If $\widehat{a}$ and $\widehat{b}$ are orthogonal unit vectors then for any non-zero vector $\overrightarrow{r}$, the vector $\left( \overrightarrow{r}\times \widehat{a} \right)$ equals:
(a) $\left[ \begin{matrix}
\overrightarrow{r} & \widehat{a} & \widehat{b} \\
\end{matrix} \right]\left( \widehat{a}\times \widehat{b} \right)$
(b) $\left[ \begin{matrix}
\overrightarrow{r} & \widehat{a} & \widehat{b} \\
\end{matrix} \right]\widehat{a}+\left( \overrightarrow{r}\centerdot \widehat{a} \right)\left( \widehat{a}\times \widehat{b} \right)$
(c) $\left[ \begin{matrix}
\overrightarrow{r} & \widehat{a} & \widehat{b} \\
\end{matrix} \right]\widehat{b}+\left( \overrightarrow{r}\centerdot \widehat{b} \right)\left( \widehat{b}\times \widehat{a} \right)$
(d) $\left[ \begin{matrix}
\overrightarrow{r} & \widehat{a} & \widehat{b} \\
\end{matrix} \right]\widehat{b}+\left( \overrightarrow{r}\centerdot \widehat{a} \right)\left( \widehat{a}\times \widehat{b} \right)$
Answer
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Hint: We start solving the problem by finding the vector perpendicular to the vectors $\widehat{a}$ and $\widehat{b}$ by using the fact that $\widehat{a}\times \widehat{b}$ is perpendicular to the vectors $\widehat{a}$ and $\widehat{b}$. We then find use the fact that any vector in the universe can be written in terms of three mutually orthogonal unit vectors to assume the vector $\overrightarrow{r}$. We then first calculate $\left( \overrightarrow{r}\times \widehat{a} \right)$, then $\left[ \begin{matrix}
\overrightarrow{r} & \widehat{a} & \widehat{b} \\
\end{matrix} \right]$ and $\overrightarrow{r}\centerdot \widehat{b}$. We then make necessary arrangements in $\left( \overrightarrow{r}\times \widehat{a} \right)$ to get the final answer.
Complete step-by-step solution
According to the problem, we are given that $\widehat{a}$ and $\widehat{b}$ are orthogonal unit vectors. We need to find which option equals $\left( \overrightarrow{r}\times \widehat{a} \right)$ for any non-zero vector $\overrightarrow{r}$.
We know that the cross product of any two given vectors is perpendicular to those two given vectors.
So, $\widehat{a}\times \widehat{b}$ is perpendicular to the vectors $\widehat{a}$ and $\widehat{b}$. Let us find the magnitude of the vector $\widehat{a}\times \widehat{b}$.
For two vectors $\overrightarrow{p}$ and $\overrightarrow{q}$ we know that $\left| \overrightarrow{p}\times \overrightarrow{q} \right|=\left| \overrightarrow{p} \right|\left| \overrightarrow{q} \right|\sin \alpha $, where $\alpha $ is the angle between the vectors $\overrightarrow{p}$ and $\overrightarrow{q}$.
So, we have $\left| \widehat{a}\times \widehat{b} \right|=\left| \widehat{a} \right|\left| \widehat{b} \right|\sin \left( {{90}^{\circ }} \right)$.
$\Rightarrow \left| \widehat{a}\times \widehat{b} \right|=1\times 1\times 1=1$.
So, we have three mutually orthogonal unit vectors $\widehat{a}$, $\widehat{b}$ and $\widehat{a}\times \widehat{b}$.
We know that any vector in the universe can be written in terms of three mutually orthogonal unit vectors.
So, let us assume $\overrightarrow{r}=x\widehat{a}+y\widehat{b}+z\left( \widehat{a}\times \widehat{b} \right)$ ---(1).
Now, let us find $\left( \overrightarrow{r}\times \widehat{a} \right)$.
