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Step-by-Step Solution
Step 1: Identify the Given Vectors
The two vectors defining the parallelogram are:
\vec{A} = 5\,\vec{p} + 2\,\vec{q}
\vec{B} = \vec{p} - 3\,\vec{q}
We also have:
|\vec{p}| = 2
|\vec{q}| = 5
The angle between \vec{p} and \vec{q} is \frac{\pi}{3} .
Step 2: Write Down the Parallelogram Diagonals
For a parallelogram formed by vectors \vec{A} and \vec{B} , the diagonals are:
\vec{A} + \vec{B}
\vec{A} - \vec{B}
Step 3: Express Each Diagonal in Terms of \vec{p} and \vec{q}
\vec{A} + \vec{B} \;=\; (5\,\vec{p} + 2\,\vec{q}) + (\vec{p} - 3\,\vec{q}) \;=\; 6\,\vec{p} - \vec{q}.
\vec{A} - \vec{B} \;=\; (5\,\vec{p} + 2\,\vec{q}) - (\vec{p} - 3\,\vec{q}) \;=\; 4\,\vec{p} + 5\,\vec{q}.
Step 4: Compute the Dot Product \vec{p} \cdot \vec{q}
By definition:
\vec{p} \cdot \vec{q} \;=\; |\vec{p}|\;|\vec{q}|\;\cos\left(\frac{\pi}{3}\right)
\;=\; 2 \times 5 \times \cos\left(\frac{\pi}{3}\right)
\;=\; 10 \times \frac{1}{2}
\;=\; 5.
Step 5: Find the Magnitude of Each Diagonal
We use the formula |\alpha\,\vec{p} + \beta\,\vec{q}|^2 = \alpha^2\,|\vec{p}|^2 + \beta^2\,|\vec{q}|^2 + 2\,\alpha\,\beta\;(\vec{p} \cdot \vec{q}) , noting a sign change if we have subtraction.
(a) Magnitude of \vec{A} + \vec{B} = 6\,\vec{p} - \vec{q}
|6\,\vec{p} - \vec{q}|^2
= 36\,|\vec{p}|^2 + |\vec{q}|^2 - 2 \times 6 \times 1 \times (\vec{p} \cdot \vec{q}).
Substitute known values:
|\vec{p}|^2 = 2^2 = 4
|\vec{q}|^2 = 5^2 = 25
\vec{p} \cdot \vec{q} = 5
Therefore:
|6\,\vec{p} - \vec{q}|^2
= 36 \times 4 + 25 - 12 \times 5
= 144 + 25 - 60
= 109.
Hence, |6\,\vec{p} - \vec{q}| = \sqrt{109}.
(b) Magnitude of \vec{A} - \vec{B} = 4\,\vec{p} + 5\,\vec{q}
|4\,\vec{p} + 5\,\vec{q}|^2
= 16\,|\vec{p}|^2 + 25\,|\vec{q}|^2 + 2 \times 4 \times 5 \times (\vec{p} \cdot \vec{q}).
Substitute:
16 \times 4 = 64
25 \times 25 = 625
2 \times 4 \times 5 \times 5 = 200
Hence,
|4\,\vec{p} + 5\,\vec{q}|^2
= 64 + 625 + 200
= 889.
So |4\,\vec{p} + 5\,\vec{q}| = \sqrt{889}.
Step 6: Identify the Smaller Diagonal
We now see:
|\vec{A} + \vec{B}| = \sqrt{109}
|\vec{A} - \vec{B}| = \sqrt{889}
Since \sqrt{109} < \sqrt{889} , the smaller diagonal has length \sqrt{109} .
Step 7: Determine k^2
Given that the smaller diagonal has length k , we have:
k = \sqrt{109}
\quad\Longrightarrow\quad
k^2 = 109.
Answer: k^2 = 109.
