Given: p×b=c×b
⇒(p−c)×b=0
⇒(p−c)∣∣b
⇒(p−c)=kb
⇒p=kb+c
⇒p=(4k+1)i^+(k−3)j^+(7k+4)k^
Now, p.a=0
⇒3(4k+1)+(k−3)−2(7k+4)=0
⇒k=−8
⇒p=−31i^−11j^−52k^
⇒p.(i^−j^−k^)=−31+11+52
⇒p.(i^−j^−k^)=32
JEE Main 2024 — Mathematics Vectors & 3D Geometry
Let a=3i^+j^−2k^,b=4i^+j^+7k^ and c=i^−3j^+4k^ be three vectors. If a vectors p satisfies p×b=c×b and p⋅a=0, then p⋅(i^−j^−k^) is equal to
Held on 31 Jan 2024 · Verified 6 Jul 2026.
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