IDNLearn.com provides a user-friendly platform for finding answers to your questions. Get accurate answers to your questions from our community of experts who are always ready to provide timely and relevant solutions.
The vectors A and B span a parallelogram. The area of this parallelogram is equal to the magnitude of the cross product, A × B. Cut this area in half and you get the area of the triangle of interest.
Recall that for any two vectors x and y, we have
||x × y|| = ||x|| ||y|| sin(θ)
where θ is the angle between the two vectors; also, recall that
x • y = ||x|| ||y|| cos(θ)
Use the dot-product identity to find the angle between A and B. Compute the dot product and magnitudes:
A • B = (3i + 10j - k) • (-i + 2j + 11k) = -3 + 20 - 11 = 6
||A|| = √(3² + 10² + (-1)²) = √110
||B|| = √((-1)² + 2² + 11²) = √126 = 3√14
Solve for the angle:
cos(θ) = 6/(√110 • 3√14) = 1/√385
θ = arccos(1/√385)
Now using the cross-product identity, we have
||A × B|| = √110 • 3√14 sin(arccos(1/√385)) = 3√1540 • √(384/385) = 48√6
and this is the area of the parallelogram. The area of the triangle in question is half of this, 24√6.
Alternatively, you can compute the cross product directly.
Recall that for any two vectors x and y,
x × x = 0
x × y = - (y × x)
and that the cross product is defined by the following rules:
i × j = k
j × k = i
k × i = j
Then the cross product is
A × B = (3i + 10j - k) × (-i + 2j + 11k)
A × B = -3 (i × i) - 10 (j × i) + (k × i)
… … … … … + 6 (i × j) + 20 (j × j) - 2 (k × j)
… … … … … + 33 (i × k) + 110 (j × k) - 11 (k × k)
A × B = 10k + j + 6k + 2i - 33j + 110i
A × B = 112i - 32j + 16k
Compute its magnitude:
||A × B|| = √(112² + (-32)² + 16²) = √288 = 48√6
and cut it in half to get the area of the triangle; again, you end up with 24√6.