Let \[\vec a,\vec b\]and $\vec c$ be three non-coplanar unit vectors such that the angle between every pair of them is $\dfrac{\pi }{3}$. If $\vec a \times \vec b + \vec b \times \vec c = p\vec a + q\vec b + r\vec c$ , where $p,q$ and $r$ are scalars, then the value of $\dfrac{{{p^2} + 2{q^2} + {r^2}}}{{{q^2}}}$ is
A) 1
B) 2
C) 3
D) 4
Answer
Verified604.2k+ views
Hint: For the equations like $\vec a \times \vec b + \vec b \times \vec c = p\vec a + q\vec b + r\vec c$ in the given question, perform dot product of all the given vectors with given equation one by one.
You’ll get the relation between scalars present in the equation.
Unit vectors are those vectors whose magnitude is 1.
Cross product (or vector product) of two nonzero vectors $\vec a$ and $\vec b$ is the product of the magnitude of both vectors $\vec a$ and $\vec b$, and sine of the angle between them. i.e.
$\hat a \times \hat b = \left| {\hat a} \right|\left| {\hat b} \right|\sin \theta {\text{ }}\hat n$, where $\theta $is the acute angle between vectors $\vec a$ and $\vec b$. Here $\hat n$is the unit vector perpendicular to the plane containing vectors $\vec a$ and $\vec b$.
Dot product (or scalar product) between two non-zero vectors $\vec a$ and $\vec b$ is the product of the magnitude of each vector and cosine of the angle between them:
\[\vec a \cdot \vec b = \left| a \right|\left| b \right|\cos \theta \] , where $\theta $ is the acute angle between vectors $\vec a$ and $\vec b$.
Complete step-by-step answer:
Step 1: Given that:
$\vec a \times \vec b + \vec b \times \vec c = p\vec a + q\vec b + r\vec c$
Where vectors \[\vec a,\vec b\]and $\vec c$ are unit vector,
Hence, $\left| {\vec a} \right| = 1$ ,
$\left| {\vec b} \right| = 1$ ,
$\left| {\vec c} \right| = 1$
The angle between every pair of them is $\dfrac{\pi }{3}$.
Step 2: Dot product of a vector \[\vec a\] with both sides on:
$\vec a \times \vec b + \vec b \times \vec c = p\vec a + q\vec b + r\vec c$
$\vec a \cdot \left( {\vec a \times \vec b + \vec b \times \vec c} \right) = \vec a \cdot \left( {p\vec a + q\vec b + r\vec c} \right)$
$ \Rightarrow \vec a \cdot \left( {\vec a \times \vec b} \right) + \vec a \cdot \left( {\vec b \times \vec c} \right) = p\left( {\vec a \cdot \vec a} \right) + q\left( {\vec a \cdot \vec b} \right) + r\left( {\vec a \cdot \vec c} \right)$
The resultant vector \[\left( {\vec a \times \vec b} \right)\] is perpendicular to the plane containing $\vec a$ and $\vec b$ i.e. the angle between vectors \[\left( {\vec a \times \vec b} \right)\] and $\vec a$ is ${90^ \circ }$. Also the angle between vectors \[\left( {\vec a \times \vec b} \right)\] and $\vec b$ is ${90^ \circ }$.
