The vertices of a triangle are \(A(2, 0, 0)\), \(B(0, 6, 0)\) and \(C(0, 0, 4)\). If \(AD\), \(BE\) and \(CF\) are the medians of the triangle, then what is \(AD^2 + BE^2 + CF^2\) equal to?
\(84\)
For any triangle, the sum of the squares of the medians equals three-fourths the sum of the squares of the sides: \(AD^2+BE^2+CF^2=\dfrac{3}{4}(AB^2+BC^2+CA^2)\). Here \(AB^2=40\), \(BC^2=52\) and \(CA^2=20\), so the sum is \(112\), giving \(AD^2+BE^2+CF^2=\dfrac{3}{4}\times112=84\).
Important Questions from Distance between points
Determine the shortest distance between the lines l1 and l2 whose vector equations are
\(\vec{r}=2 \hat{\imath}+\hat{\jmath}+\lambda(2 \hat{\imath}-\hat{\jmath}+\hat{k}) \quad \ldots(1)\)
\(\vec{r}=3 \hat{\imath}+\hat{\jmath}-\hat{k}+\mu(3 \hat{\imath}-5 \hat{\jmath}+2 \hat{k}) \quad \ldots(2)\)
Find the distance between the points P (2, -1, 3) and Q (-5, 2, 1)?
The sum of distances from origin to (0, 5, 5) and (5, 8, 6) is:
The distance between the points (2, 3) and (4, 1) is.
The distance of a point from the x-axis is called that point.