So, we have $\overrightarrow{r}\times \widehat{a}=\left( x\widehat{a}+y\widehat{b}+z\left( \widehat{a}\times \widehat{b} \right) \right)\times \widehat{a}$.
$\Rightarrow \overrightarrow{r}\times \widehat{a}=x\left( \widehat{a}\times \widehat{a} \right)+y\left( \widehat{b}\times \widehat{a} \right)+z\left( \widehat{a}\times \widehat{b} \right)\times \widehat{a}$.
We know that for a vector $\overrightarrow{q}$, the cross product $\overrightarrow{q}\times \overrightarrow{q}=\overrightarrow{0}$. We also know that the vector triple product of $\overrightarrow{p}$, $\overrightarrow{q}$ and $\overrightarrow{s}$ is defined as $\left( \overrightarrow{p}\times \overrightarrow{q} \right)\times \overrightarrow{s}=\left( \overrightarrow{p}\centerdot \overrightarrow{s} \right)\overrightarrow{q}-\left( \overrightarrow{q}\centerdot \overrightarrow{s} \right)\overrightarrow{p}$.
$\Rightarrow \overrightarrow{r}\times \widehat{a}=y\left( \widehat{b}\times \widehat{a} \right)+z\left( \left( \widehat{a}\centerdot \widehat{a} \right)\widehat{b}-\left( \widehat{b}\centerdot \widehat{a} \right)\widehat{a} \right)$.
We know that the dot product of two orthogonal vectors is 0 and $\overrightarrow{p}\centerdot \overrightarrow{p}={{\left| \overrightarrow{p} \right|}^{2}}$.
\[\Rightarrow \overrightarrow{r}\times \widehat{a}=y\left( \widehat{b}\times \widehat{a} \right)+z\left( {{\left| \widehat{a} \right|}^{2}}\widehat{b}-\left( 0 \right)\widehat{a} \right)\].
\[\Rightarrow \overrightarrow{r}\times \widehat{a}=y\left( \widehat{b}\times \widehat{a} \right)+z\times 1\times \widehat{b}\].
\[\Rightarrow \overrightarrow{r}\times \widehat{a}=y\left( \widehat{b}\times \widehat{a} \right)+z\widehat{b}\] ---(2).
Now, let us find $\left[ \begin{matrix}
\overrightarrow{r} & \widehat{a} & \widehat{b} \\
\end{matrix} \right]$. We know that scalar triple product of $\overrightarrow{p}$, $\overrightarrow{q}$ and $\overrightarrow{s}$ is defined as $\left[ \begin{matrix}
\overrightarrow{p} & \overrightarrow{q} & \overrightarrow{s} \\
\end{matrix} \right]=\overrightarrow{p}\centerdot \left( \overrightarrow{q}\times \overrightarrow{s} \right)$.
$\Rightarrow \left[ \begin{matrix}
\overrightarrow{r} & \widehat{a} & \widehat{b} \\
\end{matrix} \right]=\overrightarrow{r}\centerdot \left( \widehat{a}\times \widehat{b} \right)$.
$\Rightarrow \left[ \begin{matrix}
\overrightarrow{r} & \widehat{a} & \widehat{b} \\
\end{matrix} \right]=\left( x\widehat{a}+y\widehat{b}+z\left( \widehat{a}\times \widehat{b} \right) \right)\centerdot \left( \widehat{a}\times \widehat{b} \right)$.
\[\Rightarrow \left[ \begin{matrix}
\overrightarrow{r} & \widehat{a} & \widehat{b} \\
\end{matrix} \right]=x\widehat{a}\centerdot \left( \widehat{a}\times \widehat{b} \right)+y\widehat{b}\centerdot \left( \widehat{a}\times \widehat{b} \right)+z\left( \widehat{a}\times \widehat{b} \right)\centerdot \left( \widehat{a}\times \widehat{b} \right)\].
\[\Rightarrow \left[ \begin{matrix}
\overrightarrow{r} & \widehat{a} & \widehat{b} \\
\end{matrix} \right]=x\left[ \begin{matrix}
\widehat{a} & \widehat{a} & \widehat{b} \\
\end{matrix} \right]+y\left[ \begin{matrix}
\widehat{b} & \widehat{a} & \widehat{b} \\
\end{matrix} \right]+z{{\left| \widehat{a}\times \widehat{b} \right|}^{2}}\].
