Question 247 of 379MediumMCQ
Let a=2i^+j^−2k^, b=i^+j^. If c satisfies a⋅c=∣c∣, ∣c−a∣=22, and the angle between a×b and c is 6π, then ∣(a×b)×c∣ is: Solutions · 1
Official solution
∣a∣=3. Let c=∣c∣: a⋅c=c. From ∣c−a∣2=8: c2+9−2c=8⇒(c−1)2=0⇒c=1.
\[\vec a\times\vec b = \begin{vmatrix}\hat i&\hat j&\hat k\\2&1&-2\\1&1&0\end{vmatrix} = (2,-2,1),\qquad |\vec a\times\vec b|=3\]
\[|(\vec a\times\vec b)\times\vec c| = |\vec a\times\vec b|\,|\vec c|\sin\frac\pi6 = 3\times1\times\frac12=\frac32\] Answer the question to unlock