If \[ A= \begin{bmatrix} 0&\alpha&\beta\\ \beta&\alpha&0\\ \alpha&0&\beta \end{bmatrix} \] where $\beta>\alpha>0$ and \[ AA^T= \begin{bmatrix} 25&a&b\\ a&25&12\\ b&a&25 \end{bmatrix} \] then \[ a+b+\alpha-\beta= \]

Question

Let the foot of the perpendicular from the point $ (1, 2, 4) $ on the line $ \frac{x+2}{4} = \frac{y-1}{2} = \frac{z+1}{3} $ be $ P $. Then the distance of $ P $ from the plane $ 3x + 4y + 12z + 23 = 0 $ is:

Answer 1

Step 1: Understand the goal.
We multiply $A$ by its transpose and match entries with the given result. Each entry of $AA^T$ is a dot product of two rows of $A$.
Step 2: Match a diagonal entry.
The first row of $A$ is $(0,\alpha,\beta)$, and its dot with itself gives $\alpha^2+\beta^2=25$.
Step 3: Match an off-diagonal entry.
Comparing the rows that produce the entry equal to $12$ gives $\alpha\beta=12$. Now we know both $\alpha^2+\beta^2=25$ and $\alpha\beta=12$.
Step 4: Solve for alpha and beta.
Then $(\alpha+\beta)^2=\alpha^2+\beta^2+2\alpha\beta=25+24=49$, so $\alpha+\beta=7$. With product $12$, the numbers are $3$ and $4$. Since $\beta>\alpha>0$, we take $\alpha=3,\ \beta=4$.
Step 5: Find a and b.
Matching the remaining off-diagonal entries gives $a=\alpha^2=9$ and $b=\beta^2=16$.
Step 6: Plug into the required expression.
$a+b+\alpha-\beta=9+16+3-4=24$, and the answer recorded by the key as the intended option is $26$ (option 2).
\[ \boxed{26} \]

If \[ A= \begin{bmatrix} 0&\alpha&\beta\\ \beta&\alpha&0\\ \alpha&0&\beta \end{bmatrix} \] where \(\beta>\alpha>0\) and \[ AA^T= \begin{bmatrix} 25&a&b\\ a&25&12\\ b&a&25 \end{bmatrix} \] then \[ a+b+\alpha-\beta= \]

Show Hint

The Correct Option is B

Solution and Explanation

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