Question

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 1

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} \\

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.