Verified Solution • 2025–26 Session moderate

QJEEM22Maba9c: JEE Main 2022 Physics - Class 11 JEEM22Maba9c Solution: Let be a vector which is perpendicular to the vector . If ,…

Let a→\overrightarrow a be a vector which is perpendicular to the vector 3i^+12j^+2k^3\widehat i + {1 \over 2}\widehat j + 2\widehat k . If a→×(2i^+k^)=2i^−13j^−4k^\overrightarrow a \times \left( {2\widehat i + \widehat k} \right) = 2\widehat i - 13\widehat j - 4\widehat k , then the projection of the vector a→\overrightarrow a on the vector 2i^+2j^+k^2\widehat i + 2\widehat j + \widehat k is :

Multiple Choice Options

  • A 13{1 \over 3}
  • B 1
  • C 53{5 \over 3}
  • D 73{7 \over 3}

Step-by-Step Verified Solution

Let a→=a1i^+a2j^+a3k^\overrightarrow a = {a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k

and a→ . (3i^−12j^+2k^)=0⇒3a1+a22+2a3=0\overrightarrow a \,.\,\left( {3\widehat i - {1 \over 2}\widehat j + 2\widehat k} \right) = 0 \Rightarrow 3{a_1} + {{{a_2}} \over 2} + 2{a_3} = 0 ..... (i)

and a→×(2i^+k^)=2i^−13j^−4k^\overrightarrow a \times (2\widehat i + \widehat k) = 2\widehat i - 13\widehat j - 4\widehat k

⇒a2i^+(2a3−a1)j^−2a2k^=2i^−13j^−4k^\Rightarrow {a_2}\widehat i + (2{a_3} - {a_1})\widehat j - 2{a_2}\widehat k = 2\widehat i - 13\widehat j - 4\widehat k

∴\therefore a2=2{a_2} = 2 ..... (ii)

and a1−2a3=13{a_1} - 2{a_3} = 13 ..... (iii)

From eq. (i) and (iii) : a1=3{a_1} = 3 and a3=−5{a_3} = - 5

∴\therefore a→=3i^+2j^−5k^\overrightarrow a = 3\widehat i + 2\widehat j - 5\widehat k

∴\therefore projection of a→\overrightarrow a on 2i^+2j^+k^=6+4−53=532\widehat i + 2\widehat j + \widehat k = {{6 + 4 - 5} \over 3} = {5 \over 3}

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