Answer
Verified
496.8k+ views
Hint: Scalar triple product can directly be applied on the given sides of parallelepiped. Drawing sketches for parallelepiped with given coterminous edges might lead to an error as they are given in the form of sum of two vectors.
Complete step-by-step answer:
Here, we have a parallelepiped with $\overrightarrow{a}+\overrightarrow{b}$, $\overrightarrow{b}+\overrightarrow{c}$ and $\overrightarrow{c}+\overrightarrow{a}$ as their coterminous edges.
And coterminous edges mean the edges of a figure having or sharing the same boundaries.
Now, volume of a parallelepiped with their edges as let’s say $\overrightarrow{x},\overrightarrow{y}$ and $\overrightarrow{z}$ is defined as the area of the base times the height., $V=\left[ \overrightarrow{x}\overrightarrow{y}\overrightarrow{z} \right]$,
\[\begin{align}
& \Rightarrow V=\left[ \overrightarrow{x}\overrightarrow{y}\overrightarrow{z} \right] \\
& \Rightarrow V=\overrightarrow{x}\cdot \left( \overrightarrow{y}\times \overrightarrow{z} \right)=\left( \overrightarrow{x}\times \overrightarrow{y} \right)\cdot \overrightarrow{z}...\text{ }\left( 1 \right) \\
\end{align}\]
Which is also known as the scalar-triple product and it is further defined as,
\[\Rightarrow \left( \overrightarrow{x}\times \overrightarrow{y} \right)\cdot \overrightarrow{z}=\left| \begin{matrix}
{{z}_{1}} & {{z}_{2}} & {{z}_{3}} \\
{{x}_{1}} & {{x}_{2}} & {{x}_{3}} \\
{{y}_{1}} & {{y}_{2}} & {{y}_{3}} \\
\end{matrix} \right|...\text{ }\left( 2 \right)\]
where $\overrightarrow{x}=\left( {{x}_{1}},{{x}_{2}},{{x}_{3}} \right)$, $\overrightarrow{y}=\left( {{y}_{1}},{{y}_{2}},{{y}_{3}} \right)$ and \[\overrightarrow{z}=\left( {{z}_{1}},{{z}_{2}},{{z}_{3}} \right)\] defined in vector form.
Thus, from given conditions, we have $\overrightarrow{x}=\overrightarrow{a}+\overrightarrow{b}$, $\overrightarrow{y}=\overrightarrow{b}+\overrightarrow{c}$ and $\overrightarrow{z}=\overrightarrow{c}+\overrightarrow{a}$.
Substituting these values in equation (2), we get
\[\Rightarrow \left( \overrightarrow{x}\times \overrightarrow{y} \right)\cdot \overrightarrow{z}=\left( \left( \overrightarrow{a}+\overrightarrow{b} \right)\times \left( \overrightarrow{b}+\overrightarrow{c} \right) \right)\cdot \left( \overrightarrow{c}+\overrightarrow{a} \right)\]
Applying the product of vectors, we get
\[\begin{align}
& \Rightarrow \left( \overrightarrow{x}\times \overrightarrow{y} \right)\cdot \overrightarrow{z}=\left( \overrightarrow{a}\times \left( \overrightarrow{b}+\overrightarrow{c} \right)+\overrightarrow{b}\times \left( \overrightarrow{b}+\overrightarrow{c} \right) \right)\cdot \left( \overrightarrow{c}+\overrightarrow{a} \right) \\
& \Rightarrow \left( \overrightarrow{x}\times \overrightarrow{y} \right)\cdot \overrightarrow{z}=\left( \overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{a}\times \overrightarrow{c}+\overrightarrow{b}\times \overrightarrow{b}+\overrightarrow{b}\times \overrightarrow{c} \right)\cdot \left( \overrightarrow{c}+\overrightarrow{a} \right) \\
\end{align}\]
Now, on applying properties of cross product of two parallel vectors, i.e.,
\[\Rightarrow \overrightarrow{b}\times \overrightarrow{b}=\left| \overrightarrow{b} \right|\left| \overrightarrow{b} \right|\sin {{0}^{\circ }}=0\]
Substituting this value in above equation, we get
\[\begin{align}
& \Rightarrow \left( \overrightarrow{x}\times \overrightarrow{y} \right)\cdot \overrightarrow{z}=\left( \overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{a}\times \overrightarrow{c}+\overrightarrow{b}\times \overrightarrow{b}+\overrightarrow{b}\times \overrightarrow{c} \right)\cdot \left( \overrightarrow{c}+\overrightarrow{a} \right) \\
& \Rightarrow \left( \overrightarrow{x}\times \overrightarrow{y} \right)\cdot \overrightarrow{z}=\left( \overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{a}\times \overrightarrow{c}+0+\overrightarrow{b}\times \overrightarrow{c} \right)\cdot \left( \overrightarrow{c}+\overrightarrow{a} \right) \\
\end{align}\]
Now, applying scalar product of vectors, we get
\[\begin{align}