We know that $\left[ \begin{matrix}
\overrightarrow{p} & \overrightarrow{q} & \overrightarrow{p} \\
\end{matrix} \right]=0$.
\[\Rightarrow \left[ \begin{matrix}
\overrightarrow{r} & \widehat{a} & \widehat{b} \\
\end{matrix} \right]=x\left( 0 \right)+y\left( 0 \right)+z{{\left( 1 \right)}^{2}}\].
\[\Rightarrow \left[ \begin{matrix}
\overrightarrow{r} & \widehat{a} & \widehat{b} \\
\end{matrix} \right]=0+0+z\left( 1 \right)\].
\[\Rightarrow \left[ \begin{matrix}
\overrightarrow{r} & \widehat{a} & \widehat{b} \\
\end{matrix} \right]=z\] ---(3).
From equation (1), we can see that $\overrightarrow{r}\centerdot \widehat{b}=y$ ---(4).
Let us substitute equations (3) and (4) in equation (2).
\[\Rightarrow \overrightarrow{r}\times \widehat{a}=\left( \overrightarrow{r}\centerdot \widehat{b} \right)\left( \widehat{b}\times \widehat{a} \right)+\left[ \begin{matrix}
\overrightarrow{r} & \widehat{a} & \widehat{b} \\
\end{matrix} \right]\widehat{b}\].
\[\Rightarrow \overrightarrow{r}\times \widehat{a}=\left[ \begin{matrix}
\overrightarrow{r} & \widehat{a} & \widehat{b} \\
\end{matrix} \right]\widehat{b}+\left( \overrightarrow{r}\centerdot \widehat{b} \right)\left( \widehat{b}\times \widehat{a} \right)\].
∴ The correct option for the given problem is (c).
Note: We can see that the given problem involves a lot of calculations, so we need to perform each step carefully in order to avoid mistakes. We should confuse the vector triple product with the scalar triple product while solving this problem. We can also assume $\widehat{i}$ and $\widehat{j}$ instead of $\widehat{a}$ and $\widehat{b}$, as they were also the mutually orthogonal unit vectors. Similarly, we can expect problems to find $\overrightarrow{r}$ if \[\left[ \begin{matrix}
\overrightarrow{r} & \widehat{a} & \widehat{b} \\
\end{matrix} \right]=6\], \[\overrightarrow{r}\centerdot \widehat{a}=4\] and \[\overrightarrow{r}\centerdot \widehat{b}=8\].
\overrightarrow{r} & \widehat{a} & \widehat{b} \\
\end{matrix} \right]$ and $\overrightarrow{r}\centerdot \widehat{b}$. We then make necessary arrangements in $\left( \overrightarrow{r}\times \widehat{a} \right)$ to get the final answer.
Complete step-by-step solution
According to the problem, we are given that $\widehat{a}$ and $\widehat{b}$ are orthogonal unit vectors. We need to find which option equals $\left( \overrightarrow{r}\times \widehat{a} \right)$ for any non-zero vector $\overrightarrow{r}$.
We know that the cross product of any two given vectors is perpendicular to those two given vectors.
So, $\widehat{a}\times \widehat{b}$ is perpendicular to the vectors $\widehat{a}$ and $\widehat{b}$. Let us find the magnitude of the vector $\widehat{a}\times \widehat{b}$.
