Verified Solution • 2025–26 Session moderate

QJEEM19Ma328c: JEE Main 2019 Physics - Class 11 JEEM19Ma328c Solution: Let and , for some real x. Then = r is possible if :

Let a→=3i∧+2j∧+xk∧\mathop a\limits^ \to = 3\mathop i\limits^ \wedge + 2\mathop j\limits^ \wedge + x\mathop k\limits^ \wedge and b→=i∧−j∧+k∧\mathop b\limits^ \to = \mathop i\limits^ \wedge - \mathop j\limits^ \wedge + \mathop k\limits^ \wedge , for some real x. Then ∣a→×b→∣\left| {\mathop a\limits^ \to \times \mathop b\limits^ \to } \right| = r is possible if :

Multiple Choice Options

  • A 0 < r < 32\sqrt {{3 \over 2}}
  • B 332<r<5323\sqrt {{3 \over 2}} < r < 5\sqrt {{3 \over 2}}
  • C r≥532r \ge 5\sqrt {{3 \over 2}}
  • D 32<r≤332\sqrt {{3 \over 2}} < r \le 3\sqrt {{3 \over 2}}

Step-by-Step Verified Solution

a→×b→=|i^j^k^32x1−11|

= (2 + x) i^{\widehat i} + (3 - x) j^{\widehat j} - 5 k^{\widehat k}

∣a→×b→∣\left| {\overrightarrow a \times \overrightarrow b } \right| = r

= (2+x)2+(x−3)2+(−5)2\sqrt {{{\left( {2 + x} \right)}^2} + {{\left( {x - 3} \right)}^2} + {{\left( { - 5} \right)}^2}}

⇒\Rightarrow r = 4+4x+x2+x2+9−6x+25\sqrt {4 + 4x + {x^2} + {x^2} + 9 - 6x + 25}

= 2x2−2x+38\sqrt {2{x^2} - 2x + 38}

= 2(x2−x+14)+38−12\sqrt {2\left( {{x^2} - x + {1 \over 4}} \right) + 38 - {1 \over 2}}

= 2(x−12)2+752\sqrt {2{{\left( {x - {1 \over 2}} \right)}^2} + {{75} \over 2}}

⇒\Rightarrow r ≥\ge 752\sqrt {{{75} \over 2}}

⇒\Rightarrow r≥532r \ge 5\sqrt {{3 \over 2}}

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