& \Rightarrow \left( \overrightarrow{x}\times \overrightarrow{y} \right)\cdot \overrightarrow{z}=\left( \overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{a}\times \overrightarrow{c}+0+\overrightarrow{b}\times \overrightarrow{c} \right)\cdot \left( \overrightarrow{c}+\overrightarrow{a} \right) \\
& \Rightarrow \left( \overrightarrow{x}\times \overrightarrow{y} \right)\cdot \overrightarrow{z}=\left( \overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{a}\times \overrightarrow{c}+0+\overrightarrow{b}\times \overrightarrow{c} \right)\cdot \overrightarrow{c}+\left( \overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{a}\times \overrightarrow{c}+0+\overrightarrow{b}\times \overrightarrow{c} \right)\cdot \overrightarrow{a} \\
& \Rightarrow \left( \overrightarrow{x}\times \overrightarrow{y} \right)\cdot \overrightarrow{z}=\left( \overrightarrow{a}\times \overrightarrow{b} \right)\cdot \overrightarrow{c}+\left( \overrightarrow{a}\times \overrightarrow{c} \right)\cdot \overrightarrow{c}+0\cdot \overrightarrow{c}+\left( \overrightarrow{b}\times \overrightarrow{c} \right)\cdot \overrightarrow{c}+\left( \overrightarrow{a}\times \overrightarrow{b} \right)\cdot \overrightarrow{a}+\left( \overrightarrow{a}\times \overrightarrow{c} \right)\cdot \overrightarrow{a}+0\cdot \overrightarrow{a}+\left( \overrightarrow{b}\times \overrightarrow{c} \right)\cdot \overrightarrow{a}
\end{align}\]
Using properties of scalar triple product of vectors, we have
\[\Rightarrow \left( \overrightarrow{a}\times \overrightarrow{b} \right)\cdot \overrightarrow{c}=\left[ \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right]\]
And, \[\Rightarrow \left( \overrightarrow{a}\times \overrightarrow{c} \right)\cdot \overrightarrow{c}=\left| \overrightarrow{a}\times \overrightarrow{c} \right|\left| \overrightarrow{c} \right|\cos {{90}^{\circ }}=0\]
Thus, from above equation, we have
\[\begin{align}
& \Rightarrow \left( \overrightarrow{x}\times \overrightarrow{y} \right)\cdot \overrightarrow{z}=\left( \overrightarrow{a}\times \overrightarrow{b} \right)\cdot \overrightarrow{c}+\left( \overrightarrow{a}\times \overrightarrow{c} \right)\cdot \overrightarrow{c}+0\cdot \overrightarrow{c}+\left( \overrightarrow{b}\times \overrightarrow{c} \right)\cdot \overrightarrow{c}+\left( \overrightarrow{a}\times \overrightarrow{b} \right)\cdot \overrightarrow{a}+\left( \overrightarrow{a}\times \overrightarrow{c} \right)\cdot \overrightarrow{a}+0\cdot \overrightarrow{a}+\left( \overrightarrow{b}\times \overrightarrow{c} \right)\cdot \overrightarrow{a} \\
& \Rightarrow \left( \overrightarrow{x}\times \overrightarrow{y} \right)\cdot \overrightarrow{z}=\left[ \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right]+0+0+0+0+0+0+\left[ \overrightarrow{b}\overrightarrow{c}\overrightarrow{a} \right]
\end{align}\]
Also, from properties of scalar triple product, we have \[\left[ \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right]=\left[ \overrightarrow{b}\overrightarrow{c}\overrightarrow{a} \right]\]
\[\begin{align}
& \Rightarrow \left( \overrightarrow{x}\times \overrightarrow{y} \right)\cdot \overrightarrow{z}=\left[ \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right]+0+0+0+0+0+0+\left[ \overrightarrow{b}\overrightarrow{c}\overrightarrow{a} \right] \\
& \Rightarrow \left( \overrightarrow{x}\times \overrightarrow{y} \right)\cdot \overrightarrow{z}=2\left[ \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right]
\end{align}\]
Hence, the volume of parallelepiped = $V=2\left[ \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right]$, thus option [C] is correct.
Note: As per question, the edges of parallelepiped are given as the sum of two vectors. So, calculation of scalar triple product for volume of parallelepiped becomes quite complex. Keeping the angle between the vectors in mind might ease the calculations, in a product.
Complete step-by-step answer:
Here, we have a parallelepiped with $\overrightarrow{a}+\overrightarrow{b}$, $\overrightarrow{b}+\overrightarrow{c}$ and $\overrightarrow{c}+\overrightarrow{a}$ as their coterminous edges.
