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 \[ \alpha-2\beta+3\gamma= \]

Question

A vector parallel to the line of intersection of the planes \[ \overrightarrow{r} \cdot (3\hat{i} - \hat{j} + \hat{k}) = 1 \quad \text{and} \quad \overrightarrow{r} \cdot (\hat{i} + 4\hat{j} - 2\hat{k}) = 2 \] is:

Answer 1

Step 1: Identify the perpendicular direction.
A vector perpendicular to the plane of two vectors is their cross product, $\vec n=(2\vec a+\vec b)\times(\vec b-\vec a)$.
Step 2: Build the two plane vectors.
$2\vec a+\vec b=2(1,-2,2)+(2,3,-6)=(4,-1,-2)$ and $\vec b-\vec a=(2-1,3+2,-6-2)=(1,5,-8)$.
Step 3: Take the cross product.
$\vec n=\begin{vmatrix} i & j & k\\ 4 & -1 & -2\\ 1 & 5 & -8 \end{vmatrix}$. The $i$ part is $(-1)(-8)-(-2)(5)=8+10=18$; the $j$ part is $-\big(4(-8)-(-2)(1)\big)=-(-32+2)=30$; the $k$ part is $4(5)-(-1)(1)=20+1=21$. So $\vec n=(18,30,21)$.
Step 4: Scale to meet the sum condition.
Let $(\alpha,\beta,\gamma)=k(18,30,21)$. Then $\alpha+\beta+\gamma=k(18+30+21)=69k=46$, so $k=\frac{2}{3}$.
Step 5: Find the components.
$\alpha=\frac23\cdot 18=12$, $\beta=\frac23\cdot 30=20$, $\gamma=\frac23\cdot 21=14$.
Step 6: Evaluate the target expression.
$\alpha-2\beta+3\gamma=12-40+42=14$.
\[ \boxed{14} \]

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 \[ \alpha-2\beta+3\gamma= \]

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The Correct Option is B

Solution and Explanation

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