Hence, $\vec a \cdot \left( {\vec a \times \vec b} \right) = 0$ …… (1)
\[
\left( {\vec a \cdot \vec a} \right) = \left| {\vec a} \right|\left| {\vec a} \right|\cos 0^\circ \\
\Rightarrow {\left| {\vec a} \right|^2} \\
\Rightarrow 1 \\
\] $\left( {\because \cos 0^\circ = 1} \right)$ …… (2)
$
\left( {\vec a \cdot \vec b} \right) = \left| {\vec a} \right|\left| {\vec b} \right|\cos \dfrac{\pi }{3} \\
\Rightarrow \dfrac{1}{2}\left| {\vec a} \right|\left| {\vec b} \right| \\
\Rightarrow \dfrac{1}{2} \\
$ $\left( {\because \cos \dfrac{\pi }{3} = \dfrac{1}{2}} \right)$ …… (3)
$
\left( {\vec a \cdot \vec c} \right) = \left| {\vec a} \right|\left| {\vec c} \right|\cos \dfrac{\pi }{3} \\
\Rightarrow \dfrac{1}{2}\left| {\vec a} \right|\left| {\vec c} \right| \\
\Rightarrow \dfrac{1}{2} \\
$ $\left( {\because \cos \dfrac{\pi }{3} = \dfrac{1}{2}} \right)$ …… (4)
We have, $\vec a \cdot \left( {\vec a \times \vec b} \right) + \vec a \cdot \left( {\vec b \times \vec c} \right) = p\left( {\vec a \cdot \vec a} \right) + q\left( {\vec a \cdot \vec b} \right) + r\left( {\vec a \cdot \vec c} \right)$
From (1), (2), (3), and (4)
$0 + \vec a \cdot \left( {\vec b \times \vec c} \right) = p\left( 1 \right) + q\left( {\dfrac{1}{2}} \right) + r\left( {\dfrac{1}{2}} \right)$
$ \Rightarrow \vec a \cdot \left( {\vec b \times \vec c} \right) = \dfrac{{2p + q + r}}{2}$ …… (5)
Step 3: Similarly, Dot product of a vector \[\vec b\] with both sides on:
$\vec a \times \vec b + \vec b \times \vec c = p\vec a + q\vec b + r\vec c$
$ \Rightarrow \vec b \cdot \left( {\vec a \times \vec b + \vec b \times \vec c} \right) = \vec b \cdot \left( {p\vec a + q\vec b + r\vec c} \right)$
\[ \Rightarrow \vec b \cdot \left( {\vec a \times \vec b} \right) + \vec b \cdot \left( {\vec b \times \vec c} \right) = p\left( {\vec b \cdot \vec a} \right) + q\left( {\vec b \cdot \vec b} \right) + r\left( {\vec b \cdot \vec c} \right)\]
We know the angle between vectors \[\left( {\vec a \times \vec b} \right)\] and $\vec b$ is ${90^ \circ }$.
Hence, $\vec b \cdot \left( {\vec a \times \vec b} \right) = 0$ …… (6)
The resultant vector \[\left( {\vec b \times \vec c} \right)\] is perpendicular to the plane containing $\vec b$ and $\vec c$ i.e. the angle between vectors \[\left( {\vec b \times \vec c} \right)\] and $\vec b$ is ${90^ \circ }$. Also the angle between vectors \[\left( {\vec b \times \vec c} \right)\] and $\vec c$ is ${90^ \circ }$.
Hence, \[\vec b \cdot \left( {\vec b \times \vec c} \right) = 0\] …… (7)
$
\left( {\vec b \cdot \vec a} \right) = \left| {\vec b} \right|\left| {\vec a} \right|\cos \dfrac{\pi }{3} \\
\Rightarrow \dfrac{1}{2}\left| {\vec b} \right|\left| {\vec a} \right| \\
\Rightarrow \dfrac{1}{2} \\
$ $\left( {\because \cos \dfrac{\pi }{3} = \dfrac{1}{2}} \right)$ …… (8)
\[
\left( {\vec b \cdot \vec b} \right) = \left| {\vec b} \right|\left| {\vec b} \right|\cos 0^\circ \\
\Rightarrow {\left| {\vec b} \right|^2} \\
\Rightarrow 1 \\
\] $\left( {\because \cos 0^\circ = 1} \right)$ …… (9)
$