For two vectors $\overrightarrow{p}$ and $\overrightarrow{q}$ we know that $\left| \overrightarrow{p}\times \overrightarrow{q} \right|=\left| \overrightarrow{p} \right|\left| \overrightarrow{q} \right|\sin \alpha $, where $\alpha $ is the angle between the vectors $\overrightarrow{p}$ and $\overrightarrow{q}$.
So, we have $\left| \widehat{a}\times \widehat{b} \right|=\left| \widehat{a} \right|\left| \widehat{b} \right|\sin \left( {{90}^{\circ }} \right)$.
$\Rightarrow \left| \widehat{a}\times \widehat{b} \right|=1\times 1\times 1=1$.
So, we have three mutually orthogonal unit vectors $\widehat{a}$, $\widehat{b}$ and $\widehat{a}\times \widehat{b}$.
We know that any vector in the universe can be written in terms of three mutually orthogonal unit vectors.
So, let us assume $\overrightarrow{r}=x\widehat{a}+y\widehat{b}+z\left( \widehat{a}\times \widehat{b} \right)$ ---(1).
Now, let us find $\left( \overrightarrow{r}\times \widehat{a} \right)$.
So, we have $\overrightarrow{r}\times \widehat{a}=\left( x\widehat{a}+y\widehat{b}+z\left( \widehat{a}\times \widehat{b} \right) \right)\times \widehat{a}$.
$\Rightarrow \overrightarrow{r}\times \widehat{a}=x\left( \widehat{a}\times \widehat{a} \right)+y\left( \widehat{b}\times \widehat{a} \right)+z\left( \widehat{a}\times \widehat{b} \right)\times \widehat{a}$.
We know that for a vector $\overrightarrow{q}$, the cross product $\overrightarrow{q}\times \overrightarrow{q}=\overrightarrow{0}$. We also know that the vector triple product of $\overrightarrow{p}$, $\overrightarrow{q}$ and $\overrightarrow{s}$ is defined as $\left( \overrightarrow{p}\times \overrightarrow{q} \right)\times \overrightarrow{s}=\left( \overrightarrow{p}\centerdot \overrightarrow{s} \right)\overrightarrow{q}-\left( \overrightarrow{q}\centerdot \overrightarrow{s} \right)\overrightarrow{p}$.
$\Rightarrow \overrightarrow{r}\times \widehat{a}=y\left( \widehat{b}\times \widehat{a} \right)+z\left( \left( \widehat{a}\centerdot \widehat{a} \right)\widehat{b}-\left( \widehat{b}\centerdot \widehat{a} \right)\widehat{a} \right)$.
We know that the dot product of two orthogonal vectors is 0 and $\overrightarrow{p}\centerdot \overrightarrow{p}={{\left| \overrightarrow{p} \right|}^{2}}$.
\[\Rightarrow \overrightarrow{r}\times \widehat{a}=y\left( \widehat{b}\times \widehat{a} \right)+z\left( {{\left| \widehat{a} \right|}^{2}}\widehat{b}-\left( 0 \right)\widehat{a} \right)\].
\[\Rightarrow \overrightarrow{r}\times \widehat{a}=y\left( \widehat{b}\times \widehat{a} \right)+z\times 1\times \widehat{b}\].
\[\Rightarrow \overrightarrow{r}\times \widehat{a}=y\left( \widehat{b}\times \widehat{a} \right)+z\widehat{b}\] ---(2).
Now, let us find $\left[ \begin{matrix}
\overrightarrow{r} & \widehat{a} & \widehat{b} \\
\end{matrix} \right]$. We know that scalar triple product of $\overrightarrow{p}$, $\overrightarrow{q}$ and $\overrightarrow{s}$ is defined as $\left[ \begin{matrix}
\overrightarrow{p} & \overrightarrow{q} & \overrightarrow{s} \\
\end{matrix} \right]=\overrightarrow{p}\centerdot \left( \overrightarrow{q}\times \overrightarrow{s} \right)$.