And coterminous edges mean the edges of a figure having or sharing the same boundaries.
Now, volume of a parallelepiped with their edges as let’s say $\overrightarrow{x},\overrightarrow{y}$ and $\overrightarrow{z}$ is defined as the area of the base times the height., $V=\left[ \overrightarrow{x}\overrightarrow{y}\overrightarrow{z} \right]$,
\[\begin{align}
& \Rightarrow V=\left[ \overrightarrow{x}\overrightarrow{y}\overrightarrow{z} \right] \\
& \Rightarrow V=\overrightarrow{x}\cdot \left( \overrightarrow{y}\times \overrightarrow{z} \right)=\left( \overrightarrow{x}\times \overrightarrow{y} \right)\cdot \overrightarrow{z}...\text{ }\left( 1 \right) \\
\end{align}\]
Which is also known as the scalar-triple product and it is further defined as,
\[\Rightarrow \left( \overrightarrow{x}\times \overrightarrow{y} \right)\cdot \overrightarrow{z}=\left| \begin{matrix}
{{z}_{1}} & {{z}_{2}} & {{z}_{3}} \\
{{x}_{1}} & {{x}_{2}} & {{x}_{3}} \\
{{y}_{1}} & {{y}_{2}} & {{y}_{3}} \\
\end{matrix} \right|...\text{ }\left( 2 \right)\]
where $\overrightarrow{x}=\left( {{x}_{1}},{{x}_{2}},{{x}_{3}} \right)$, $\overrightarrow{y}=\left( {{y}_{1}},{{y}_{2}},{{y}_{3}} \right)$ and \[\overrightarrow{z}=\left( {{z}_{1}},{{z}_{2}},{{z}_{3}} \right)\] defined in vector form.
Thus, from given conditions, we have $\overrightarrow{x}=\overrightarrow{a}+\overrightarrow{b}$, $\overrightarrow{y}=\overrightarrow{b}+\overrightarrow{c}$ and $\overrightarrow{z}=\overrightarrow{c}+\overrightarrow{a}$.
Substituting these values in equation (2), we get
\[\Rightarrow \left( \overrightarrow{x}\times \overrightarrow{y} \right)\cdot \overrightarrow{z}=\left( \left( \overrightarrow{a}+\overrightarrow{b} \right)\times \left( \overrightarrow{b}+\overrightarrow{c} \right) \right)\cdot \left( \overrightarrow{c}+\overrightarrow{a} \right)\]
Applying the product of vectors, we get
\[\begin{align}
& \Rightarrow \left( \overrightarrow{x}\times \overrightarrow{y} \right)\cdot \overrightarrow{z}=\left( \overrightarrow{a}\times \left( \overrightarrow{b}+\overrightarrow{c} \right)+\overrightarrow{b}\times \left( \overrightarrow{b}+\overrightarrow{c} \right) \right)\cdot \left( \overrightarrow{c}+\overrightarrow{a} \right) \\
& \Rightarrow \left( \overrightarrow{x}\times \overrightarrow{y} \right)\cdot \overrightarrow{z}=\left( \overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{a}\times \overrightarrow{c}+\overrightarrow{b}\times \overrightarrow{b}+\overrightarrow{b}\times \overrightarrow{c} \right)\cdot \left( \overrightarrow{c}+\overrightarrow{a} \right) \\
\end{align}\]
Now, on applying properties of cross product of two parallel vectors, i.e.,
\[\Rightarrow \overrightarrow{b}\times \overrightarrow{b}=\left| \overrightarrow{b} \right|\left| \overrightarrow{b} \right|\sin {{0}^{\circ }}=0\]
Substituting this value in above equation, we get
\[\begin{align}
& \Rightarrow \left( \overrightarrow{x}\times \overrightarrow{y} \right)\cdot \overrightarrow{z}=\left( \overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{a}\times \overrightarrow{c}+\overrightarrow{b}\times \overrightarrow{b}+\overrightarrow{b}\times \overrightarrow{c} \right)\cdot \left( \overrightarrow{c}+\overrightarrow{a} \right) \\
& \Rightarrow \left( \overrightarrow{x}\times \overrightarrow{y} \right)\cdot \overrightarrow{z}=\left( \overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{a}\times \overrightarrow{c}+0+\overrightarrow{b}\times \overrightarrow{c} \right)\cdot \left( \overrightarrow{c}+\overrightarrow{a} \right) \\
\end{align}\]
Now, applying scalar product of vectors, we get
\[\begin{align}
& \Rightarrow \left( \overrightarrow{x}\times \overrightarrow{y} \right)\cdot \overrightarrow{z}=\left( \overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{a}\times \overrightarrow{c}+0+\overrightarrow{b}\times \overrightarrow{c} \right)\cdot \left( \overrightarrow{c}+\overrightarrow{a} \right) \\
& \Rightarrow \left( \overrightarrow{x}\times \overrightarrow{y} \right)\cdot \overrightarrow{z}=\left( \overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{a}\times \overrightarrow{c}+0+\overrightarrow{b}\times \overrightarrow{c} \right)\cdot \overrightarrow{c}+\left( \overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{a}\times \overrightarrow{c}+0+\overrightarrow{b}\times \overrightarrow{c} \right)\cdot \overrightarrow{a} \\