\left( {\vec b \cdot \vec c} \right) = \left| {\vec b} \right|\left| {\vec c} \right|\cos \dfrac{\pi }{3} \\
\Rightarrow \dfrac{1}{2}\left| {\vec b} \right|\left| {\vec c} \right| \\
\Rightarrow \dfrac{1}{2} \\
$ $\left( {\because \cos \dfrac{\pi }{3} = \dfrac{1}{2}} \right)$ …… (10)
We have, \[\vec b \cdot \left( {\vec a \times \vec b} \right) + \vec b \cdot \left( {\vec b \times \vec c} \right) = p\left( {\vec b \cdot \vec a} \right) + q\left( {\vec b \cdot \vec b} \right) + r\left( {\vec b \cdot \vec c} \right)\]
From (6), (7), (8), (9), and (10)
\[
0 + 0 = p\left( {\dfrac{1}{2}} \right) + q\left( 1 \right) + r\left( {\dfrac{1}{2}} \right) \\
\Rightarrow 0 = \dfrac{{p + 2q + r}}{2} \\
\Rightarrow p + 2q + r = 0 \\
\] …… (11)
Step 4: Similarly, Dot product of a vector \[\vec c\] with both sides on:
$\vec a \times \vec b + \vec b \times \vec c = p\vec a + q\vec b + r\vec c$
$ \Rightarrow \vec c \cdot \left( {\vec a \times \vec b + \vec b \times \vec c} \right) = \vec c \cdot \left( {p\vec a + q\vec b + r\vec c} \right)$
\[ \Rightarrow \vec c \cdot \left( {\vec a \times \vec b} \right) + \vec c \cdot \left( {\vec b \times \vec c} \right) = p\left( {\vec c \cdot \vec a} \right) + q\left( {\vec c \cdot \vec b} \right) + r\left( {\vec c \cdot \vec c} \right)\]
We know the angle between vectors \[\left( {\vec b \times \vec c} \right)\] and $\vec c$ is ${90^ \circ }$.
Hence, \[\vec c \cdot \left( {\vec b \times \vec c} \right) = 0\] …… (12)
\[
\left( {\vec a \cdot \vec c} \right) = \left| {\vec a} \right|\left| {\vec c} \right|\cos \dfrac{\pi }{3} \\
\Rightarrow \dfrac{1}{2}\left| {\vec a} \right|\left| {\vec c} \right| \\
\Rightarrow \dfrac{1}{2} \\
\] $\left( {\because \cos \dfrac{\pi }{3} = \dfrac{1}{2}} \right)$ …… (13)
$
\left( {\vec c \cdot \vec b} \right) = \left| {\vec c} \right|\left| {\vec b} \right|\cos \dfrac{\pi }{3} \\
\Rightarrow \dfrac{1}{2}\left| {\vec c} \right|\left| {\vec b} \right| \\
\Rightarrow \dfrac{1}{2} \\
$ $\left( {\because \cos \dfrac{\pi }{3} = \dfrac{1}{2}} \right)$ …… (14)
\[
\left( {\vec c \cdot \vec c} \right) = \left| {\vec c} \right|\left| {\vec c} \right|\cos 0^\circ \\
\Rightarrow {\left| {\vec c} \right|^2} \\
\Rightarrow 1 \\
\] $\left( {\because \cos 0^\circ = 1} \right)$ …… (15)
We have, \[\vec c \cdot \left( {\vec a \times \vec b} \right) + \vec c \cdot \left( {\vec b \times \vec c} \right) = p\left( {\vec c \cdot \vec a} \right) + q\left( {\vec c \cdot \vec b} \right) + r\left( {\vec c \cdot \vec c} \right)\]
From (12), (13), (14), and (15)
\[
\vec c \cdot \left( {\vec a \times \vec b} \right) + 0 = p\left( {\dfrac{1}{2}} \right) + q\left( {\dfrac{1}{2}} \right) + r\left( 1 \right) \\
\Rightarrow \vec c \cdot \left( {\vec a \times \vec b} \right) = \dfrac{{p + q + 2r}}{2} \\
\] …… (16)
Step 5: Use the property of scalar triple product:
For any three vectors \[\vec a,\vec b\]and $\vec c$, the scalar product of $\vec a$ and \[\left( {\vec b \times \vec c} \right)\] , i.e.,
\[\vec a \cdot \left( {\vec b \times \vec c} \right)\] is called the scalar triple product.
