Vectors & 3D GeometrymediumPYQ · JEE Main · 6 Apr 2026 · Shift 1 (Morning)Free

Vector Triple-Product Value = 3 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
If a⃗=i^+j^+k^\vec a=\hat i+\hat j+\hat k, b⃗=j^−k^\vec b=\hat j-\hat k and c⃗\vec c be three vectors such that a⃗×c⃗=b⃗\vec a\times\vec c=\vec b and a⃗⋅c⃗=3\vec a\cdot\vec c=3, then c⃗⋅(a⃗−2b⃗)\vec c\cdot(\vec a-2\vec b) is equal to
Solution
Answer: 3 (± 0.01)
Step 1: From a⃗×c⃗=b⃗\vec a\times\vec c=\vec b, take a⃗×\vec a\times both sides:
a⃗×(a⃗×c⃗)=a⃗×b⃗⇒(a⃗⋅c⃗)a⃗−(a⃗⋅a⃗)c⃗=a⃗×b⃗.\vec a\times(\vec a\times\vec c)=\vec a\times\vec b\Rightarrow(\vec a\cdot\vec c)\vec a-(\vec a\cdot\vec a)\vec c=\vec a\times\vec b.
 Step 2: a⃗=(1,1,1)\vec a=(1,1,1), b⃗=(0,1,−1)\vec b=(0,1,-1):
a⃗×b⃗=∣i^j^k^11101−1∣.\vec a\times\vec b=\begin{vmatrix}\hat i&\hat j&\hat k\\1&1&1\\0&1&-1\end{vmatrix}.
 i^:(1)(−1)−(1)(1)=−2\hat i:(1)(-1)-(1)(1)=-2; j^:−[(1)(−1)−(1)(0)]=1\hat j:-[(1)(-1)-(1)(0)]=1; k^:(1)(1)−(1)(0)=1\hat k:(1)(1)-(1)(0)=1.
⇒a⃗×b⃗=−2i^+j^+k^.\Rightarrow\vec a\times\vec b=-2\hat i+\hat j+\hat k.
 Step 3: a⃗⋅c⃗=3\vec a\cdot\vec c=3, a⃗⋅a⃗=1+1+1=3\vec a\cdot\vec a=1+1+1=3:
3a⃗−3c⃗=−2i^+j^+k^.3\vec a-3\vec c=-2\hat i+\hat j+\hat k.
3c⃗=(3,3,3)−(−2,1,1)=(5,2,2)⇒c⃗=13(5i^+2j^+2k^).3\vec c=(3,3,3)-(-2,1,1)=(5,2,2)\Rightarrow\vec c=\dfrac13(5\hat i+2\hat j+2\hat k).
 Step 4: a⃗−2b⃗=(1,1,1)−2(0,1,−1)=(1,−1,3)\vec a-2\vec b=(1,1,1)-2(0,1,-1)=(1,-1,3).
∴c⃗⋅(a⃗−2b⃗)=13(5,2,2)⋅(1,−1,3)=13(5−2+6)=93=3.\therefore\vec c\cdot(\vec a-2\vec b)=\dfrac13(5,2,2)\cdot(1,-1,3)=\dfrac13(5-2+6)=\dfrac{9}{3}=3.
 Correct answer: 3
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