Q37 / 90 · JEE Main 2023 (30 Jan Shift 2) · Mathematics

A vector v→ in the first octant is inclined to the x axis at 60°, to the y-axis at 45° and to the z-axis at an acute angle. If a plane passing through the points 2,-1,1 and a,b,c, is normal to v→, then

  • A
    2a+b+c=1
  • B
    a+b+2c=1
  • Correct answer
    C
    a+2b+c=1
  • D
    2a-b+c=1

Solution

Direction cosines l,m and n of given vector v→,

l=cos60∘=12

m=cos45∘=12

n=cosγ

v→=cos60°i^+cos45°j^+cosγk^....(1)

We know,

l2+m2+n2=1

⇒14+12+cos2γ=1  (γ→Acute)

⇒cosγ=12

⇒ γ=60°

Equation of plane is 12x-2+12y+1+12z-1=0

⇒x2+y2+z2=12

⇒  x+2y+z=1

a,b,c lies on it,

⇒  a+2b+c=1

Hence, the correct answer is option: 'C'.

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