Name: Cross Product
Uploaded: 2019-07-19
Duration: 4 min 16 s
Description: The cross product a × b is defined as a vector c that is perpendicular (orthogonal) to both a and b, with a particular direction and magnitude. Watch this vi...

Question

Find a unit vector perpendicular to each of the vectors

(\vec{a} + \vec{b})

and

(\vec{a} - \vec{b})

, where

\vec{a} = \hat{i} + \hat{j} + \hat{k}

and

\vec{b} = \hat{i} + 2 \hat{j} + 3 \hat{k}

Accepted Answer

The given two vectors are $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2 \hat{j} + 3 \hat{k}$ .

Now we need to find $(\vec{a} + \vec{b})$ and $(\vec{a} - \vec{b})$ .

So,

$(\vec{a} + \vec{b}) = (\hat{i} + \hat{j} + \hat{k}) + (\hat{i} + 2 \hat{j} + 3 \hat{k})$

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

And,

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

$(\vec{a} - \vec{b}) = - \hat{j} - 2 \hat{k}$

Now the cross product of these vectors will be

$(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b}) = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 0 & - 1 & - 2 \end{matrix}|$

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

And the corresponding magnitude will be

$|(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b})| = \sqrt{(- 2)^{2} + (4)^{2} + (- 2)^{2}}$

$\Rightarrow |(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b})| = \sqrt{4 + 16 + 4}$

$\Rightarrow |(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b})| = \sqrt{24}$

$\Rightarrow |(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b})| = \sqrt{6 \times 4}$

$\Rightarrow |(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b})| = 2 \sqrt{6}$

Now we know vector the product $(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b})$ will be perpendicular to the plane of $(\vec{a} + \vec{b})$ and $(\vec{a} - \vec{b})$ .

So the unit vector perpendicular to the plane of $(\vec{a} + \vec{b})$ and $(\vec{a} - \vec{b})$ is given as

$= \frac{(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b})}{|(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b})|}$

Now substituting the corresponding value, we get

$= \frac{- 2 \hat{i} + 4 \hat{j} - 2 \hat{k}}{2 \sqrt{6}}$

$= \frac{- \hat{i} + 2 \hat{j} - \hat{k}}{\sqrt{6}}$

Or, the unit vector can be written as

$= - \frac{1}{\sqrt{6}} \hat{i} + \frac{2}{\sqrt{6}} \hat{j} - \frac{1}{\sqrt{6}} \hat{k} .$

Find a unit vector perpendicular to each of the vectors $(\vec{a} + \vec{b})$ and $(\vec{a} - \vec{b})$ , where $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2 \hat{j} + 3 \hat{k}$

Important Questions on Vector Algebra

Learn and Prepare for any exam you want !

Find a unit vector perpendicular to each of the vectors a→+b→ and a→-b→, where a→=i^+j^+k^ and b→=i^+2j^+3k^

Important Questions on Vector Algebra

Learn and Prepare for any exam you want !

Find a unit vector perpendicular to each of the vectors $(\vec{a} + \vec{b})$ and $(\vec{a} - \vec{b})$ , where $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2 \hat{j} + 3 \hat{k}$