Question

Let \[ \vec a=\vec i-2\vec j+2\vec k \] and \[ \vec b=2\vec i+3\vec j-6\vec k \] If \[ \alpha\vec i+\beta\vec j+\gamma\vec k \] is perpendicular to plane of \[ 2\vec a+\vec b \] and \[ \vec b-\vec a \] such that \[ \alpha+\beta+\gamma=46 \] then \[ \alpha-2\beta+3\gamma= \]

• A. 12
• B. 14
• C. 0
• D. 1

Hint:

If a vector is perpendicular to a plane formed by two vectors, immediately think cross product.

Answer 1

The Correct Option is B

Concept: A vector perpendicular to plane formed by two vectors equals their cross product. \[ \vec n= (2\vec a+\vec b)\times(\vec b-\vec a) \]

Step 1: Calculate vectors.
\[ 2\vec a+\vec b = (4,-1,-2) \] \[ \vec b-\vec a = (1,5,-8) \]

Step 2: Cross product.
\[ \vec n= \begin{vmatrix} i&j&k\\ 4&-1&-2\\ 1&5&-8 \end{vmatrix} \] \[ =(18,30,21) \] Required vector proportional: \[ (\alpha,\beta,\gamma)=k(18,30,21) \]

Step 3: Find constant.
\[ 18k+30k+21k=46 \] \[ 69k=46 \] \[ k=\frac23 \] Hence \[ \alpha=12,\qquad\beta=20,\qquad\gamma=14 \]

Step 4: Required expression.
\[ \alpha-2\beta+3\gamma \] \[ =12-40+42 \] \[ =14 \] Therefore \[ \boxed{14} \]