Question

If $\hat{i}+\hat{j}+\hat{k}$, $a_1\hat{i}+b_1\hat{j}+c_1\hat{k}$, $a_2\hat{i}+b_2\hat{j}+c_2\hat{k}$, $a_3\hat{i}+b_3\hat{j}+c_3\hat{k}$ are the position vectors of the points A, B, C, D respectively, $\frac{2}{3}(\hat{i}+\hat{j}+\hat{k})$ is the position vector of the centroid of the triangular face BCD of the tetrahedron ABCD, and if $\alpha\hat{i}+\beta\hat{j}+\gamma\hat{k}$ is the position vector of the centroid of the tetrahedron, then $2\alpha+\beta+\gamma =$

• A. 3
• B. 2
• C. $2/3$
• D. $3/4$

Hint:

Remember the formulas for centroids. For a triangle with vertices A, B, C, the centroid is $\frac{\vec{a}+\vec{b}+\vec{c}}{3}$. For a tetrahedron with vertices A, B, C, D, the centroid is $\frac{\vec{a}+\vec{b}+\vec{c}+\vec{d}}{4}$.

Answer 1

The Correct Option is A

Let the position vectors of the vertices be $\vec{a}, \vec{b}, \vec{c}, \vec{d}$.
We are given $\vec{a} = \hat{i}+\hat{j}+\hat{k}$.
The position vector of the centroid of the triangular face BCD is given by $\frac{\vec{b}+\vec{c}+\vec{d}}{3}$.
It is given that this centroid is at $\frac{2}{3}(\hat{i}+\hat{j}+\hat{k})$.
So, $\frac{\vec{b}+\vec{c}+\vec{d}}{3} = \frac{2}{3}(\hat{i}+\hat{j}+\hat{k})$.
Multiplying both sides by 3, we get $\vec{b}+\vec{c}+\vec{d} = 2(\hat{i}+\hat{j}+\hat{k})$.
The position vector of the centroid of the tetrahedron ABCD is given by $\vec{G} = \frac{\vec{a}+\vec{b}+\vec{c}+\vec{d}}{4}$.
We are given that this centroid is $\vec{G} = \alpha\hat{i}+\beta\hat{j}+\gamma\hat{k}$.
Substitute the known vectors into the centroid formula:
$\vec{G} = \frac{(\hat{i}+\hat{j}+\hat{k}) + (\vec{b}+\vec{c}+\vec{d})}{4}$.
$\vec{G} = \frac{(\hat{i}+\hat{j}+\hat{k}) + 2(\hat{i}+\hat{j}+\hat{k})}{4} = \frac{3(\hat{i}+\hat{j}+\hat{k})}{4}$.
$\alpha\hat{i}+\beta\hat{j}+\gamma\hat{k} = \frac{3}{4}\hat{i}+\frac{3}{4}\hat{j}+\frac{3}{4}\hat{k}$.
By comparing the coefficients, we find $\alpha = \frac{3}{4}$, $\beta = \frac{3}{4}$, and $\gamma = \frac{3}{4}$.
The required value is $2\alpha+\beta+\gamma$.
$2\alpha+\beta+\gamma = 2\left(\frac{3}{4}\right) + \frac{3}{4} + \frac{3}{4} = \frac{6}{4} + \frac{3}{4} + \frac{3}{4} = \frac{12}{4} = 3$.