Let a=i^+2j^+3k^, b=2i^−j^+k^, and c=i^−2j^+4k^. Find a vector r that is perpendicular to both a and b, has magnitude 53, and makes an obtuse angle with c. This kind of problem appeared in JEE Main 2023 with slightly different numbers.
Solution — Step by Step
A vector perpendicular to a and b is parallel to a×b:
The cross product a×b gives a vector perpendicular to both — that’s its defining property. There are two such directions (positive and negative), and the additional constraint (here, the obtuse angle) pins down which sign to take.
The magnitude ∣a×b∣ also has geometric meaning: it’s the area of the parallelogram spanned by a and b. In this problem, ∣a×b∣=53, which happens to be the magnitude we want — coincidence-friendly setup.
Alternative Method
You can solve via simultaneous equations: write r=xi^+yj^+zk^ and impose r⋅a=0, r⋅b=0, and ∣r∣=53. Three equations, three unknowns. The cross product is just a quicker way to encode the first two perpendicularity conditions in one shot.
Whenever a problem says “perpendicular to two vectors”, reach for the cross product immediately. It’s faster than solving two scalar equations and gives you direction directly. Just remember to check the sign with any further constraint (obtuse/acute angle, specific component positive, etc.).
Common Mistake
The most common error is computing the cross product with a sign mistake in the determinant expansion. The cofactor expansion has alternating signs (+,−,+), and forgetting the minus sign on the j^ term flips one component. Always double-check by computing r⋅a and r⋅b — both should be exactly zero.
The other slip: assuming the cross product vector has the same magnitude as the answer wanted. Almost never true. Compute the magnitude, then scale.
Final answer:r=5i^+5j^−5k^.
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