If vectors a = i + j + k, b = i - j + k and c = i + 2j - k, then the area of the triangle with a,b, and c as vertices is
Correct: B
Area of triangle ABC = 1/2 |AB x AC|. AB = b - a = (i - j + k) - (i + j + k) = -2j. AC = c - a = (i + 2j - k) - (i + j + k) = j - 2k. AB x AC = (-2j) x (j - 2k) = -2j x j + 4j x k = 0 + 4i = 4i. So |AB x AC| = |4i| = 4. Area = 1/2 * 4 = 2. We are asked to compute area with vertices a,b and c. Area = 1/2 |(b-a)x(c-a)| = 1/2|(-2j)x(j-2k)| =1/2|4i| = 2. This is wrong, the answer is supposed to be in terms of √11. There must be an error. Need to re-calculate cross-product: b-a = 0i-2j+0k, c-a= 0i+j-2k so (b-a)x(c-a) = [[i,j,k],[0,-2,0],[0,1,-2]] = i(4-0)-j(0-0)+k(0) = 4i+0j+0k. The magnitude of it should be |4i|=4, hence area = 1/2(4)=2. It still does not get to any kind of √11. The question might be about area of Parallelogram.