Question

If $\vec a,\vec b,\vec c$ are $3$ vectors such that \[ \vec b=2\hat i-\hat j,\qquad \vec c=\hat j+2\hat k, \] \[ |\vec a+\vec b|=3,\qquad |\vec a\times(\vec b\times\vec c)|=3\sqrt2 \] and \[ (\vec a,\vec b\times\vec c)=\frac{\pi}{3}, \] then $\vec a\cdot\vec b=$

• A. $\dfrac{3}{2}$
• B. $\sqrt6$
• C. $\dfrac{\sqrt3}{\sqrt2}$
• D. $6$

Hint:

Use $|\vec a+\vec b|^2=|\vec a|^2+|\vec b|^2+2\vec a\cdot\vec b$ whenever a dot product is required.

Answer 1

The Correct Option is A

Step 1: Concept
Use vector triple product and angle relations.

Step 2: Meaning
First compute \[ \vec b\times\vec c= \begin{vmatrix} \hat i&\hat j&\hat k\\ 2&-1&0\\ 0&1&2 \end{vmatrix} = -2\hat i-4\hat j+2\hat k. \] Hence \[ |\vec b\times\vec c| = \sqrt{4+16+4} = 2\sqrt6. \]

Step 3: Analysis
Since \[ |\vec a\times(\vec b\times\vec c)| = |\vec a||\vec b\times\vec c| \sin\frac{\pi}{3}, \] \[ 3\sqrt2 = |\vec a|(2\sqrt6)\left(\frac{\sqrt3}{2}\right). \] Therefore \[ |\vec a|=\sqrt2. \] Also \[ |\vec b|=\sqrt5. \] Using \[ |\vec a+\vec b|^2 = |\vec a|^2+|\vec b|^2+2\vec a\cdot\vec b, \] \[ 9=2+5+2\vec a\cdot\vec b. \] Thus \[ 2\vec a\cdot\vec b=2 \] and \[ \vec a\cdot\vec b=\frac32. \]

Step 4: Conclusion
Therefore, \[ \vec a\cdot\vec b=\frac32. \]

Final Answer: (A)