Question

If $\overrightarrow{a}$ and $\overrightarrow{b}$ are two, unit vectors such that $\overrightarrow{a}+(\overrightarrow{a}\times \overrightarrow{b})=\overrightarrow{c}$, such that $\left|\overrightarrow{c}\right|=2$, then find the value of $\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]$ is
(a) 0
(b) $\pm 1$
(c) 3
(d) -3

Answer 1

Hint: Using the given equation, we must find the values of $(\overrightarrow{a}\times \overrightarrow{b})$ and $\overrightarrow{a}\cdot \overrightarrow{c}$. Then, with the help of these values, and the expansion of scalar triple product as $(\overrightarrow{a}\times \overrightarrow{b})\cdot \overrightarrow{c}$, we can find the value of this triple product $\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]$.

Complete step-by-step solution:
Here, we are given that $\overrightarrow{a}+(\overrightarrow{a}\times \overrightarrow{b})=\overrightarrow{c}$.
Let us subtract $\overrightarrow{a}$ from both sides of the above equation. Hence, we write
$\overrightarrow{a}+(\overrightarrow{a}\times \overrightarrow{b})-\overrightarrow{a}=\overrightarrow{c}-\overrightarrow{a}$.
Thus, we can also write the above equation as $(\overrightarrow{a}\times \overrightarrow{b})=\overrightarrow{c}-\overrightarrow{a}...\left(i\right)$
We need to find the value of $\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]$. We know that $\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]$ is the scalar triple product of $\overrightarrow{a}$, $\overrightarrow{b}$ and $\overrightarrow{c}$, and this scalar triple product is defined as $\overrightarrow{a}\cdot (\overrightarrow{b}\times \overrightarrow{c})$ or $(\overrightarrow{a}\times \overrightarrow{b})\cdot \overrightarrow{c}$.
Thus, we can write this mathematically, as
$\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]=(\overrightarrow{a}\times \overrightarrow{b})\cdot \overrightarrow{c}$
Using the value of $(\overrightarrow{a}\times \overrightarrow{b})$ from equation (i), we can write,
$\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]=(\overrightarrow{c}-\overrightarrow{a})\cdot \overrightarrow{c}$
We know that the dot product is distributive. Hence, using the distributive property, we can write
$\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]=\overrightarrow{c}\cdot \overrightarrow{c}-\overrightarrow{a}\cdot \overrightarrow{c}$
Thus, we have
$\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]={\left|\overrightarrow{c}\right|}^2-\overrightarrow{a}\cdot \overrightarrow{c}...\left(ii\right)$
Now, we need to find the value of $\overrightarrow{a}\cdot \overrightarrow{c}$.
We are given that $\overrightarrow{a}+(\overrightarrow{a}\times \overrightarrow{b})=\overrightarrow{c}$. Hence, we can also write
$\overrightarrow{a}\cdot \{\overrightarrow{a}+(\overrightarrow{a}\times \overrightarrow{b})\}=\overrightarrow{a}\cdot \overrightarrow{c}$
Again, using the distributive property, we can write
$\overrightarrow{a}\cdot \overrightarrow{a}+\overrightarrow{a}\cdot (\overrightarrow{a}\times \overrightarrow{b})=\overrightarrow{a}\cdot \overrightarrow{c}$
Thus, we have
${\left|\overrightarrow{a}\right|}^2+\left[\overrightarrow{a}\overrightarrow{a}\overrightarrow{b}\right]=\overrightarrow{a}\cdot \overrightarrow{c}$
We know that if any two vectors in the scalar triple product are the same, then its value becomes 0. Thus, we have
$1+0=\overrightarrow{a}\cdot \overrightarrow{c}$
Hence, $\overrightarrow{a}\cdot \overrightarrow{c}=1$.
Using the above value in equation (ii), we get
$\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]={\left(2\right)}^2-1$
And so, $\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]=3$.
Hence, option (c) is the correct answer.

Note: We can see that $\left[\overrightarrow{a}\overrightarrow{a}\overrightarrow{b}\right]$ can be expressed as $\left[\overrightarrow{a}\overrightarrow{a}\overrightarrow{c}\right]=(\overrightarrow{a}\times \overrightarrow{a})\cdot \overrightarrow{c}$, and since $(\overrightarrow{a}\times \overrightarrow{a})=0$, we can write $\left[\overrightarrow{a}\overrightarrow{a}\overrightarrow{b}\right]=0$. We must, also, remember that the scalar triple product $\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]$ can be expressed in multiple forms, like $\left[\overrightarrow{b}\overrightarrow{c}\overrightarrow{a}\right]$ and $\left[\overrightarrow{c}\overrightarrow{a}\overrightarrow{b}\right]$.