& \Rightarrow \left( \overrightarrow{x}\times \overrightarrow{y} \right)\cdot \overrightarrow{z}=\left( \overrightarrow{a}\times \overrightarrow{b} \right)\cdot \overrightarrow{c}+\left( \overrightarrow{a}\times \overrightarrow{c} \right)\cdot \overrightarrow{c}+0\cdot \overrightarrow{c}+\left( \overrightarrow{b}\times \overrightarrow{c} \right)\cdot \overrightarrow{c}+\left( \overrightarrow{a}\times \overrightarrow{b} \right)\cdot \overrightarrow{a}+\left( \overrightarrow{a}\times \overrightarrow{c} \right)\cdot \overrightarrow{a}+0\cdot \overrightarrow{a}+\left( \overrightarrow{b}\times \overrightarrow{c} \right)\cdot \overrightarrow{a}
\end{align}\]
Using properties of scalar triple product of vectors, we have
\[\Rightarrow \left( \overrightarrow{a}\times \overrightarrow{b} \right)\cdot \overrightarrow{c}=\left[ \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right]\]
And, \[\Rightarrow \left( \overrightarrow{a}\times \overrightarrow{c} \right)\cdot \overrightarrow{c}=\left| \overrightarrow{a}\times \overrightarrow{c} \right|\left| \overrightarrow{c} \right|\cos {{90}^{\circ }}=0\]
Thus, from above equation, we have
\[\begin{align}
& \Rightarrow \left( \overrightarrow{x}\times \overrightarrow{y} \right)\cdot \overrightarrow{z}=\left( \overrightarrow{a}\times \overrightarrow{b} \right)\cdot \overrightarrow{c}+\left( \overrightarrow{a}\times \overrightarrow{c} \right)\cdot \overrightarrow{c}+0\cdot \overrightarrow{c}+\left( \overrightarrow{b}\times \overrightarrow{c} \right)\cdot \overrightarrow{c}+\left( \overrightarrow{a}\times \overrightarrow{b} \right)\cdot \overrightarrow{a}+\left( \overrightarrow{a}\times \overrightarrow{c} \right)\cdot \overrightarrow{a}+0\cdot \overrightarrow{a}+\left( \overrightarrow{b}\times \overrightarrow{c} \right)\cdot \overrightarrow{a} \\
& \Rightarrow \left( \overrightarrow{x}\times \overrightarrow{y} \right)\cdot \overrightarrow{z}=\left[ \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right]+0+0+0+0+0+0+\left[ \overrightarrow{b}\overrightarrow{c}\overrightarrow{a} \right]
\end{align}\]
Also, from properties of scalar triple product, we have \[\left[ \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right]=\left[ \overrightarrow{b}\overrightarrow{c}\overrightarrow{a} \right]\]
\[\begin{align}
& \Rightarrow \left( \overrightarrow{x}\times \overrightarrow{y} \right)\cdot \overrightarrow{z}=\left[ \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right]+0+0+0+0+0+0+\left[ \overrightarrow{b}\overrightarrow{c}\overrightarrow{a} \right] \\
& \Rightarrow \left( \overrightarrow{x}\times \overrightarrow{y} \right)\cdot \overrightarrow{z}=2\left[ \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right]
\end{align}\]
Hence, the volume of parallelepiped = $V=2\left[ \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right]$, thus option [C] is correct.
Note: As per question, the edges of parallelepiped are given as the sum of two vectors. So, calculation of scalar triple product for volume of parallelepiped becomes quite complex. Keeping the angle between the vectors in mind might ease the calculations, in a product.
Recently Updated Pages
A particle is undergoing a horizontal circle of radius class 11 physics CBSE
A particle is thrown vertically upwards with a velocity class 11 physics CBSE
A particle is rotated in a vertical circle by connecting class 11 physics CBSE
A particle is projected with a velocity v such that class 11 physics CBSE
A particle is projected with a velocity u making an class 11 physics CBSE
A particle is projected vertically upwards and it reaches class 11 physics CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
Who was the leader of the Bolshevik Party A Leon Trotsky class 9 social science CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Which is the largest saltwater lake in India A Chilika class 8 social science CBSE
Ghatikas during the period of Satavahanas were aHospitals class 6 social science CBSE