Let a⃗=6 î+ĵ-k̂ and b⃗=î+ĵ. If c⃗ is a is vector such that |c⃗| ≥ 6, a⃗ · c⃗=6|c⃗|,|c⃗-a⃗|=2 √2 and the angle between a⃗ × b⃗ and c⃗ is 60°, then |(a⃗ × b⃗) × c⃗| is equal to:

Solution

Maths sir · 4 steps

Turn |c⃗-a⃗| into a quadratic in |c⃗|, then the cross product only needs a sine.

1. Square the distance. Expand |c⃗-a⃗|² = |c⃗|² - 2 a⃗·c⃗ + |a⃗|². Here |a⃗|² = 36+1+1 = 38 and a⃗·c⃗ = 6|c⃗|.

   |c⃗|² - 12|c⃗| + 38 = 8 ⇒ |c⃗|² - 12|c⃗| + 30 = 0

2. Pick the root. The roots are |c⃗| = 6 ± √6. Since |c⃗| ≥ 6, only |c⃗| = 6+√6 is allowed.
3. Find a cross b. a⃗×b⃗ = |î, ĵ, k̂; 6, 1, -1; 1, 1, 0| = î - ĵ + 5k̂, so |a⃗×b⃗| = √27 = 3√3.
4. So. |(a⃗×b⃗)×c⃗| = 3√3 (6+√6)sin 60° = 3√3×√3/2(6+√6) = 9/2(6+√6), which is option (c).

The idea to keep. When you know a dot product and a distance, square the distance: it gives a quadratic in the unknown length. Then the magnitude of a cross product is just |u⃗||v⃗|sin θ.

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