If we are given an equation as \[\overrightarrow{d}=\lambda \left( \overrightarrow{a}\times \overrightarrow{b} \right)+\mu \left( \overrightarrow{b}\times \overrightarrow{c} \right)+\nu \left( \overrightarrow{c}\times \overrightarrow{a} \right)\] is equal to and \[\left[ \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right]=\dfrac{1}{8}\], then \[\lambda +\mu +\nu \]
(a) \[\overrightarrow{d}.\left( \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right)\]
(b) \[2\overrightarrow{d}.\left( \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right)\]
(c) \[4\overrightarrow{d}.\left( \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right)\]
(d) \[8\overrightarrow{d}.\left( \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right)\]
Answer
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Hint: We start solving the problem by recalling the definition of scalar triple product of given three vectors. We then equate this definition with the given value of the scalar triple product in the problem. We then take dot product on both sides to the equation \[\overrightarrow{d}=\lambda \left( \overrightarrow{a}\times \overrightarrow{b} \right)+\mu \left( \overrightarrow{b}\times \overrightarrow{c} \right)+\nu \left( \overrightarrow{c}\times \overrightarrow{a} \right)\] with the three vectors \[\overrightarrow{a},\overrightarrow{b}\]and \[\overrightarrow{c}\] one after other to find the value of $\lambda $, $\mu $ and $\nu $. We then add all these results in order to get the result.
Complete step-by-step solution:
According to the problem, we are given that \[\overrightarrow{d}=\lambda \left( \overrightarrow{a}\times \overrightarrow{b} \right)+\mu \left( \overrightarrow{b}\times \overrightarrow{c} \right)+\nu \left( \overrightarrow{c}\times \overrightarrow{a} \right)\]and \[\left[ \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right]=\dfrac{1}{8}\].
We know that for three vectors \[\overrightarrow{a},\overrightarrow{b}\]and \[\overrightarrow{c}\]the scalar triple product is defined as \[\left[ \overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c} \right]=\overrightarrow{a}.\left( \overrightarrow{b}\times \overrightarrow{c} \right)=\overrightarrow{b}.\left( \overrightarrow{c}\times \overrightarrow{a} \right)=\overrightarrow{c}.\left( \overrightarrow{a}\times \overrightarrow{b} \right)\] ---(1).
From the problem, we have \[\left[ \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right]=\dfrac{1}{8}\].
So, we get \[\overrightarrow{a}.\left( \overrightarrow{b}\times \overrightarrow{c} \right)=\overrightarrow{b}.\left( \overrightarrow{c}\times \overrightarrow{a} \right)=\overrightarrow{c}.\left( \overrightarrow{a}\times \overrightarrow{b} \right)=\dfrac{1}{8}\] ---(2).
According to the problem, we have given an equation \[\overrightarrow{d}=\lambda \left( \overrightarrow{a}\times \overrightarrow{b} \right)+\mu \left( \overrightarrow{b}\times \overrightarrow{c} \right)+\nu \left( \overrightarrow{c}\times \overrightarrow{a} \right)\] ---(3).
Let us take the dot product on both sides with the vector $\overrightarrow{a}$ in the equation (3).
$\Rightarrow \overrightarrow{d}\bullet \overrightarrow{a}=\lambda \left( \overrightarrow{a}\times \overrightarrow{b} \right)\bullet \overrightarrow{a}+\mu \left( \overrightarrow{b}\times \overrightarrow{c} \right)\bullet \overrightarrow{a}+\nu \left( \overrightarrow{c}\times \overrightarrow{a} \right)\bullet \overrightarrow{a}$.
We know that if a vector is repeated in the scalar triple then its value is zero. i.e., $\left[ \overrightarrow{a}\text{ }\overrightarrow{a}\text{ }\overrightarrow{b} \right]=0$. We use this and result from the equation (2).
$\Rightarrow \overrightarrow{d}\bullet \overrightarrow{a}=\lambda \left( 0 \right)+\mu \left( \dfrac{1}{8} \right)+\nu \left( 0 \right)$.