In the scalar triple product \[\vec a \cdot \left( {\vec b \times \vec c} \right)\] , the dot and cross can be interchanged. Indeed,
\[\vec a \cdot \left( {\vec b \times \vec c} \right) = \vec c \cdot \left( {\vec a \times \vec b} \right) = \left( {\vec a \times \vec b} \right) \cdot \vec c\]
Thus, L.H.S of equation (5) and (16) are equal hence,
\[
\dfrac{{2p + q + r}}{2} = \dfrac{{p + q + 2r}}{2} \\
\Rightarrow 2p + q + r = p + q + 2r \\
\Rightarrow 2p + r = p + 2r \\
\Rightarrow 2p - p = 2r - r \\
\]
$\because p = r$ …… (17)
Using equation (17) in (11) we get,
\[ \Rightarrow p + 2q + r = 0\]
\[
\Rightarrow p + 2q + p = 0 \\
\Rightarrow 2p + 2q = 0 \\
\Rightarrow p = - q \\
\]
Or $q = - p$ …… (18)
Step 6: Evaluate using equation (17) and (18)
$\dfrac{{{p^2} + 2{q^2} + {r^2}}}{{{q^2}}}$
$
\Rightarrow \dfrac{{{p^2} + 2{{\left( { - p} \right)}^2} + {p^2}}}{{{{\left( { - p} \right)}^2}}} \\
\Rightarrow \dfrac{{{p^2} + 2{p^2} + {p^2}}}{{{p^2}}} \\
\Rightarrow \dfrac{{4{p^2}}}{{{p^2}}} \\
\Rightarrow 4 \\
$
The value of $\dfrac{{{p^2} + 2{q^2} + {r^2}}}{{{q^2}}}$ is 4.
Note: The dot product \[\left( {\vec b \cdot \vec c} \right) = \left( {\vec c \cdot \vec b} \right)\] . And the cross product $\hat a \times \hat b = - \left( {\hat b \times \hat a} \right)$.
The scalar triple product is denoted by $\left[ {\vec a,\vec b,\vec c} \right]$, or $\left( {\vec a,\vec b,\vec c} \right)$. We thus have
\[\left[ {\vec a,\vec b,\vec c} \right]\] \[ = \vec a \cdot \left( {\vec b \times \vec c} \right)\]
If you had done a cross product one by one of each vector involved in the given equation, it would not give the desired result. As a result of the cross product is a vector, it will only complicate the calculation.
You’ll get the relation between scalars present in the equation.
Unit vectors are those vectors whose magnitude is 1.
Cross product (or vector product) of two nonzero vectors $\vec a$ and $\vec b$ is the product of the magnitude of both vectors $\vec a$ and $\vec b$, and sine of the angle between them. i.e.
$\hat a \times \hat b = \left| {\hat a} \right|\left| {\hat b} \right|\sin \theta {\text{ }}\hat n$, where $\theta $is the acute angle between vectors $\vec a$ and $\vec b$. Here $\hat n$is the unit vector perpendicular to the plane containing vectors $\vec a$ and $\vec b$.
Dot product (or scalar product) between two non-zero vectors $\vec a$ and $\vec b$ is the product of the magnitude of each vector and cosine of the angle between them:
\[\vec a \cdot \vec b = \left| a \right|\left| b \right|\cos \theta \] , where $\theta $ is the acute angle between vectors $\vec a$ and $\vec b$.
Complete step-by-step answer:
Step 1: Given that:
$\vec a \times \vec b + \vec b \times \vec c = p\vec a + q\vec b + r\vec c$
Where vectors \[\vec a,\vec b\]and $\vec c$ are unit vector,
Hence, $\left| {\vec a} \right| = 1$ ,
$\left| {\vec b} \right| = 1$ ,
$\left| {\vec c} \right| = 1$
The angle between every pair of them is $\dfrac{\pi }{3}$.
Step 2: Dot product of a vector \[\vec a\] with both sides on:
$\vec a \times \vec b + \vec b \times \vec c = p\vec a + q\vec b + r\vec c$
$\vec a \cdot \left( {\vec a \times \vec b + \vec b \times \vec c} \right) = \vec a \cdot \left( {p\vec a + q\vec b + r\vec c} \right)$
$ \Rightarrow \vec a \cdot \left( {\vec a \times \vec b} \right) + \vec a \cdot \left( {\vec b \times \vec c} \right) = p\left( {\vec a \cdot \vec a} \right) + q\left( {\vec a \cdot \vec b} \right) + r\left( {\vec a \cdot \vec c} \right)$
The resultant vector \[\left( {\vec a \times \vec b} \right)\] is perpendicular to the plane containing $\vec a$ and $\vec b$ i.e. the angle between vectors \[\left( {\vec a \times \vec b} \right)\] and $\vec a$ is ${90^ \circ }$. Also the angle between vectors \[\left( {\vec a \times \vec b} \right)\] and $\vec b$ is ${90^ \circ }$.