$\Rightarrow \left[ \begin{matrix}
\overrightarrow{r} & \widehat{a} & \widehat{b} \\
\end{matrix} \right]=\overrightarrow{r}\centerdot \left( \widehat{a}\times \widehat{b} \right)$.
$\Rightarrow \left[ \begin{matrix}
\overrightarrow{r} & \widehat{a} & \widehat{b} \\
\end{matrix} \right]=\left( x\widehat{a}+y\widehat{b}+z\left( \widehat{a}\times \widehat{b} \right) \right)\centerdot \left( \widehat{a}\times \widehat{b} \right)$.
\[\Rightarrow \left[ \begin{matrix}
\overrightarrow{r} & \widehat{a} & \widehat{b} \\
\end{matrix} \right]=x\widehat{a}\centerdot \left( \widehat{a}\times \widehat{b} \right)+y\widehat{b}\centerdot \left( \widehat{a}\times \widehat{b} \right)+z\left( \widehat{a}\times \widehat{b} \right)\centerdot \left( \widehat{a}\times \widehat{b} \right)\].
\[\Rightarrow \left[ \begin{matrix}
\overrightarrow{r} & \widehat{a} & \widehat{b} \\
\end{matrix} \right]=x\left[ \begin{matrix}
\widehat{a} & \widehat{a} & \widehat{b} \\
\end{matrix} \right]+y\left[ \begin{matrix}
\widehat{b} & \widehat{a} & \widehat{b} \\
\end{matrix} \right]+z{{\left| \widehat{a}\times \widehat{b} \right|}^{2}}\].
We know that $\left[ \begin{matrix}
\overrightarrow{p} & \overrightarrow{q} & \overrightarrow{p} \\
\end{matrix} \right]=0$.
\[\Rightarrow \left[ \begin{matrix}
\overrightarrow{r} & \widehat{a} & \widehat{b} \\
\end{matrix} \right]=x\left( 0 \right)+y\left( 0 \right)+z{{\left( 1 \right)}^{2}}\].
\[\Rightarrow \left[ \begin{matrix}
\overrightarrow{r} & \widehat{a} & \widehat{b} \\
\end{matrix} \right]=0+0+z\left( 1 \right)\].
\[\Rightarrow \left[ \begin{matrix}
\overrightarrow{r} & \widehat{a} & \widehat{b} \\
\end{matrix} \right]=z\] ---(3).
From equation (1), we can see that $\overrightarrow{r}\centerdot \widehat{b}=y$ ---(4).
Let us substitute equations (3) and (4) in equation (2).
\[\Rightarrow \overrightarrow{r}\times \widehat{a}=\left( \overrightarrow{r}\centerdot \widehat{b} \right)\left( \widehat{b}\times \widehat{a} \right)+\left[ \begin{matrix}
\overrightarrow{r} & \widehat{a} & \widehat{b} \\
\end{matrix} \right]\widehat{b}\].
\[\Rightarrow \overrightarrow{r}\times \widehat{a}=\left[ \begin{matrix}
\overrightarrow{r} & \widehat{a} & \widehat{b} \\
\end{matrix} \right]\widehat{b}+\left( \overrightarrow{r}\centerdot \widehat{b} \right)\left( \widehat{b}\times \widehat{a} \right)\].
∴ The correct option for the given problem is (c).
Note: We can see that the given problem involves a lot of calculations, so we need to perform each step carefully in order to avoid mistakes. We should confuse the vector triple product with the scalar triple product while solving this problem. We can also assume $\widehat{i}$ and $\widehat{j}$ instead of $\widehat{a}$ and $\widehat{b}$, as they were also the mutually orthogonal unit vectors. Similarly, we can expect problems to find $\overrightarrow{r}$ if \[\left[ \begin{matrix}
\overrightarrow{r} & \widehat{a} & \widehat{b} \\
\end{matrix} \right]=6\], \[\overrightarrow{r}\centerdot \widehat{a}=4\] and \[\overrightarrow{r}\centerdot \widehat{b}=8\].
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