Verified Solution • 2025–26 Session moderate

QJEEM19Mac41a: JEE Main 2019 Mathematics - Class 12 JEEM19Mac41a Solution: If the length of the perpendicular from the point ( , 0, ) ( 0) …

If the length of the perpendicular from the point ( β\beta , 0, β\beta ) ( β\beta ≠\ne 0) to the line,
x1=y−10=z+1−1{x \over 1} = {{y - 1} \over 0} = {{z + 1} \over { - 1}} is 32\sqrt {{3 \over 2}} , then β\beta is equal to :

Multiple Choice Options

  • A 2
  • B 1
  • C -2
  • D -1

Step-by-Step Verified Solution

x1=y−10=z+1−1=p  P(β,0,β){x \over 1} = {{y - 1} \over 0} = {{z + 1} \over { - 1}} = p\,\,P\left( {\beta ,0,\beta } \right)

any point on line A = (p, 1, – p – 1)

Now, DR of AP ≡\equiv < p – β\beta , 1 – 0, – p – 1 – β\beta >

Which is perpendicular to line so

(p – β\beta ). 1 + 0.1 – 1(– p – 1 – β\beta ) = 0

⇒\Rightarrow p – β\beta + p + 1 + β\beta = 0

p=−12p = {{ - 1} \over 2}

Point A(−12,1−12)A\left( {{{ - 1} \over 2},1 - {1 \over 2}} \right)

Now, distance AP = 32\sqrt {{3 \over 2}}

⇒AP2=32\Rightarrow A{P^2} = {3 \over 2}

⇒(β+12)2+1+(β+12)2=32\Rightarrow {\left( {\beta + {1 \over 2}} \right)^2} + 1 + {\left( {\beta + {1 \over 2}} \right)^2} = {3 \over 2}

⇒2(β+12)2=12\Rightarrow 2{\left( {\beta + {1 \over 2}} \right)^2} = {1 \over 2}

⇒(β+12)2=14\Rightarrow {\left( {\beta + {1 \over 2}} \right)^2} = {1 \over 4}

⇒β=0,−1,(β≠0)\Rightarrow \beta = 0, - 1,\left( {\beta \ne 0} \right)

∴β=−1\therefore \beta = - 1

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