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Step-by-Step Solution
Step 1: Understand the Given Vectors
We have three vectors:
\overrightarrow{a} = \hat{i} + 5 \hat{j} + \alpha \hat{k} \, ,
\overrightarrow{b} = \hat{i} + 3 \hat{j} + \beta \hat{k} \, ,
\overrightarrow{c} = -\hat{i} + 2 \hat{j} - 3 \hat{k} \, .
The problem gives two conditions:
|\overrightarrow{b} \times \overrightarrow{c}| = 5 \sqrt{3} , which implies |\overrightarrow{b} \times \overrightarrow{c}|^2 = 75 .
\overrightarrow{a} is perpendicular to \overrightarrow{b} , i.e., \overrightarrow{a} \cdot \overrightarrow{b} = 0 .
We wish to find the greatest possible value of |\overrightarrow{a}|^2 .
Step 2: Apply the Perpendicularity Condition
Since \overrightarrow{a} \cdot \overrightarrow{b} = 0 , use the dot product:
(\hat{i} + 5 \hat{j} + \alpha \hat{k}) \cdot (\hat{i} + 3 \hat{j} + \beta \hat{k}) = 0 .
Compute this dot product:
1 \cdot 1 + 5 \cdot 3 + \alpha \cdot \beta = 1 + 15 + \alpha \beta = 0.
Hence,
\alpha \beta = -16 \quad (1)
Step 3: Compute \overrightarrow{b} \times \overrightarrow{c}
We can find the cross product of \overrightarrow{b} and \overrightarrow{c} :
\overrightarrow{b} \times \overrightarrow{c} =
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
1 & 3 & \beta \\
-1 & 2 & -3
\end{vmatrix}.
This determinant expands to:
\overrightarrow{b} \times \overrightarrow{c} =
\bigl(3 \cdot (-3) - \beta \cdot 2 \bigr)\hat{i}
- \bigl(1 \cdot (-3) - \beta \cdot (-1)\bigr)\hat{j}
+ \bigl(1 \cdot 2 - 3 \cdot (-1)\bigr)\hat{k}.
So:
\overrightarrow{b} \times \overrightarrow{c} =
(-9 - 2\beta)\hat{i} + (3 - \beta)\hat{j} + 5\hat{k}.
Step 4: Use the Magnitude Condition |\overrightarrow{b} \times \overrightarrow{c}|^2 = 75
Now we calculate the square of the magnitude:
|\overrightarrow{b} \times \overrightarrow{c}|^2
= (-9 - 2\beta)^2 + (3 - \beta)^2 + 5^2.
Compute each part:
(-9 - 2\beta)^2 = 81 + 36\beta + 4\beta^2,
\quad
(3 - \beta)^2 = 9 - 6\beta + \beta^2,
\quad
5^2 = 25.
Summing these,
(-9 - 2\beta)^2 + (3 - \beta)^2 + 25
= (81 + 36\beta + 4\beta^2) + (9 - 6\beta + \beta^2) + 25
= 81 + 9 + 25 + (36\beta - 6\beta) + (4\beta^2 + \beta^2).
So we get
115 + 30\beta + 5\beta^2.
We know this must be 75 , so:
115 + 30\beta + 5\beta^2 = 75.
Rearrange:
5\beta^2 + 30\beta + 40 = 0.
Divide by 5:
\beta^2 + 6\beta + 8 = 0.
Factor:
(\beta + 2)(\beta + 4) = 0 \implies \beta = -2 \text{ or } \beta = -4.
Step 5: Determine Possible Pairs (\alpha, \beta)
From equation (1) \alpha \beta = -16 :
If \beta = -4 , then \alpha = 4 because 4 \cdot (-4) = -16.
If \beta = -2 , then \alpha = 8 because 8 \cdot (-2) = -16.
Step 6: Calculate |\overrightarrow{a}|^2 for Each Pair
Recall that \overrightarrow{a} = \hat{i} + 5\hat{j} + \alpha \hat{k}.
The magnitude squared is:
|\overrightarrow{a}|^2
= 1^2 + 5^2 + \alpha^2
= 1 + 25 + \alpha^2
= 26 + \alpha^2.
Now check each possible \alpha :
\alpha = 4 \implies |\overrightarrow{a}|^2 = 26 + 4^2 = 26 + 16 = 42.
\alpha = 8 \implies |\overrightarrow{a}|^2 = 26 + 8^2 = 26 + 64 = 90.
Hence, the greatest value among these is
|\overrightarrow{a}|_{\max}^2 = 90.
Final Answer
The greatest possible value of |\overrightarrow{a}|^2 is \boxed{90} .
