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Three Dimensional Geometry

JEE Main 2024 (Online) 29th January Evening ShiftNumericalMedium📈 Topic in 43% papers+4
Ques. 1JEE Main 2024, 29 January (Evening Shift)

Let O be the origin, and M and N be the points on the lines x-54= y-41= z-53 and x+812= y+25= z+119 respectively such that MN is the shortest distance between the given lines. Then OM ON is equal to .

Answer

9

Solution

M(4 +5, +4, 3 +5), N(12 -8, 5 -2, 9 -11)

MN=(12 -4 -13, 5 - -6, 9 -3 -16)

b1 b2= vmatrix i& j& k\\4&1&3\\12&5&9 vmatrix=-6 i+0 j+8 k

MN both lines ⇒ MN b1 b2 (-3, 0, 4)

y-component =0 =5 -6

MN=(11-8 , 0, 2-6 ); 11-8 -3= 2-6 4 =1, =-1

M(1, 3, 2), N(4, 3, -2)

OM ON=4+9-4=9

Answer: 9

Keep practising — Three Dimensional Geometry

Ques. 2JEE Main 2024, 27 January (Morning Shift)

The distance of the point (7,-2,11) from the line x-61= y-40= z-83 along the line x-52= y-1-3= z-56, is :

  1. 12
  2. 18
  3. 21
  4. 14
Ques. 3JEE Main 2024, 27 January (Morning Shift)

If the shortest distance between the lines x-41= y+12= z-3 and x- 2= y+14= z-2-5 is 6 5, then the sum of all possible values of is :

  1. 10
  2. 5
  3. 7
  4. 8
Ques. 4JEE Main 2024, 27 January (Evening Shift)

Let the position vectors of the vertices A, B and C of a triangle be 2 i+2 j+ k, i+2 j+2 k and 2 i+ j+2 k respectively. Let l1, l2 and l3 be the lengths of perpendiculars drawn from the orthocentre of the triangle on the sides AB, BC and CA respectively, then l12+l22+l32 equals:

  1. 14
  2. 15
  3. 13
  4. 12
Ques. 5JEE Main 2024, 27 January (Evening Shift)

Let the image of the point (1,0,7) in the line x1= y-12= z-23 be the point ( , , ). Then which one of the following points lies on the line passing through ( , , ) and making angles 2 3 and 3 4 with y-axis and z-axis respectively and an acute angle with x-axis ?

  1. (1,-2,1+ 2 )
  2. (3,-4,3+2 2 )
  3. (3,4,3-2 2 )
  4. (1,2,1- 2 )
Ques. 6JEE Main 2024, 27 January (Evening Shift)

The lines x-22= y-2= z-716 and x+34= y+23= z+21 intersect at the point P. If the distance of P from the line x+12= y-13= z-11 is l, then 14l2 is equal to .

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