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

Last updated date: 02nd Aug 2024
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Hint: Using the given equation, we must find the values of $\left( \overrightarrow{a}\times \overrightarrow{b} \right)$ and $\overrightarrow{a}\cdot \overrightarrow{c}$. Then, with the help of these values, and the expansion of scalar triple product as $\left( \overrightarrow{a}\times \overrightarrow{b} \right)\cdot \overrightarrow{c}$, we can find the value of this triple product $\left[ \overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c} \right]$.

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

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