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Find the equation of the plane through (3,4,-1) which is parallel to the plane `vecr.(2hati-3hatj+5hatk)+7=0` |
Answer» The equation of any plane which is parallel to `vecr*(2hati-3hatj+5hatk)+7=0` is `" "vecr*(2hati-3hatj+5hatk)+lamda=0" "` (i) or `" "2x-3y+5z+lamda=0` Further (i) will pass through (3, 4, -1) if `(2)(3)+(-3)(4)+5(-1)+lamda=0 or -11+lamda=0 or lamda=0 or lamda=11` Thus, equation of the required plane is `vecr*(2hati-3hatj+5hatk)+11=0`. |
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