$\Rightarrow \overrightarrow{d}\bullet \overrightarrow{a}=\mu \left( \dfrac{1}{8} \right)$.
$\Rightarrow 8\left( \overrightarrow{d}\bullet \overrightarrow{a} \right)=\mu $ ---(4).
Let us take the dot product on both sides with the vector $\overrightarrow{b}$ in the equation (3).
$\Rightarrow \overrightarrow{d}\bullet \overrightarrow{b}=\lambda \left( \overrightarrow{a}\times \overrightarrow{b} \right)\bullet \overrightarrow{b}+\mu \left( \overrightarrow{b}\times \overrightarrow{c} \right)\bullet \overrightarrow{b}+\nu \left( \overrightarrow{c}\times \overrightarrow{a} \right)\bullet \overrightarrow{b}$.
We know that if a vector is repeated in the scalar triple then its value is zero. i.e., $\left[ \overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{b} \right]=0$. We use this and result from the equation (2).
$\Rightarrow \overrightarrow{d}\bullet \overrightarrow{b}=\lambda \left( 0 \right)+\mu \left( 0 \right)+\nu \left( \dfrac{1}{8} \right)$.
$\Rightarrow \overrightarrow{d}\bullet \overrightarrow{b}=\nu \left( \dfrac{1}{8} \right)$.
$\Rightarrow 8\left( \overrightarrow{d}\bullet \overrightarrow{b} \right)=\nu $ ---(5).
Let us take the dot product on both sides with the vector $\overrightarrow{c}$ in the equation (3).
$\Rightarrow \overrightarrow{d}\bullet \overrightarrow{b}=\lambda \left( \overrightarrow{a}\times \overrightarrow{b} \right)\bullet \overrightarrow{c}+\mu \left( \overrightarrow{b}\times \overrightarrow{c} \right)\bullet \overrightarrow{c}+\nu \left( \overrightarrow{c}\times \overrightarrow{a} \right)\bullet \overrightarrow{c}$.
We know that if a vector is repeated in the scalar triple then its value is zero. i.e., $\left[ \overrightarrow{b}\text{ }\overrightarrow{c}\text{ }\overrightarrow{c} \right]=0$. We use this and result from the equation (2).
\[\Rightarrow \overrightarrow{d}\bullet \overrightarrow{c}=\lambda \left( \dfrac{1}{8} \right)+\mu \left( 0 \right)+\nu \left( 0 \right)\].
$\Rightarrow \overrightarrow{d}\bullet \overrightarrow{c}=\lambda \left( \dfrac{1}{8} \right)$.
$\Rightarrow 8\left( \overrightarrow{d}\bullet \overrightarrow{c} \right)=\lambda $ ---(6).
Let us add all the equations (4), (5), and (6).
$\Rightarrow 8\left( \overrightarrow{d}\bullet \overrightarrow{a} \right)+8\left( \overrightarrow{d}\bullet \overrightarrow{b} \right)+8\left( \overrightarrow{d}\bullet \overrightarrow{c} \right)=\mu +\nu +\lambda $.
$\Rightarrow 8\left( \left( \overrightarrow{d}\bullet \overrightarrow{a} \right)+\left( \overrightarrow{d}\bullet \overrightarrow{b} \right)+\left( \overrightarrow{d}\bullet \overrightarrow{c} \right) \right)=\mu +\nu +\lambda $ ---(7).
We know that $\overrightarrow{x}.\left( \overrightarrow{y}+\overrightarrow{z} \right)=\overrightarrow{x}.\overrightarrow{y}+\overrightarrow{x}.\overrightarrow{z}$. We use this in equation (7).
$\Rightarrow 8\overrightarrow{d}\bullet \left( \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right)=\mu +\nu +\lambda $.
So, we have found the value of \[\lambda +\mu +\nu \] as \[8\overrightarrow{d}.\left( \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right)\].
$\therefore$ The correct option for the given problem is (d).