Hence, $\vec a \cdot \left( {\vec a \times \vec b} \right) = 0$ …… (1)
\[
\left( {\vec a \cdot \vec a} \right) = \left| {\vec a} \right|\left| {\vec a} \right|\cos 0^\circ \\
\Rightarrow {\left| {\vec a} \right|^2} \\
\Rightarrow 1 \\
\] $\left( {\because \cos 0^\circ = 1} \right)$ …… (2)
$
\left( {\vec a \cdot \vec b} \right) = \left| {\vec a} \right|\left| {\vec b} \right|\cos \dfrac{\pi }{3} \\
\Rightarrow \dfrac{1}{2}\left| {\vec a} \right|\left| {\vec b} \right| \\
\Rightarrow \dfrac{1}{2} \\
$ $\left( {\because \cos \dfrac{\pi }{3} = \dfrac{1}{2}} \right)$ …… (3)
$
\left( {\vec a \cdot \vec c} \right) = \left| {\vec a} \right|\left| {\vec c} \right|\cos \dfrac{\pi }{3} \\
\Rightarrow \dfrac{1}{2}\left| {\vec a} \right|\left| {\vec c} \right| \\
\Rightarrow \dfrac{1}{2} \\
$ $\left( {\because \cos \dfrac{\pi }{3} = \dfrac{1}{2}} \right)$ …… (4)
We have, $\vec a \cdot \left( {\vec a \times \vec b} \right) + \vec a \cdot \left( {\vec b \times \vec c} \right) = p\left( {\vec a \cdot \vec a} \right) + q\left( {\vec a \cdot \vec b} \right) + r\left( {\vec a \cdot \vec c} \right)$
From (1), (2), (3), and (4)
$0 + \vec a \cdot \left( {\vec b \times \vec c} \right) = p\left( 1 \right) + q\left( {\dfrac{1}{2}} \right) + r\left( {\dfrac{1}{2}} \right)$
$ \Rightarrow \vec a \cdot \left( {\vec b \times \vec c} \right) = \dfrac{{2p + q + r}}{2}$ …… (5)
Step 3: Similarly, Dot product of a vector \[\vec b\] with both sides on:
$\vec a \times \vec b + \vec b \times \vec c = p\vec a + q\vec b + r\vec c$
$ \Rightarrow \vec b \cdot \left( {\vec a \times \vec b + \vec b \times \vec c} \right) = \vec b \cdot \left( {p\vec a + q\vec b + r\vec c} \right)$
\[ \Rightarrow \vec b \cdot \left( {\vec a \times \vec b} \right) + \vec b \cdot \left( {\vec b \times \vec c} \right) = p\left( {\vec b \cdot \vec a} \right) + q\left( {\vec b \cdot \vec b} \right) + r\left( {\vec b \cdot \vec c} \right)\]
We know the angle between vectors \[\left( {\vec a \times \vec b} \right)\] and $\vec b$ is ${90^ \circ }$.
Hence, $\vec b \cdot \left( {\vec a \times \vec b} \right) = 0$ …… (6)
The resultant vector \[\left( {\vec b \times \vec c} \right)\] is perpendicular to the plane containing $\vec b$ and $\vec c$ i.e. the angle between vectors \[\left( {\vec b \times \vec c} \right)\] and $\vec b$ is ${90^ \circ }$. Also the angle between vectors \[\left( {\vec b \times \vec c} \right)\] and $\vec c$ is ${90^ \circ }$.
