arthham

JEE Advanced 2007 · Vector Algebra

Let the vectors PQ→\overrightarrow{PQ}, QR→\overrightarrow{QR}, RS→\overrightarrow{RS}, ST→\overrightarrow{ST}, TU→\overrightarrow{TU} and UP→\overrightarrow{UP} represent the sides of a regular hexagon. STATEMENT I: PQ→×(RS→+ST→)≠0⃗\overrightarrow{PQ} \times \left(\overrightarrow{RS} + \overrightarrow{ST}\right) \neq \vec{0} . because \ STATEMENT II: PQ→×RS→=0⃗\overrightarrow{PQ} \times \overrightarrow{RS} = \vec{0} and PQ→×ST→≠0⃗\overrightarrow{PQ} \times \overrightarrow{ST} \neq \vec{0} .

Answer: C. Statement I is True, Statement II is False.

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Written solution

Given: PQ→, QR→, RS→, ST→, TU→, UP→\text{Given: } \overrightarrow{PQ},\ \overrightarrow{QR},\ \overrightarrow{RS},\ \overrightarrow{ST},\ \overrightarrow{TU},\ \overrightarrow{UP}

are the sides of a regular hexagon.\text{are the sides of a regular hexagon.}

To find: whether Statement I and Statement II are true,\text{To find: whether Statement I and Statement II are true,}

and whether Statement II correctly explains Statement I.\text{and whether Statement II correctly explains Statement I.}

Regular hexagon PQRSTU (vertices in order)

equal sides, each interior angle =120∘\text{equal sides, each interior angle } = 120^\circ

PQ⃗, QR⃗, RS⃗, ST⃗, TU⃗, UP⃗\vec{PQ},\ \vec{QR},\ \vec{RS},\ \vec{ST},\ \vec{TU},\ \vec{UP}

∣PQ⃗∣=∣QR⃗∣=⋯=∣UP⃗∣=s|\vec{PQ}| = |\vec{QR}| = \dots = |\vec{UP}| = s

each side direction rotates 60∘ going round\text{each side direction rotates } 60^\circ \text{ going round}

PQ⃗∥ST⃗,∠(PQ⃗,RS⃗)=120∘\vec{PQ} \parallel \vec{ST}, \qquad \angle(\vec{PQ}, \vec{RS}) = 120^\circ

PQ⃗=−ST⃗,QR⃗=−TU⃗,RS⃗=−UP⃗\vec{PQ} = -\vec{ST}, \quad \vec{QR} = -\vec{TU}, \quad \vec{RS} = -\vec{UP}

1. expand PQ⃗×(RS⃗+ST⃗)1.\ \text{expand } \vec{PQ} \times (\vec{RS} + \vec{ST})

2. find which cross products vanish2.\ \text{find which cross products vanish}

3. test both statements3.\ \text{test both statements}

PQ⃗×(RS⃗+ST⃗)=PQ⃗×RS⃗+PQ⃗×ST⃗\vec{PQ} \times (\vec{RS} + \vec{ST}) = \vec{PQ} \times \vec{RS} + \vec{PQ} \times \vec{ST}

a⃗∥b⃗ ⇒ a⃗×b⃗=0⃗\vec{a} \parallel \vec{b} \ \Rightarrow \ \vec{a} \times \vec{b} = \vec{0}

PQ⃗∥ST⃗ ⇒ PQ⃗×ST⃗=0⃗\vec{PQ} \parallel \vec{ST} \ \Rightarrow \ \vec{PQ} \times \vec{ST} = \vec{0}

PQ⃗∦RS⃗ ⇒ PQ⃗×RS⃗≠0⃗\vec{PQ} \not\parallel \vec{RS} \ \Rightarrow \ \vec{PQ} \times \vec{RS} \neq \vec{0}

∣PQ⃗×RS⃗∣=∣PQ⃗∣∣RS⃗∣sin⁡120∘=s2⋅32≠0|\vec{PQ} \times \vec{RS}| = |\vec{PQ}||\vec{RS}|\sin 120^\circ = s^2 \cdot \dfrac{\sqrt{3}}{2} \neq 0

PQ⃗×(RS⃗+ST⃗)=PQ⃗×RS⃗+0⃗≠0⃗\vec{PQ} \times (\vec{RS} + \vec{ST}) = \vec{PQ} \times \vec{RS} + \vec{0} \neq \vec{0}

⇒Statement I is TRUE\Rightarrow \text{Statement I is TRUE}

Statement II claims PQ⃗×RS⃗=0⃗\text{Statement II claims } \vec{PQ} \times \vec{RS} = \vec{0}

But PQ⃗×RS⃗≠0⃗ ⇒ this claim is FALSE\text{But } \vec{PQ} \times \vec{RS} \neq \vec{0} \ \Rightarrow \ \text{this claim is FALSE}

Statement II claims PQ⃗×ST⃗≠0⃗\text{Statement II claims } \vec{PQ} \times \vec{ST} \neq \vec{0}

But PQ⃗×ST⃗=0⃗ ⇒ this claim is FALSE\text{But } \vec{PQ} \times \vec{ST} = \vec{0} \ \Rightarrow \ \text{this claim is FALSE}

Both claims of Statement II are false\text{Both claims of Statement II are false}

⇒Statement II is FALSE\Rightarrow \text{Statement II is FALSE}

Statement II is false⇒it cannot explain Statement I\text{Statement II is false} \Rightarrow \text{it cannot explain Statement I}

Statement I: True\text{Statement I: True}

Statement II: False\text{Statement II: False}

Statement I True, Statement II False\text{Statement I True, Statement II False}

Answer: (3)\boxed{\text{Answer: (3)}}

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