Question

If $\overrightarrow a ,\overrightarrow b ,\overrightarrow c$ are unit vectors such that $\overrightarrow a .\overrightarrow b = 0 = \overrightarrow a .\overrightarrow c$ and the angle between $\overrightarrow b$and $\overrightarrow c$is $\dfrac{\pi }{3}$, then find the value of $\left| {\overrightarrow a \times \overrightarrow b - \overrightarrow a \times \overrightarrow c } \right|$.

Answer 1

In order to find the values of the given terms, first simplify the terms given, for example unit vectors, which means value is one, that is $\overrightarrow {\left| a \right|} ,\left| {\overrightarrow b } \right|,\left| {\overrightarrow c } \right| = 1$, then $\overrightarrow a .\overrightarrow b = 0$, which gives that $\overrightarrow a ,\overrightarrow b$ are perpendicular to each other. Then using the scalar and dot product solves the values.

Formula used:

1. Dot product: $\overrightarrow a .\overrightarrow b = \left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\cos \theta$
2. Cross product: $\overrightarrow a \times \overrightarrow b = \left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\sin \theta$

Complete step by step solution:
We are given with unit vectors $\overrightarrow a ,\overrightarrow b ,\overrightarrow c$, that means:

$$\overrightarrow {\left| a \right|} = 1 \\\ \left| {\overrightarrow b } \right| = 1 \\\ \left| {\overrightarrow c } \right| = 1 \\\$$

Next, given $\overrightarrow a .\overrightarrow b = 0 = \overrightarrow a .\overrightarrow c$. From dot product formula we know that, $\overrightarrow a .\overrightarrow b = \left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\cos \theta$.
Substituting the value of $\overrightarrow {\left| a \right|} = 1,\left| {\overrightarrow b } \right| = 1$ and $\overrightarrow a .\overrightarrow b = 0$ in $\overrightarrow a .\overrightarrow b = \left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\cos \theta$, where $\theta$ is the angle between $\overrightarrow a ,\overrightarrow b$, we get:
$\overrightarrow a .\overrightarrow b = \left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\cos \theta \\\ 0 = 1.1.\cos \theta \\\ \cos \theta = 0 \\\ \theta = \dfrac{\pi }{2} \\\$
That means $\overrightarrow a ,\overrightarrow b$ are perpendicular to each other.
Similarly, for the $\overrightarrow a .\overrightarrow c = 0$.
Substituting the value of $\overrightarrow {\left| a \right|} = 1,\left| {\overrightarrow c } \right| = 1$ and $\overrightarrow a .\overrightarrow c = 0$ in $\overrightarrow a .\overrightarrow c = \left| {\overrightarrow a } \right|\left| {\overrightarrow c } \right|\cos \phi$ where $\phi$ is the angle between $\overrightarrow a ,\overrightarrow c$, we get:
$\overrightarrow a .\overrightarrow c = \left| {\overrightarrow a } \right|\left| {\overrightarrow c } \right|\cos \phi \\\ 0 = 1.1.\cos \phi \\\ \cos \phi = 0 \\\ \cos \phi = \dfrac{\pi }{2} \\\$
Which also depicts that $\overrightarrow a ,\overrightarrow c$ are perpendicular to each other.

We need to find the value of $\left| {\overrightarrow a \times \overrightarrow b - \overrightarrow a \times \overrightarrow c } \right|$.
From cross product, we know that $\overrightarrow a \times \overrightarrow b = \left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\sin \theta$ and $\overrightarrow a \times \overrightarrow c = \left| {\overrightarrow a } \right|\left| {\overrightarrow c } \right|\sin \phi$
Substituting the value of $\overrightarrow {\left| a \right|} = 1,\left| {\overrightarrow b } \right| = 1$ and $\theta = \dfrac{\pi }{2}$ in $\overrightarrow a \times \overrightarrow b = \left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\sin \theta$, we get:

$$\overrightarrow a \times \overrightarrow b = \left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\sin \theta \\\ \overrightarrow a \times \overrightarrow b = 1.1.\sin \dfrac{\pi }{2} \\\ \overrightarrow a \times \overrightarrow b = 1 \\\$$

Similarly, Substituting the value of $\overrightarrow {\left| a \right|} = 1,\left| {\overrightarrow c } \right| = 1$ and $\phi = \dfrac{\pi }{2}$ in $\overrightarrow a \times \overrightarrow c = \left| {\overrightarrow a } \right|\left| {\overrightarrow c } \right|\sin \phi$, we get:

$$\overrightarrow a \times \overrightarrow c = \left| {\overrightarrow a } \right|\left| {\overrightarrow c } \right|\sin \phi \\\ \overrightarrow a \times \overrightarrow c = 1.1.\sin \dfrac{\pi }{2} \\\ \overrightarrow a \times \overrightarrow c = 1 \\\$$

Substituting the above obtained value in $\left| {\overrightarrow a \times \overrightarrow b - \overrightarrow a \times \overrightarrow c } \right|$, we get:
$\left| {\overrightarrow a \times \overrightarrow b - \overrightarrow a \times \overrightarrow c } \right| = \left| {1 - 1} \right| = 0$
Therefore, the value of $\left| {\overrightarrow a \times \overrightarrow b - \overrightarrow a \times \overrightarrow c } \right| = 0$.

Note:

Since, there was no use of the angle between $b$ and $c$, so no need to use any of the steps of the solving process.
We cannot use $\theta$, in every angle as different angles between different vectors is represented uniquely.
Always prefer solving step by step for ease rather than solving at once.