Question

Let $\overrightarrow{a}$, $\overrightarrow{b}$ and $\overrightarrow{c}$ be three non-coplanar unit vectors such that the angle between every pair of them is $\frac{\pi}{3}$. If $\overrightarrow{a} \times \overrightarrow{b} + \overrightarrow{b} \times \overrightarrow{c} = p\overrightarrow{a} + q\overrightarrow{b} + r\overrightarrow{c}$, where $p$, $q$ and $r$ are scalars. Match the entries of List-I with the correct entries of List-II.

• A. $\frac{p+q}{[\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{c}]}$ is equal to
• B. $\frac{q+r}{[\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{c}]}$ is equal to
• C. $\frac{r+p}{[\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{c}]}$ is equal to
• D. $2[\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{c}]^2 + 1$

Answer 1

Answer: (A), (B)

(P) - (1), (Q) - (1), (R) - (4), (S) - (4)

Answer 2

We are given three non-coplanar unit vectors $\overrightarrow{a}$, $\overrightarrow{b}$, $\overrightarrow{c}$ with $|\overrightarrow{a}| = |\overrightarrow{b}| = |\overrightarrow{c}| = 1$ and the angle between any pair is $\frac{\pi}{3}$. This means $\overrightarrow{a} \cdot \overrightarrow{b} = \overrightarrow{b} \cdot \overrightarrow{c} = \overrightarrow{c} \cdot \overrightarrow{a} = \cos(\frac{\pi}{3}) = \frac{1}{2}$.

The square of the scalar triple product is given by the determinant of the Gram matrix: $[\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{c}]^2 = \begin{vmatrix} \overrightarrow{a}\cdot\overrightarrow{a} & \overrightarrow{a}\cdot\overrightarrow{b} & \overrightarrow{a}\cdot\overrightarrow{c} \\ \overrightarrow{b}\cdot\overrightarrow{a} & \overrightarrow{b}\cdot\overrightarrow{b} & \overrightarrow{b}\cdot\overrightarrow{c} \\ \overrightarrow{c}\cdot\overrightarrow{a} & \overrightarrow{c}\cdot\overrightarrow{b} & \overrightarrow{c}\cdot\overrightarrow{c} \end{vmatrix} = \begin{vmatrix} 1 & 1/2 & 1/2 \\ 1/2 & 1 & 1/2 \\ 1/2 & 1/2 & 1 \end{vmatrix}$ $[\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{c}]^2 = 1(1 - 1/4) - 1/2(1/2 - 1/4) + 1/2(1/4 - 1/2) = 3/4 - 1/8 - 1/8 = 3/4 - 1/4 = 1/2$ Let $S = [\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{c}]$. So, $S^2 = 1/2$.

We are given $\overrightarrow{a} \times \overrightarrow{b} + \overrightarrow{b} \times \overrightarrow{c} = p\overrightarrow{a} + q\overrightarrow{b} + r\overrightarrow{c}$. Taking dot product with $\overrightarrow{a}$, $\overrightarrow{b}$, $\overrightarrow{c}$:

1. $(\overrightarrow{a} \times \overrightarrow{b} + \overrightarrow{b} \times \overrightarrow{c}) \cdot \overrightarrow{a} = p|\overrightarrow{a}|^2 + q(\overrightarrow{b}\cdot\overrightarrow{a}) + r(\overrightarrow{c}\cdot\overrightarrow{a})$ $0 + [\overrightarrow{b}\ \overrightarrow{c}\ \overrightarrow{a}] = p(1) + q(1/2) + r(1/2) \implies S = p + q/2 + r/2$
2. $(\overrightarrow{a} \times \overrightarrow{b} + \overrightarrow{b} \times \overrightarrow{c}) \cdot \overrightarrow{b} = p(\overrightarrow{a}\cdot\overrightarrow{b}) + q|\overrightarrow{b}|^2 + r(\overrightarrow{c}\cdot\overrightarrow{b})$ $[\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{b}] + 0 = p(1/2) + q(1) + r(1/2) \implies 0 = p/2 + q + r/2$
3. $(\overrightarrow{a} \times \overrightarrow{b} + \overrightarrow{b} \times \overrightarrow{c}) \cdot \overrightarrow{c} = p(\overrightarrow{a}\cdot\overrightarrow{c}) + q(\overrightarrow{b}\cdot\overrightarrow{c}) + r|\overrightarrow{c}|^2$ $[\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{c}] + 0 = p(1/2) + q(1/2) + r(1) \implies S = p/2 + q/2 + r$

From (2), $p + 2q + r = 0 \implies p+r = -2q$. Adding (1) and (3): $2S = (p+p/2) + (q/2+q/2) + (r/2+r) = 3p/2 + q + 3r/2$. $2S = \frac{3}{2}(p+r) + q$. Substitute $p+r = -2q$: $2S = \frac{3}{2}(-2q) + q = -3q + q = -2q \implies q = -S$. Since $p+r = -2q$, we have $p+r = -2(-S) = 2S$.

Substitute $q=-S$ into (1): $S = p - S/2 + r/2 \implies p+r/2 = 3S/2$. We have: $p+r = 2S$ $p+r/2 = 3S/2$ Subtracting the second from the first: $r/2 = S/2 \implies r = S$. Substituting $r=S$ into $p+r=2S$: $p+S=2S \implies p=S$. So, $p=S, q=-S, r=S$.

Now we evaluate the options: (P) $\frac{p+q}{S} = \frac{S+(-S)}{S} = \frac{0}{S} = 0$. This matches List-II (1). (Q) $\frac{q+r}{S} = \frac{-S+S}{S} = \frac{0}{S} = 0$. This matches List-II (1). (R) $\frac{r+p}{S} = \frac{S+S}{S} = \frac{2S}{S} = 2$. This matches List-II (4). (S) $2[\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{c}]^2 + 1 = 2S^2 + 1 = 2(1/2) + 1 = 1+1=2$. This matches List-II (4).

The correct matching is: (P) - (1), (Q) - (1), (R) - (4), (S) - (4).