\hat{k},\quad \vec{y} = 2\hat{i} + 3\hat{j} + \hat{k},\quad \vec{z} = 3\hat{i} + \hat{j} + 2\hat{k}. $$ For two distinct positive real numbers $ \alpha $ and $ \beta $, define $$ \vec{X} = \alpha \vec{x} + \beta \vec{y} - \vec{z},\quad \vec{Y} = \alpha \vec{y} + \beta \vec{z} - \vec{x},\quad \vec{Z} = \alpha \vec{z} + \beta \vec{x} - \vec{y}. $$ If the vectors $ \vec{X}, \vec{Y}, \vec{Z} $ lie in a plane, then the value of $ \alpha + \beta - 3 $ is ________.
Let $ \vec{w} = \hat{i} + \hat{j} - 2\hat{k} $, and $ \vec{u} $ and $ \vec{v} $ be two vectors, such that $ \vec{u} \times \vec{v} = \vec{w} $ and $ \vec{v} \times \vec{w} = \vec{u} $. Let $ \alpha, \beta, \gamma $, and $ t $ be real numbers such that: $$ \vec{u} = \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}, $$ and the system of equations is: $$ -t\alpha + \beta + \gamma = 0 \quad \cdots (1) $$ $$ \alpha - t\beta + \gamma = 0 \quad \cdots (2) $$ $$ \alpha + \beta - t\gamma = 0 \quad \cdots (3) $$ Match each entry in List-I to the correct entry in List-II and choose the correct option.
List-I- [(P)] $ |\vec{v}|^2 $ is equal to
- [(Q)] If $ \alpha = \sqrt{3} $, then $ \gamma^2 $ is equal to
- [(R)] If $ \alpha = \sqrt{3} $, then $ (\beta + \gamma)^2 $ is equal to
- [(S)] If $ \alpha = \sqrt{2} $, then $ t + 3 $ is equal to
List-II
- [(1)] 0
- [(2)] 1
- [(3)] 2
- [(4)] 3
- [(5)]
- Find the shortest distance between the lines: \[ \frac{x+1}{2} = \frac{y-1}{1} = \frac{z-9}{-3} \] and \[ \frac{x-3}{2} = \frac{y+15}{-7} = \frac{z-9}{5}. \]
Let \( \vec{p} \) and \( \vec{q} \) be two unit vectors and \( \alpha \) be the angle between them. Then \( (\vec{p} + \vec{q}) \) will be a unit vector for what value of \( \alpha \)?
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