Question:medium

Find the area of a triangle with vertices \( A(1,2,3), B(2,3,4), \) and \( C(3,4,5) \).

Show Hint

If two side vectors are scalar multiples, the triangle is degenerate and its area is zero.
Show Solution

Solution and Explanation

To Find the Area of the Triangle with vertices A(1,2,3), B(2,3,4) and C(3,4,5):

Step 1: Use the Formula
Area of triangle = (1/2) |AB × AC|

First, find vectors AB and AC.

AB = B − A = (2−1, 3−2, 4−3) = (1,1,1)

AC = C − A = (3−1, 4−2, 5−3) = (2,2,2)

Step 2: Compute Cross Product AB × AC

AB × AC = | i    j    k |
            | 1    1    1 |
            | 2    2    2 |

Expanding determinant:

= i(1×2 − 1×2) − j(1×2 − 1×2) + k(1×2 − 1×2)

= i(2−2) − j(2−2) + k(2−2)

= 0i − 0j + 0k = (0,0,0)

Step 3: Find Magnitude

|AB × AC| = 0

Therefore, Area = (1/2) × 0 = 0

Conclusion:
The area of the triangle is 0.
This means the three points are collinear (they lie on a straight line).

Was this answer helpful?
0