Note: Here we are told that $\left[ \overrightarrow{b}\text{ }\overrightarrow{c}\text{ }\overrightarrow{c} \right]=0$. This can be verified by using the standard definition of scalar triple product. So, we have $\left[ \overrightarrow{b}\text{ }\overrightarrow{c}\text{ }\overrightarrow{c} \right]=\left( \overrightarrow{b}\times \overrightarrow{c} \right)\bullet \overrightarrow{c}$. We know that the vectors $\overrightarrow{b}$ and $\overrightarrow{c}$ is perpendicular to the vector $\overrightarrow{b}\times \overrightarrow{c}$. We also know that the dot product of two perpendicular vectors is 0 which gives our result. We should keep in mind that swapping any two vectors in a scalar triple product will lead to change in direction of the product which gives us the negative of the given value. We should know that a scalar triple product is different from a vector triple product.
Complete step-by-step solution:
According to the problem, we are given that \[\overrightarrow{d}=\lambda \left( \overrightarrow{a}\times \overrightarrow{b} \right)+\mu \left( \overrightarrow{b}\times \overrightarrow{c} \right)+\nu \left( \overrightarrow{c}\times \overrightarrow{a} \right)\]and \[\left[ \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right]=\dfrac{1}{8}\].
We know that for three vectors \[\overrightarrow{a},\overrightarrow{b}\]and \[\overrightarrow{c}\]the scalar triple product is defined as \[\left[ \overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c} \right]=\overrightarrow{a}.\left( \overrightarrow{b}\times \overrightarrow{c} \right)=\overrightarrow{b}.\left( \overrightarrow{c}\times \overrightarrow{a} \right)=\overrightarrow{c}.\left( \overrightarrow{a}\times \overrightarrow{b} \right)\] ---(1).
From the problem, we have \[\left[ \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right]=\dfrac{1}{8}\].
So, we get \[\overrightarrow{a}.\left( \overrightarrow{b}\times \overrightarrow{c} \right)=\overrightarrow{b}.\left( \overrightarrow{c}\times \overrightarrow{a} \right)=\overrightarrow{c}.\left( \overrightarrow{a}\times \overrightarrow{b} \right)=\dfrac{1}{8}\] ---(2).
According to the problem, we have given an equation \[\overrightarrow{d}=\lambda \left( \overrightarrow{a}\times \overrightarrow{b} \right)+\mu \left( \overrightarrow{b}\times \overrightarrow{c} \right)+\nu \left( \overrightarrow{c}\times \overrightarrow{a} \right)\] ---(3).
Let us take the dot product on both sides with the vector $\overrightarrow{a}$ in the equation (3).
$\Rightarrow \overrightarrow{d}\bullet \overrightarrow{a}=\lambda \left( \overrightarrow{a}\times \overrightarrow{b} \right)\bullet \overrightarrow{a}+\mu \left( \overrightarrow{b}\times \overrightarrow{c} \right)\bullet \overrightarrow{a}+\nu \left( \overrightarrow{c}\times \overrightarrow{a} \right)\bullet \overrightarrow{a}$.
We know that if a vector is repeated in the scalar triple then its value is zero. i.e., $\left[ \overrightarrow{a}\text{ }\overrightarrow{a}\text{ }\overrightarrow{b} \right]=0$. We use this and result from the equation (2).
$\Rightarrow \overrightarrow{d}\bullet \overrightarrow{a}=\lambda \left( 0 \right)+\mu \left( \dfrac{1}{8} \right)+\nu \left( 0 \right)$.
$\Rightarrow \overrightarrow{d}\bullet \overrightarrow{a}=\mu \left( \dfrac{1}{8} \right)$.
$\Rightarrow 8\left( \overrightarrow{d}\bullet \overrightarrow{a} \right)=\mu $ ---(4).
Let us take the dot product on both sides with the vector $\overrightarrow{b}$ in the equation (3).
$\Rightarrow \overrightarrow{d}\bullet \overrightarrow{b}=\lambda \left( \overrightarrow{a}\times \overrightarrow{b} \right)\bullet \overrightarrow{b}+\mu \left( \overrightarrow{b}\times \overrightarrow{c} \right)\bullet \overrightarrow{b}+\nu \left( \overrightarrow{c}\times \overrightarrow{a} \right)\bullet \overrightarrow{b}$.
