Let the three sides of a triangle ABC be given by the vectors 2i^−j^+k^, i^−3j^−5k^ and 3i^−4j^−4k^. Let G be the centroid of the triangle ABC. Then 6(∣AG∣2+∣BG∣2+∣CG∣2) $ is equal to:
Get the complete, step-by-step math solution for: "Let the three sides of a triangle ABC be given by the vectors 2 {i}- {j}+ {k}, {i}-3 {j}-5 {k} and 3 {i}-4 {j}-4 {k}. Let G be the centroid of the tri...". Powered by SolveForX AI math tutor.
Step-by-Step Solution
Step 1: Verify if vectors form a triangle
First, we need to check if the given vectors can form a triangle. For three vectors to form a triangle, the sum of any two vectors must be equal to the third vector (or their sum must be the zero vector if they are taken in order around the triangle). Let the given vectors be v1=2i^−j^+k^, v2=i^−3j^−5k^, and v3=3i^−4j^−4k^. We observe that v1+v2=(2i^−j^+k^)+(i^−3j^−5k^)=(2+1)i^+(−1−3)j^+(1−5)k^=3i^−4j^−4k^. This sum is equal to v3. Therefore, these three vectors can represent the sides of a triangle.
Step 2: Recall the centroid property
For a triangle ABC with centroid G, there is a known property relating the sum of the squares of the distances from the vertices to the centroid and the sum of the squares of the side lengths. This property states that the sum of the squares of the lengths of the medians is equal to three-fourths the sum of the squares of the sides. More directly, the sum of the squares of the distances from the vertices to the centroid is one-third the sum of the squares of the side lengths.
Step 3: Calculate the square of the magnitudes of the side vectors
Now, we calculate the square of the magnitude for each given vector. The magnitude of a vector ai^+bj^+ck^ is a2+b2+c2, so its square is a2+b2+c2. We apply this formula to each of the three vectors.
Step 4: Substitute values into the centroid property
We substitute the calculated squared magnitudes of the side vectors into the centroid property formula. The sum of the squares of the side lengths is 6+35+41=82. Therefore, the sum of the squares of the distances from the vertices to the centroid is 31×82.
Step 5: Calculate the final expression
Finally, we need to find the value of 6(∣AG∣2+∣BG∣2+∣CG∣2). We multiply the result from the previous step by 6. This simplifies to 2×82=164.