Question 309 of 379MediumMCQ
Let a=2i^+j^−2k^, b=i^+j^. If c is a vector such that a⋅c=∣c∣, ∣c−a∣=22 and the angle between a×b and c is 30∘, then ∣(a×b)×c∣ equals: Solutions · 1
Official solution
∣a∣=3, a×b=2i^−2j^+k^, ∣a×b∣=3.
\[|\vec c-\vec a|^2=8 \implies |\vec c|^2+|\vec a|^2-2\vec c\cdot\vec a=8 \implies |\vec c|^2+9-2|\vec c|=8 \implies (|\vec c|-1)^2=0 \implies |\vec c|=1\]
\[|(\vec a\times\vec b)\times\vec c| = |\vec a\times\vec b||\vec c|\sin30^{\circ} = 3\cdot1\cdot\frac12=\frac32\] Answer the question to unlock