We know that if a vector is repeated in the scalar triple then its value is zero. i.e., $\left[ \overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{b} \right]=0$. We use this and result from the equation (2).
$\Rightarrow \overrightarrow{d}\bullet \overrightarrow{b}=\lambda \left( 0 \right)+\mu \left( 0 \right)+\nu \left( \dfrac{1}{8} \right)$.
$\Rightarrow \overrightarrow{d}\bullet \overrightarrow{b}=\nu \left( \dfrac{1}{8} \right)$.
$\Rightarrow 8\left( \overrightarrow{d}\bullet \overrightarrow{b} \right)=\nu $ ---(5).
Let us take the dot product on both sides with the vector $\overrightarrow{c}$ in the equation (3).
$\Rightarrow \overrightarrow{d}\bullet \overrightarrow{b}=\lambda \left( \overrightarrow{a}\times \overrightarrow{b} \right)\bullet \overrightarrow{c}+\mu \left( \overrightarrow{b}\times \overrightarrow{c} \right)\bullet \overrightarrow{c}+\nu \left( \overrightarrow{c}\times \overrightarrow{a} \right)\bullet \overrightarrow{c}$.
We know that if a vector is repeated in the scalar triple then its value is zero. i.e., $\left[ \overrightarrow{b}\text{ }\overrightarrow{c}\text{ }\overrightarrow{c} \right]=0$. We use this and result from the equation (2).
\[\Rightarrow \overrightarrow{d}\bullet \overrightarrow{c}=\lambda \left( \dfrac{1}{8} \right)+\mu \left( 0 \right)+\nu \left( 0 \right)\].
$\Rightarrow \overrightarrow{d}\bullet \overrightarrow{c}=\lambda \left( \dfrac{1}{8} \right)$.
$\Rightarrow 8\left( \overrightarrow{d}\bullet \overrightarrow{c} \right)=\lambda $ ---(6).
Let us add all the equations (4), (5), and (6).
$\Rightarrow 8\left( \overrightarrow{d}\bullet \overrightarrow{a} \right)+8\left( \overrightarrow{d}\bullet \overrightarrow{b} \right)+8\left( \overrightarrow{d}\bullet \overrightarrow{c} \right)=\mu +\nu +\lambda $.
$\Rightarrow 8\left( \left( \overrightarrow{d}\bullet \overrightarrow{a} \right)+\left( \overrightarrow{d}\bullet \overrightarrow{b} \right)+\left( \overrightarrow{d}\bullet \overrightarrow{c} \right) \right)=\mu +\nu +\lambda $ ---(7).
We know that $\overrightarrow{x}.\left( \overrightarrow{y}+\overrightarrow{z} \right)=\overrightarrow{x}.\overrightarrow{y}+\overrightarrow{x}.\overrightarrow{z}$. We use this in equation (7).
$\Rightarrow 8\overrightarrow{d}\bullet \left( \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right)=\mu +\nu +\lambda $.
So, we have found the value of \[\lambda +\mu +\nu \] as \[8\overrightarrow{d}.\left( \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right)\].
$\therefore$ The correct option for the given problem is (d).
Note: Here we are told that $\left[ \overrightarrow{b}\text{ }\overrightarrow{c}\text{ }\overrightarrow{c} \right]=0$. This can be verified by using the standard definition of scalar triple product. So, we have $\left[ \overrightarrow{b}\text{ }\overrightarrow{c}\text{ }\overrightarrow{c} \right]=\left( \overrightarrow{b}\times \overrightarrow{c} \right)\bullet \overrightarrow{c}$. We know that the vectors $\overrightarrow{b}$ and $\overrightarrow{c}$ is perpendicular to the vector $\overrightarrow{b}\times \overrightarrow{c}$. We also know that the dot product of two perpendicular vectors is 0 which gives our result. We should keep in mind that swapping any two vectors in a scalar triple product will lead to change in direction of the product which gives us the negative of the given value. We should know that a scalar triple product is different from a vector triple product.
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