Verified Solution • 2025–26 Session moderate

QJEEM19Maffe7: JEE Main 2019 Physics - Class 11 JEEM19Maffe7 Solution: Let       ,   …

Let   a→=i^+j^+2k^,\overrightarrow a = \widehat i + \widehat j + \sqrt 2 \widehat k,     b→=b1i^+b2j^+2k^\overrightarrow b = {b_1}\widehat i + {b_2}\widehat j + \sqrt 2 \widehat k ,    c→=5i^+j^+2k^\overrightarrow c = 5\widehat i + \widehat j + \sqrt 2 \widehat k   be three vectors such that the projection vector of b→\overrightarrow b on a→\overrightarrow a is a→\overrightarrow a .
If    a→+b→\overrightarrow a + \overrightarrow b    is perpendicular to c→\overrightarrow c , then ∣b→∣\left| {\overrightarrow b } \right| is equal to :

Multiple Choice Options

  • A 32\sqrt {32}
  • B 6
  • C 22\sqrt {22}
  • D 4

Step-by-Step Verified Solution

Projection of b→\overrightarrow b on a→\overrightarrow a is a→\overrightarrow a

∴\therefore     b→⋅a→∣a→∣=∣a→∣{{\overrightarrow b \cdot \overrightarrow a } \over {\left| {\overrightarrow a } \right|}} = \left| {\overrightarrow a } \right|

⇒\Rightarrow    b1+b2+22=2{{{b_1} + {b_2} + 2} \over 2} = 2

⇒\Rightarrow    b1+b2=2           ....(1){b_1} + {b_2} = 2\,\,\,\,\,\,\,\,\,\,\,....(1)

and a→\overrightarrow a + b→\overrightarrow b is perpendicular to c→\overrightarrow c

⇒\Rightarrow    (a→+b→)⋅c→=0\left( {\overrightarrow a + \overrightarrow b } \right) \cdot \overrightarrow c = 0

⇒\Rightarrow    5(b1+1)+(b2+1)+2(22)=05\left( {{b_1} + 1} \right) + \left( {{b_2} + 1} \right) + \sqrt 2 \left( {2\sqrt 2 } \right) = 0

⇒\Rightarrow    5b1+b2+10=0          ......(2)5{b_1} + {b_2} + 10 = 0\,\,\,\,\,\,\,\,\,\,......(2)

solving (1) & (2)

b1 == −- 3 and b2 == 5

⇒\Rightarrow    ∣b→∣=9+25+2=6\left| {\overrightarrow b } \right| = \sqrt {9 + 25 + 2} = 6

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