Let the vectors a⃗, b⃗, c⃗ represent three coterminous edges of a parallelopiped of volume V. Then the volume of the parallelopiped, whose coterminous edges are represented by a⃗, b⃗+c⃗ and a⃗+2 b⃗+3 c⃗ is equal to :

Solution

Maths sir · 4 steps

Write each new edge in terms of a, b, c. The determinant of those coefficients scales the volume.

1. Volume as triple product. The volume of a parallelopiped is the absolute value of the scalar triple product of its edges. Here V = |[a⃗ b⃗ c⃗]|.
2. Coefficient determinant. Each new edge is a combination of a⃗, b⃗, c⃗. The new triple product is the determinant of their coefficients, multiplied by [a⃗ b⃗ c⃗].

   [a⃗ b⃗+c⃗ a⃗+2b⃗+3c⃗] = |1, 0, 0; 0, 1, 1; 1, 2, 3| [a⃗ b⃗ c⃗] = 1 × (3-2) [a⃗ b⃗ c⃗]

3. Check by expanding. Any term with a repeated vector is zero, so only [a⃗ b⃗ 3c⃗] + [a⃗ c⃗ 2b⃗] survives. That is 3[a⃗ b⃗ c⃗] - 2[a⃗ b⃗ c⃗] = [a⃗ b⃗ c⃗].
4. So. The new triple product equals the old one, so the new volume is V, option (d).

The idea to keep. When new edges are linear combinations of old ones, the volume gets multiplied by the absolute value of the determinant of the coefficients. Any triple product with a repeated vector is zero.

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