SATHEE: Vectors 5 Question 13

Vectors 5 Question 13

13. Let $\vec{a}, \vec{b}$ and $\vec{c}$ be non-coplanar unit vectors, equally inclined to one another at an angle $θ$ . If $\vec{a} \times \vec{b} + \vec{b} \times \vec{c} = p \vec{a} + q \vec{b} + r \vec{c}$ , then find scalars $p, q$ and $r$ in terms of $θ$ .

(1997, 5M)

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Solution:

Since, $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar vectors.

$\begin{array}{ll} \Rightarrow & [\vec{a} \vec{b} \vec{c}] \neq 0 Also, & \vec{a} \times \vec{b} + \vec{b} \times \vec{c} = p \vec{a} + q \vec{b} + r \vec{c} \end{array}$

Taking dot product with $\vec{a}, \vec{b}$ and $\vec{c}$ respectively both sides, we get

$\begin{aligned} p + q \cos θ + r \cos θ = [\vec{a} \vec{b} \vec{c}] \\ p \cos θ + q + r \cos θ = 0 \end{aligned}$

and $p \cos θ + q \cos θ + r = [\vec{a} \vec{b} \vec{c}]$

On adding above equations,

$p + q + r = \frac{2 [\vec{a} \vec{b} \vec{c}]}{2 \cos θ + 1}$

On multiplying Eq. (iv) by $\cos θ$ and subtracting Eq. (i), we get

$p (\cos θ - 1) = \frac{2 [\vec{a} \vec{b} \vec{c}] \cos θ}{2 \cos θ + 1} - [\vec{a} \vec{b} \vec{c}]$

$\begin{array}{ll} \Rightarrow & p = \frac{[\vec{a} \vec{b} \vec{c}]}{(1 - \cos θ) (2 \cos θ + 1)} \\ Similarly, & q = \frac{- 2 [\vec{a} \vec{b} \vec{c}] \cos θ}{(1 + 2 \cos θ) (1 - \cos θ)} \end{array}$

and

$r = \frac{[\vec{a} \vec{b} \vec{c}]}{(1 + 2 \cos θ) (1 - \cos θ)}$

Now, $[\vec{a} \vec{b} \vec{c}]^{2} = | \begin{array}{lll} \vec{a} \cdot \vec{a} & \vec{a} \cdot \vec{b} & \vec{a} \cdot \vec{c} \vec{b} \cdot \vec{a} & \vec{b} \cdot \vec{b} & \vec{b} \cdot \vec{c} \vec{c} \cdot \vec{a} & \vec{c} \cdot \vec{b} & \vec{c} \cdot \vec{c} \end{array} | = | \begin{array}{ccc} 1 & \cos θ & \cos θ \cos θ & 1 & \cos θ \cos θ & \cos θ & 1 \end{array} |$

Applying $R_{1} \to R_{1} + R_{2} + R_{3}$

$= (1 + 2 \cos θ) | \begin{array}{ccc} 1 & 1 & 1 \\ \cos θ & 1 & \cos θ \\ \cos θ & \cos θ & 1 \end{array} |$

Applying $C_{2} \to C_{2} - C_{1}, C_{3} \to C_{3} - C_{1}$

$= (1 + 2 \cos θ) \cdot | \begin{array}{ccc} 1 & 0 & 0 \\ \cos θ & 1 - \cos θ & 0 \\ \cos θ & 0 & 1 - \cos θ \end{array} |$

$\begin{matrix} = (1 + 2 \cos θ) \cdot (1 - \cos θ)^{2} \\ \Rightarrow [\vec{a} \vec{b} \vec{c}] = (\sqrt{1 + 2 \cos θ}) \cdot (1 - \cos θ) \\ ∴ p = \frac{1}{\sqrt{1 + 2 \cos θ}}, q = \frac{- 2 \cos θ}{\sqrt{1 + 2 \cos θ}} and r = \frac{1}{\sqrt{1 + 2 \cos θ}} \end{matrix}$