Name: Scalar Triple Product
Uploaded: 2019-07-19
Duration: 6 min 56 s
Description: Have you ever heard about scalar triple products? They involve both dot and cross products! Interesting, aren't they? Watch this amazing video to learn more...

Question

The two adjacent sides of a parallelogram are represented by the vectors $2 \hat{i} - 4 \hat{j} - 5 \hat{k}$ and $2 \hat{i} + 2 \hat{j} + 3 \hat{k}$ . Find the two unit vectors parallel to its diagonals. Using the diagonal vectors, find the area of the parallelogram.

Accepted Answer

Given $2 \hat{i} - 4 \hat{j} - 5 \hat{k}$ and $\vec{b} = 2 \hat{i} + 2 \hat{j} + 3 \hat{k}$

As we know, the two diagonals of a parallelogram are represented by $(\vec{a} + \vec{b})$ and $(\vec{b} - \vec{a})$ .

$\Rightarrow (\vec{a} + \vec{b}) = (2 \hat{i} - 4 \hat{j} - 5 \hat{k}) + (2 \hat{i} + 2 \hat{j} + 3 \hat{k}) = 4 \hat{i} - 2 \hat{j} - 2 \hat{k}$ and

$\Rightarrow (\vec{b} - \vec{a}) = (2 \hat{i} + 2 \hat{j} + 3 \hat{k}) - (2 \hat{i} - 4 \hat{j} - 5 \hat{k}) = 6 \hat{j} + 8 \hat{k}$ .

Now,

$\Rightarrow \hat{n_{1}} (unit vector parallel to \vec{a} + \vec{b})$

$\Rightarrow \frac{\vec{a} + \vec{b}}{|\vec{a} + \vec{b}|} = \frac{4 \hat{i} - 2 \hat{j} - 2 \hat{k}}{\sqrt{{(4)}^{2} + {(- 2)}^{2} + {(- 2)}^{2}}} = \frac{4 \hat{i} - 2 \hat{j} - 2 \hat{k}}{2 \sqrt{6}} = \frac{1}{\sqrt{6}} (2 \hat{i} - \hat{j} - \hat{k})$ and

$\Rightarrow \hat{n_{2}} (unit vector parallel to \vec{b} - \vec{a})$

$\Rightarrow \frac{\vec{b} - \vec{a}}{|\vec{b} - \vec{a}|} = \frac{6 \hat{j} + 8 \hat{k}}{\sqrt{{(6)}^{2} + {(8)}^{2}}} = \frac{6 \hat{j} + 8 \hat{k}}{10} = \frac{1}{5} (3 \hat{j} + 4 \hat{k})$

Area of the parallelogram whose diagnols are $(\vec{a} + \vec{b})$ and $(\vec{b} - \vec{a})$ $= \frac{1}{2} |(\vec{a} + \vec{b}) \times (\vec{b} - \vec{a})|$

$\Rightarrow (\vec{a} + \vec{b}) \times (\vec{b} - \vec{a}) = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & - 2 & - 2 \\ 0 & 6 & 8 \end{matrix}| = (- 16 + 12) \hat{i} - (32 - 0) \hat{j} + (24 - 0) \hat{k} = - 4 \hat{i} - 32 \hat{j} + 24 \hat{k}$

$∴$ Magnitude of $(\vec{a} + \vec{b}) \times (\vec{b} - \vec{a}) = |(\vec{a} + \vec{b}) \times (\vec{b} - \vec{a})| = \sqrt{{(- 4)}^{2} + {(- 32)}^{2} + {(24)}^{2}} = \sqrt{16 + 1024 + 576} = \sqrt{1616} = 4 \sqrt{101}$

$\Rightarrow$ Area of parallelogram whose diagonals are $(\vec{a} + \vec{b})$ and $(\vec{b} - \vec{a}) = \frac{1}{2} (4 \sqrt{101}) = 2 \sqrt{101}$ unit.

The two adjacent sides of a parallelogram are represented by the vectors $2 \hat{i} - 4 \hat{j} - 5 \hat{k}$ and $2 \hat{i} + 2 \hat{j} + 3 \hat{k}$ . Find the two unit vectors parallel to its diagonals. Using the diagonal vectors, find the area of the parallelogram.

Important Questions on Vectors

Learn and Prepare for any exam you want !

The two adjacent sides of a parallelogram are represented by the vectors 2i^-4j^-5k^ and 2i^+2j^+3k^. Find the two unit vectors parallel to its diagonals. Using the diagonal vectors, find the area of the parallelogram.

Important Questions on Vectors

Learn and Prepare for any exam you want !

The two adjacent sides of a parallelogram are represented by the vectors $2 \hat{i} - 4 \hat{j} - 5 \hat{k}$ and $2 \hat{i} + 2 \hat{j} + 3 \hat{k}$ . Find the two unit vectors parallel to its diagonals. Using the diagonal vectors, find the area of the parallelogram.