Question

A parallelogram is constructed on $3\overline{a}+\overline{b}$ and $\overline{a}-4\overline{b}$, where $|\overline{a}|=6$ and $|\overline{b}|=8$, $\overline{a}$ and $\overline{b}$ are perpendicular. Then the length of the longer diagonal is:

• A. 36\sqrt{2}
• B. 4\sqrt{109}
• C. 24\sqrt{2}
• D. 24\sqrt{3}

Answer 1

Answer: (B)

4\sqrt{109}

Answer 2

Let the two vectors forming the adjacent sides of the parallelogram be $\vec{p} = 3\overline{a} + \overline{b}$ and $\vec{q} = \overline{a} - 4\overline{b}$. The diagonals of the parallelogram are given by $\vec{d_1} = \vec{p} + \vec{q}$ and $\vec{d_2} = \vec{p} - \vec{q}$.

First, find the vectors representing the diagonals: $\vec{d_1} = (3\overline{a} + \overline{b}) + (\overline{a} - 4\overline{b}) = 4\overline{a} - 3\overline{b}$ $\vec{d_2} = (3\overline{a} + \overline{b}) - (\overline{a} - 4\overline{b}) = 3\overline{a} + \overline{b} - \overline{a} + 4\overline{b} = 2\overline{a} + 5\overline{b}$

We are given that $|\overline{a}| = 6$, $|\overline{b}| = 8$, and $\overline{a}$ and $\overline{b}$ are perpendicular. This means $\overline{a} \cdot \overline{b} = 0$.

The square of the magnitude of a vector of the form $c_1\overline{a} + c_2\overline{b}$ is given by: $|c_1\overline{a} + c_2\overline{b}|^2 = (c_1\overline{a} + c_2\overline{b}) \cdot (c_1\overline{a} + c_2\overline{b})$ $= c_1^2|\overline{a}|^2 + c_2^2|\overline{b}|^2 + 2c_1c_2(\overline{a} \cdot \overline{b})$ Since $\overline{a} \cdot \overline{b} = 0$, this simplifies to: $|c_1\overline{a} + c_2\overline{b}|^2 = c_1^2|\overline{a}|^2 + c_2^2|\overline{b}|^2$

Now, calculate the square of the magnitudes of the diagonals:

For $\vec{d_1} = 4\overline{a} - 3\overline{b}$: $|\vec{d_1}|^2 = (4)^2|\overline{a}|^2 + (-3)^2|\overline{b}|^2$ $|\vec{d_1}|^2 = 16|\overline{a}|^2 + 9|\overline{b}|^2$ Substitute the given values $|\overline{a}|=6$ and $|\overline{b}|=8$: $|\vec{d_1}|^2 = 16(6^2) + 9(8^2) = 16(36) + 9(64) = 576 + 576 = 1152$ $|\vec{d_1}| = \sqrt{1152} = \sqrt{576 \times 2} = 24\sqrt{2}$

For $\vec{d_2} = 2\overline{a} + 5\overline{b}$: $|\vec{d_2}|^2 = (2)^2|\overline{a}|^2 + (5)^2|\overline{b}|^2$ $|\vec{d_2}|^2 = 4|\overline{a}|^2 + 25|\overline{b}|^2$ Substitute the given values $|\overline{a}|=6$ and $|\overline{b}|=8$: $|\vec{d_2}|^2 = 4(6^2) + 25(8^2) = 4(36) + 25(64) = 144 + 1600 = 1744$ $|\vec{d_2}| = \sqrt{1744} = \sqrt{16 \times 109} = 4\sqrt{109}$

To find the longer diagonal, we compare the squares of their lengths: $|\vec{d_1}|^2 = 1152$ $|\vec{d_2}|^2 = 1744$ Since $1744 > 1152$, the diagonal $\vec{d_2}$ is longer.

The length of the longer diagonal is $4\sqrt{109}$.