SATHEE: Vectors 3 Question 3

Vectors 3 Question 3

3. If $\vec{a}, \vec{b}, \vec{c}$ and $\vec{d}$ are the unit vectors such that $(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = 1$ and $\vec{a} \cdot \vec{c} = \frac{1}{2}$ , then

(2009)

(a) $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar

(b) $\vec{a}, \vec{b}, \vec{d}$ are non-coplanar

(d) $\vec{a}, \vec{d}$ are parallel and $\vec{b}, \vec{c}$ are parallel

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Answer:

Correct Answer: 3. (c)

Solution:

Let angle between $\vec{a}$ and $\vec{b}$ be $θ_{1}, \vec{c}$ and $\vec{d}$ be $θ_{2}$ and $\vec{a} \times \vec{b}$ and $\vec{b} \times \vec{d}$ be $θ$ .

Since, $(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = 1$

$\Rightarrow \sin θ_{1} \cdot \sin θ_{2} \cdot \cos θ = 1$

$\Rightarrow θ_{1} = 90^{\circ}, θ_{2} = 90^{\circ}, θ = 0^{\circ}$

$\Rightarrow \vec{a} ⊥ \vec{b}, \vec{c} ⊥ \vec{d}, (\vec{a} \times \vec{b}) | (\vec{c} \times \vec{d})$

So, $\vec{a} \times \vec{b} = k (\vec{c} \times \vec{d})$ and $\vec{a} \times \vec{b} = k (\vec{c} \times \vec{d})$

$\Rightarrow (\vec{a} \times \vec{b}) \cdot \vec{c} = k (\vec{c} \times \vec{d}) \cdot \vec{c}$

an $d (\vec{a} \times \vec{b}) \cdot \vec{d} = k (\vec{c} \times \vec{d}) \cdot \vec{d}$

$\Rightarrow [\vec{a} \vec{b} \vec{c}] = 0$ and $[\vec{a} \vec{b} \vec{d}] = 0$

$\Rightarrow \vec{a}, \vec{b}, \vec{c}$ and $\vec{a}, \vec{b}, \vec{d}$ are coplanar vectors, so options (a) and (b) are incorrect.

Let $\vec{b} | \vec{d} \Rightarrow \vec{b} = \pm \vec{d}$

$\begin{aligned} As (\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = 1 \Rightarrow (\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{b}) = \pm 1 \\ \Rightarrow [\vec{a} \times \vec{b} \vec{c} \vec{b}] = \pm 1 \Rightarrow [\vec{c} \vec{b} \vec{a} \times \vec{b}] = \pm 1 \\ \Rightarrow \vec{c} \cdot [\vec{b} \times (\vec{a} \times \vec{b})] = \pm 1 \Rightarrow \vec{c} \cdot [\vec{a} - (\vec{b} \cdot \vec{a}) \vec{b}] = \pm 1 \\ \Rightarrow \vec{c} \cdot \vec{a} = \pm 1 [∵ \vec{a} \cdot \vec{b} = 0] \end{aligned}$

which is a contradiction, so option (c) is correct.

Let option (d) is correct.

$\begin{aligned} \Rightarrow & \vec{d} = \pm \vec{a} \\ and & \vec{c} = \pm \vec{b} \\ As & (\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = 1 \\ \Rightarrow & (\vec{a} \times \vec{b}) \cdot (\vec{b} \times \vec{a}) = \pm 1 \end{aligned}$