Verified Solution • 2025–26 Session moderate

QJEEM17Ma88a6: JEE Main 2017 Physics - Class 11 JEEM17Ma88a6 Solution: Let and . Let be a vector such that , and the angle between and …

Let a→=2i^+j^−2k^\overrightarrow a = 2\widehat i + \widehat j -2 \widehat k and b→=i^+j^\overrightarrow b = \widehat i + \widehat j .

Let c→\overrightarrow c be a vector such that ∣c→−a→∣=3\left| {\overrightarrow c - \overrightarrow a } \right| = 3 ,

∣(a→×b→)×c→∣=3\left| {\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c } \right| = 3 and the angle between c→\overrightarrow c and a→×b→\overrightarrow a \times \overrightarrow b is 30∘30^\circ .

Then a→.c→\overrightarrow a .\overrightarrow c is equal to :

Multiple Choice Options

  • A 2
  • B 5
  • C 18{1 \over 8}
  • D 258{{25} \over 8}

Step-by-Step Verified Solution

Given:

a→=2i^+j^−2k^,  b→=i^+j^\overrightarrow a = 2\widehat i + \widehat j - 2\widehat k,\,\,\overrightarrow b = \widehat i + \widehat j

⇒\Rightarrow    ∣a→∣=3\left| {\overrightarrow a } \right| = 3

∴\therefore    a→×b→=2i^−2j^+k^\overrightarrow a \times \overrightarrow b = 2\widehat i - 2\widehat j + \widehat k

∣a→×b→∣=22+22+12=3\left| {\overrightarrow a \times \overrightarrow b } \right| = \sqrt {{2^2} + {2^2} + {1^2}} = 3

We have (a→×b→)×c→=∣a→×b→∣∣c→∣sin⁡30∘\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c = \left| {\overrightarrow a \times \overrightarrow b } \right|\left| {\overrightarrow c } \right|\sin 30^\circ

⇒\Rightarrow    ∣(a→×b→)×c→∣=3∣c→∣.12\left| {\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c } \right| = 3\left| {\overrightarrow c } \right|.{1 \over 2}

⇒\Rightarrow    3=3∣c→∣.123 = 3\left| {\overrightarrow c } \right|.{1 \over 2}

∴\therefore    ∣c→∣=2\left| {\overrightarrow c } \right| = 2

Now   ∣c→−a→∣=3\left| {\overrightarrow c - \overrightarrow a } \right| = 3

On squaring, we get

⇒\Rightarrow    c2+a2−2−c→.a→=9{c^2} + {a^2} - 2 - \overrightarrow c .\overrightarrow a = 9

⇒\Rightarrow    4+9−2−a→.c→=94 + 9 - 2 - \overrightarrow a .\overrightarrow c = 9

⇒\Rightarrow    a→.c→=2\overrightarrow a .\overrightarrow c = 2     [ ∵\because    c→.a→  =  a→.c→\overrightarrow c .\overrightarrow a \,\, = \,\,\overrightarrow a .\overrightarrow c ]

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