Let a⃗=2 î-3 ĵ+k̂, b⃗=3 î+2 ĵ+5 k̂ and a vector c⃗ be such that (a⃗-c⃗) × b⃗=-18 î-3 ĵ+12 k̂ and a⃗ · c⃗=3. If b⃗ × c⃗=d⃗, then |a⃗ · d⃗| is equal to :

Solution

Maths sir · 5 steps

Open the bracket and d⃗ falls out: the given vector is a⃗×b⃗+d⃗.

1. Expand the bracket. The cross product distributes, and -c⃗×b⃗ = b⃗×c⃗. So the given vector equals a⃗×b⃗ + d⃗.

   d⃗ = (-18î-3ĵ+12k̂) - a⃗×b⃗

2. Find a cross b. a⃗×b⃗ = ((-3)(5)-(1)(2))î - ((2)(5)-(1)(3))ĵ + ((2)(2)-(-3)(3))k̂ = -17î-7ĵ+13k̂.
3. Get d. d⃗ = (-18+17)î + (-3+7)ĵ + (12-13)k̂ = -î+4ĵ-k̂. As a check, b⃗·d⃗ = -3+8-5 = 0, as it must be for b⃗×c⃗.
4. Dot with a. a⃗·d⃗ = 2(-1)+(-3)(4)+1(-1) = -15. The condition a⃗·c⃗ = 3 is not needed for this.
5. So. |a⃗·d⃗| = 15, which is option (a).

The idea to keep. When a cross product of a difference is given, expand it and move the known part across. Also, a⃗·(a⃗×b⃗)=0, so dotting a⃗ with the given vector gives a⃗·d⃗ directly.

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