For three vectors A⃗=(-xî-6ĵ-2k̂),B⃗=(-î+4ĵ+3k̂) and C⃗=(-8î-ĵ+3k̂), if A⃗· (B⃗× C⃗)=0, then value of x is ________.

Solution

Physics sir · 4 steps

Find B⃗ × C⃗ first, then dot it with A⃗ and set the result to zero.

1. Cross product first. Expand the determinant with rows î, ĵ, k̂, then B⃗, then C⃗. The middle (ĵ) term always carries a minus sign.

   B⃗ × C⃗ = î(4 × 3 - 3 × (-1)) - ĵ((-1) × 3 - 3 × (-8)) + k̂((-1)(-1) - 4 × (-8))

2. Simplify. The brackets are 15, 21 and 33. So B⃗ × C⃗ = 15î - 21ĵ + 33k̂.
3. Dot with A. A⃗ · (B⃗ × C⃗) = (-x)(15) + (-6)(-21) + (-2)(33) = -15x + 126 - 66 = 60 - 15x.
4. So. Setting 60 - 15x = 0 gives x = 4.

The common slip: -12.8. Dropping the minus sign on the ĵ term makes it +21ĵ, which gives -15x - 192 = 0 and x = -12.8.

The idea to keep. A scalar triple product is a 3 × 3 determinant of the three vectors. When it is zero, the vectors are coplanar. Always put the minus sign on the middle term.

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