Are the points A $(3{,}6{,}9)$, B $(10{,}20{,}30)$ and $C(25,-41{,}5)$, the vertices of a right angled triangle?

Step 1: Calculate the square of the distance between each pair of points

To determine if the points form a right-angled triangle, we first need to calculate the square of the distance between each pair of points using the distance formula in 3D: $d^2 = (x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2$. We will calculate $AB^2$, $BC^2$, and $AC^2$.

Step 2: Compute the squared distances

Now, we perform the arithmetic for each squared distance. For $AB^2$, we get $49 + 196 + 441 = 686$. For $BC^2$, we get $225 + 3721 + 625 = 4571$. For $AC^2$, we get $484 + 2209 + 16 = 2709$.

Step 3: Apply the Pythagorean theorem

For the points to form a right-angled triangle, the sum of the squares of the two shorter sides must be equal to the square of the longest side (Pythagorean theorem). We check all three possible combinations.

Step 4: Compare the sums of squares

Comparing the sums, we find that $AB^2 + AC^2 = 3395$, which is not equal to $BC^2 = 4571$. Similarly, $AB^2 + BC^2 = 5257$, which is not equal to $AC^2 = 2709$. And $BC^2 + AC^2 = 7280$, which is not equal to $AB^2 = 686$. Since none of the combinations satisfy the Pythagorean theorem, the given points do not form a right-angled triangle.

More from this chapter

• Find the equation of set of points P such that PA²+PB²=2 k², where A and B are the points (3,4,5) and (-1,3,-7), respectively.
• The centroid of a triangle ABC is at the point (1,1,1). If the coordinates of A and B are (3,-5,7) and (-1,7,-6), respectively, find the coordinates of the poin
• Show that the points A (1, 2, 3), B (-1, -2, -1), C (2, 3, 2) and D(4,7,6) are the vertices of a parallelogram ABCD, but it is not a rectangle.
• Find the equation of the set of the points P such that its distances from the points A(3,4,-5) and B(-2,1,4) are equal.
• Show that the points P(-2,3,5), Q(1,2,3) and R(7,0,-1) are collinear.

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