Hence, \[\vec b \cdot \left( {\vec b \times \vec c} \right) = 0\] …… (7)
$
\left( {\vec b \cdot \vec a} \right) = \left| {\vec b} \right|\left| {\vec a} \right|\cos \dfrac{\pi }{3} \\
\Rightarrow \dfrac{1}{2}\left| {\vec b} \right|\left| {\vec a} \right| \\
\Rightarrow \dfrac{1}{2} \\
$ $\left( {\because \cos \dfrac{\pi }{3} = \dfrac{1}{2}} \right)$ …… (8)
\[
\left( {\vec b \cdot \vec b} \right) = \left| {\vec b} \right|\left| {\vec b} \right|\cos 0^\circ \\
\Rightarrow {\left| {\vec b} \right|^2} \\
\Rightarrow 1 \\
\] $\left( {\because \cos 0^\circ = 1} \right)$ …… (9)
$
\left( {\vec b \cdot \vec c} \right) = \left| {\vec b} \right|\left| {\vec c} \right|\cos \dfrac{\pi }{3} \\
\Rightarrow \dfrac{1}{2}\left| {\vec b} \right|\left| {\vec c} \right| \\
\Rightarrow \dfrac{1}{2} \\
$ $\left( {\because \cos \dfrac{\pi }{3} = \dfrac{1}{2}} \right)$ …… (10)
We have, \[\vec b \cdot \left( {\vec a \times \vec b} \right) + \vec b \cdot \left( {\vec b \times \vec c} \right) = p\left( {\vec b \cdot \vec a} \right) + q\left( {\vec b \cdot \vec b} \right) + r\left( {\vec b \cdot \vec c} \right)\]
From (6), (7), (8), (9), and (10)
\[
0 + 0 = p\left( {\dfrac{1}{2}} \right) + q\left( 1 \right) + r\left( {\dfrac{1}{2}} \right) \\
\Rightarrow 0 = \dfrac{{p + 2q + r}}{2} \\
\Rightarrow p + 2q + r = 0 \\
\] …… (11)
Step 4: Similarly, Dot product of a vector \[\vec c\] with both sides on:
$\vec a \times \vec b + \vec b \times \vec c = p\vec a + q\vec b + r\vec c$
$ \Rightarrow \vec c \cdot \left( {\vec a \times \vec b + \vec b \times \vec c} \right) = \vec c \cdot \left( {p\vec a + q\vec b + r\vec c} \right)$
\[ \Rightarrow \vec c \cdot \left( {\vec a \times \vec b} \right) + \vec c \cdot \left( {\vec b \times \vec c} \right) = p\left( {\vec c \cdot \vec a} \right) + q\left( {\vec c \cdot \vec b} \right) + r\left( {\vec c \cdot \vec c} \right)\]
We know the angle between vectors \[\left( {\vec b \times \vec c} \right)\] and $\vec c$ is ${90^ \circ }$.
Hence, \[\vec c \cdot \left( {\vec b \times \vec c} \right) = 0\] …… (12)
\[
\left( {\vec a \cdot \vec c} \right) = \left| {\vec a} \right|\left| {\vec c} \right|\cos \dfrac{\pi }{3} \\
\Rightarrow \dfrac{1}{2}\left| {\vec a} \right|\left| {\vec c} \right| \\
\Rightarrow \dfrac{1}{2} \\
\] $\left( {\because \cos \dfrac{\pi }{3} = \dfrac{1}{2}} \right)$ …… (13)
$
\left( {\vec c \cdot \vec b} \right) = \left| {\vec c} \right|\left| {\vec b} \right|\cos \dfrac{\pi }{3} \\
\Rightarrow \dfrac{1}{2}\left| {\vec c} \right|\left| {\vec b} \right| \\
\Rightarrow \dfrac{1}{2} \\
$ $\left( {\because \cos \dfrac{\pi }{3} = \dfrac{1}{2}} \right)$ …… (14)
\[
\left( {\vec c \cdot \vec c} \right) = \left| {\vec c} \right|\left| {\vec c} \right|\cos 0^\circ \\
\Rightarrow {\left| {\vec c} \right|^2} \\
\Rightarrow 1 \\
\] $\left( {\because \cos 0^\circ = 1} \right)$ …… (15)
We have, \[\vec c \cdot \left( {\vec a \times \vec b} \right) + \vec c \cdot \left( {\vec b \times \vec c} \right) = p\left( {\vec c \cdot \vec a} \right) + q\left( {\vec c \cdot \vec b} \right) + r\left( {\vec c \cdot \vec c} \right)\]
From (12), (13), (14), and (15)
\[
\vec c \cdot \left( {\vec a \times \vec b} \right) + 0 = p\left( {\dfrac{1}{2}} \right) + q\left( {\dfrac{1}{2}} \right) + r\left( 1 \right) \\
\Rightarrow \vec c \cdot \left( {\vec a \times \vec b} \right) = \dfrac{{p + q + 2r}}{2} \\
\] …… (16)
Step 5: Use the property of scalar triple product:
For any three vectors \[\vec a,\vec b\]and $\vec c$, the scalar product of $\vec a$ and \[\left( {\vec b \times \vec c} \right)\] , i.e.,
\[\vec a \cdot \left( {\vec b \times \vec c} \right)\] is called the scalar triple product.
