Verified Solution • 2025–26 Session moderate

QJEEM18Mabd82: JEE Main 2018 Physics - Class 11 JEEM18Mabd82 Solution: If and are unit vectors such that then is equal to :

If a→,  b→,\overrightarrow a ,\,\,\overrightarrow b , and C→\overrightarrow C are unit vectors such that a→+2b→+2c→=0→,\overrightarrow a + 2\overrightarrow b + 2\overrightarrow c = \overrightarrow 0 , then ∣a→×c→∣\left| {\overrightarrow a \times \overrightarrow c } \right| is equal to :

Multiple Choice Options

  • A 154{{\sqrt {15} } \over 4}
  • B 14{{1} \over {4}}
  • C 1516{{15} \over {16}}
  • D 1516{{\sqrt {15} } \over 16}

Step-by-Step Verified Solution

Given,

a→+2b→+2c→=0→\overrightarrow a + 2\overrightarrow b + 2\overrightarrow c = \overrightarrow 0

⇒\Rightarrow a→+2c→=−2b→\overrightarrow a + 2\overrightarrow c = - 2\overrightarrow b

Squaring both sides,

∣a→∣2+4a→.c→+4∣c→∣2=4∣b→∣2{\left| {\overrightarrow a } \right|^2} + 4\overrightarrow a .\overrightarrow c + 4{\left| {\overrightarrow c } \right|^2} = 4{\left| {\overrightarrow b } \right|^2}

⇒\Rightarrow 1 + 4a→.c→4\overrightarrow a .\overrightarrow c + 4 = 4   [as ∣a→∣2\left| {\overrightarrow a } \right|^2 = ∣b→∣2\left| {\overrightarrow b } \right|^2 = ∣c→∣2\left| {\overrightarrow c } \right|^2 = 1]

⇒\Rightarrow a→.c→=−14\overrightarrow a .\overrightarrow c = - {1 \over 4}

⇒\Rightarrow ∣a→∣∣c→∣cos⁡θ=−14\left| {\overrightarrow a } \right|\left| {\overrightarrow c } \right|\cos \theta = - {1 \over 4}

∴\therefore cos⁡θ\cos \theta = −14- {1 \over 4}

∴\therefore sin⁡2θ\sin ^2 \theta = 1 - cos⁡2θ\cos ^2 \theta

= 1 - 1161 \over 16

= 151615\over 16

∴\therefore sin θ\theta = 154{{\sqrt {15} } \over 4}

∴\therefore ∣a→×c→∣=∣a→∣∣c→∣sin⁡θ\left| {\overrightarrow a \times \overrightarrow c } \right| = \left| {\overrightarrow a } \right|\left| {\overrightarrow c } \right|\sin \theta

= 1 . 1 . 154{{\sqrt {15} } \over 4}

= 154{{\sqrt {15} } \over 4}

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