Question

If the position vector of the centroid of tetrahedron whose position vector of vertices are $\hat{i}+\hat{j}+3\hat{k},2\hat{i}-\hat{j},-3\hat{i}+2\hat{j}+8\hat{k}$ and $3\hat{i}+2\hat{j}+\hat{k}$ is $x\hat{i}+y\hat{j}+z\hat{k}.$ Then the value of $4(x-y+z)=$

• A. 11.00
• B. 11.0
• C. 11

Answer 1

Answer: (A), (B), (C)

Given vertices :
$\hat{i}+\hat{j}+3\hat{k},2\hat{i}-\hat{j},-3\hat{i}+2\hat{j}+8\hat{k}$ and $3\hat{i}+2\hat{j}+\hat{k}$ is $x\hat{i}+y\hat{j}+z\hat{k}.$
So, centroid of tetrahedron $G=\frac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}+\overrightarrow{d}}{4}$
$=\frac{(\hat{i}+\hat{j}+3\hat{k})+(2\hat{i}-\hat{j})+(-3\hat{i}+2\hat{j}+8\hat{k})+(3\hat{i}+2\hat{j}+\hat{k})}{4}\Rightarrow x\hat{i}+y\hat{j}+z\hat{k}=\frac{3\hat{i}+4\hat{j}+12\hat{k}}{4}\Rightarrow x=\frac{3}{4},y=\frac{4}{4}=1,z=\frac{12}{4}\therefore 4(x-y+z)=11$