In the scalar triple product \[\vec a \cdot \left( {\vec b \times \vec c} \right)\] , the dot and cross can be interchanged. Indeed,
\[\vec a \cdot \left( {\vec b \times \vec c} \right) = \vec c \cdot \left( {\vec a \times \vec b} \right) = \left( {\vec a \times \vec b} \right) \cdot \vec c\]
Thus, L.H.S of equation (5) and (16) are equal hence,
\[
\dfrac{{2p + q + r}}{2} = \dfrac{{p + q + 2r}}{2} \\
\Rightarrow 2p + q + r = p + q + 2r \\
\Rightarrow 2p + r = p + 2r \\
\Rightarrow 2p - p = 2r - r \\
\]
$\because p = r$ …… (17)
Using equation (17) in (11) we get,
\[ \Rightarrow p + 2q + r = 0\]
\[
\Rightarrow p + 2q + p = 0 \\
\Rightarrow 2p + 2q = 0 \\
\Rightarrow p = - q \\
\]
Or $q = - p$ …… (18)
Step 6: Evaluate using equation (17) and (18)
$\dfrac{{{p^2} + 2{q^2} + {r^2}}}{{{q^2}}}$
$
\Rightarrow \dfrac{{{p^2} + 2{{\left( { - p} \right)}^2} + {p^2}}}{{{{\left( { - p} \right)}^2}}} \\
\Rightarrow \dfrac{{{p^2} + 2{p^2} + {p^2}}}{{{p^2}}} \\
\Rightarrow \dfrac{{4{p^2}}}{{{p^2}}} \\
\Rightarrow 4 \\
$
The value of $\dfrac{{{p^2} + 2{q^2} + {r^2}}}{{{q^2}}}$ is 4.
Note: The dot product \[\left( {\vec b \cdot \vec c} \right) = \left( {\vec c \cdot \vec b} \right)\] . And the cross product $\hat a \times \hat b = - \left( {\hat b \times \hat a} \right)$.
The scalar triple product is denoted by $\left[ {\vec a,\vec b,\vec c} \right]$, or $\left( {\vec a,\vec b,\vec c} \right)$. We thus have
\[\left[ {\vec a,\vec b,\vec c} \right]\] \[ = \vec a \cdot \left( {\vec b \times \vec c} \right)\]
If you had done a cross product one by one of each vector involved in the given equation, it would not give the desired result. As a result of the cross product is a vector, it will only complicate the calculation.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
Master Class 12 Social Science: Engaging Questions & Answers for Success
Master Class 12 Physics: Engaging Questions & Answers for Success
Master Class 12 Maths: Engaging Questions & Answers for Success
Master Class 12 Economics: Engaging Questions & Answers for Success
Master Class 12 Chemistry: Engaging Questions & Answers for Success
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
Master Class 12 Social Science: Engaging Questions & Answers for Success
Master Class 12 Physics: Engaging Questions & Answers for Success
Trending doubts
Which are the Top 10 Largest Countries of the World?
Draw a labelled sketch of the human eye class 12 physics CBSE
What are the major means of transport Explain each class 12 social science CBSE
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
Sulphuric acid is known as the king of acids State class 12 chemistry CBSE
Why should a magnesium ribbon be cleaned before burning class 12 chemistry CBSE