If the volume of a parallelopiped, whose
coterminus edges are given by the
vectors a⃗ = î + ĵ + nk̂,
b⃗ = 2î + 4ĵ - nk̂ and
c⃗ = î + nĵ + 3k̂ (n ≥ 0), is 158 cu. units, then :

Solution

Maths sir · 4 steps

Set the triple-product determinant equal to 158, find n, and only then test each option.

1. Volume as determinant. The volume of the parallelopiped is |[a⃗ b⃗ c⃗]|, the absolute value of the determinant of the three component rows.

   |1, 1, n; 2, 4, -n; 1, n, 3| = (12 + n²) - (6 + n) + n(2n - 4) = 3n² - 5n + 6

2. Solve for n. 3n² - 5n + 6 = 158 gives 3n² - 5n - 152 = 0, so (n - 8)(3n + 19) = 0. Since n ≥ 0, n = 8. The other case, 3n² - 5n + 6 = -158, has no real root because its discriminant 25 - 1968 is negative.
3. Check the dot products. b⃗ · c⃗ = 2 + 4n - 3n = 2 + n = 10. Also a⃗ · c⃗ = 1 + n + 3n = 1 + 4n = 33, not 17.
4. So. n = 8, which makes b⃗ · c⃗ = 10. The answer is (b).

The idea to keep. Volume of a parallelopiped is the modulus of the scalar triple product. Expand the determinant carefully, reject roots that break the given condition, then substitute.

Answer from the NTA final answer key. Solution by PrepWiser.