If planes x+4y−2z=1, x+7y−5z=β and x+5y+αz=5 intersect in a line in R3 then α+β is equal to 

If planes x+4y2z=1, x+7y5z=β and x+5y+αz=5 intersect in a line in R3 then α+β is equal to 

  1. A
  2. B
  3. C
  4. D

    Register to Get Free Mock Test and Study Material

    +91

    Verify OTP Code (required)

    I agree to the terms and conditions and privacy policy.

    Solution:

    Given planes intersect in a line. This means that

    the system of equations x+4y2z=1, x+7y5z=β and 

    x+5y+αz=5 has infinitely many solutions.

     D=0=D1=D2=D3

    Now, D=014217515α=0

        (7α+25)4(α+5)2(57)=03α+9=0α=3    D3=014117β155=0    (355β)4(5β)+2(57)=0β=13    α+β=3+13=10

    Chat on WhatsApp Call Infinity Learn

      Register to Get Free Mock Test and Study Material

      +91

      Verify OTP Code (required)

      I agree to the terms and conditions and privacy policy.