Verified Solution • 2025–26 Session moderate

QJEEM19Ma0feb: JEE Main 2019 Physics - Class 11 JEEM19Ma0feb Solution: Let and . If , where is parallel to and is perpendicular to ,…

Let α→=3i^+j^\overrightarrow \alpha = 3\widehat i + \widehat j and β→=2i^−j^+3k^\overrightarrow \beta = 2\widehat i - \widehat j + 3 \widehat k . If β→=β→1−β2→\overrightarrow \beta = {\overrightarrow \beta _1} - \overrightarrow {{\beta _2}} , where β→1{\overrightarrow \beta _1} is parallel to α→\overrightarrow \alpha and β2→\overrightarrow {{\beta _2}} is perpendicular to α→\overrightarrow \alpha , then β→1×β2→{\overrightarrow \beta _1} \times \overrightarrow {{\beta _2}} is equal to

Multiple Choice Options

  • A 12{1 \over 2} ( −3i^+9j^+5k^- 3\widehat i + 9\widehat j + 5\widehat k )
  • B 3i^−9j^−5k^3\widehat i - 9\widehat j - 5\widehat k
  • C −3i^+9j^+5k^- 3\widehat i + 9\widehat j + 5\widehat k
  • D 12{1 \over 2} ( 3i^−9j^+5k^3\widehat i - 9\widehat j + 5\widehat k )

Step-by-Step Verified Solution

Given α→=3i^+j^\overrightarrow \alpha = 3\widehat i + \widehat j

β→=2i^−j^+3k^\overrightarrow \beta = 2\widehat i - \widehat j + 3 \widehat k

β→1{\overrightarrow \beta _1} is parallel to α→\overrightarrow \alpha

∴\therefore β→1{\overrightarrow \beta _1} = λ\lambda α→\overrightarrow \alpha

⇒\Rightarrow β→1{\overrightarrow \beta _1} = 3λi^+λj^3\lambda \widehat i + \lambda \widehat j

β2→\overrightarrow {{\beta _2}} is perpendicular to α→\overrightarrow \alpha

β2→\overrightarrow {{\beta _2}} . α→\overrightarrow \alpha = 0

Let β2→\overrightarrow {{\beta _2}} = xi^+yj^+zk^x\widehat i + y\widehat j + z\widehat k

∴\therefore ( xi^+yj^+zk^x\widehat i + y\widehat j + z\widehat k ).( 3i^+j^3\widehat i + \widehat j ) = 0

⇒\Rightarrow 3x + y = 0

⇒\Rightarrow y = -3x

∴\therefore β2→\overrightarrow {{\beta _2}} = xi^−3xj^+zk^x\widehat i -3x\widehat j + z\widehat k

Given β→=β→1−β2→\overrightarrow \beta = {\overrightarrow \beta _1} - \overrightarrow {{\beta _2}}

⇒\Rightarrow (2i^−j^+3k^(2\widehat i - \widehat j + 3 \widehat k ) = ( 3λi^+λj^3\lambda \widehat i + \lambda \widehat j ) - ( xi^−3xj^+zk^x\widehat i -3x\widehat j + z\widehat k )

= (3λ−x)i^+(λ+3x)j^−zk^\left( {3\lambda - x} \right)\widehat i + \left( {\lambda + 3x} \right)\widehat j - z\widehat k

∴\therefore 3 λ\lambda - x = 2 ...(1)

λ\lambda + 3x = -1.....(2)

z = -3

Solving (1) and (2), we get

λ\lambda = 12{1 \over 2} and x = −12-{1 \over 2}

∴\therefore β→1{\overrightarrow \beta _1} = 32i^+12j^{3 \over 2}\widehat i + {1 \over 2}\widehat j

and β2→\overrightarrow {{\beta _2}} = −12i^+32j^−3k^- {1 \over 2}\widehat i + {3 \over 2}\widehat j - 3\widehat k

β→1×β→2=|i^j^k^32120−1232−3|

= 12{1 \over 2} ( −3i^+9j^+5k^- 3\widehat i + 9\widehat j + 5\widehat k )

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