Let the plane P : 4x - y + z = 10 be rotated by an angle π2 about its line of intersection with the plane x + y - z = 4. If α is the distance of the point (2, 3, -4) from the new position of the plan…

Question

Let the plane P : 4x - y + z = 10 be rotated by an angle π2 about its line of intersection with the plane x + y - z = 4. If α is the distance of the point (2, 3, -4) from the new position of the plane P. then 35α is equal to

Accepted Answer

Correct answer: C. (c) We know that if a plane passes through the line of intersection of planes P_1 and P_2 then the family of planes is given by P_1+ P_2 where ∈R. Equation of plane after rotation : (4x-y+z-10 )+ (x+y-z-4 )=0 ⇒ (4+ )x+y ( -1 )+z (1- )-4 -10=0 Now angle between the new plane and 4x-y+z-10 is 90^ . ⇒n_1 n_2=0 ⇒ (4+ )4+ ( -1 ) (-1 )+ (1- )1=0 ⇒16+4 - +1+1- =0 ⇒2 =-18 ⇒ =-9 ∴ Equation of plane : -5x-10y+10z+26=0 Distance of plane from (2 ,3 ,-4 ) = ( -10-30-40+26 25+100+100 )= 5415= 35 =35 5415=7 543=7 18=126 Hence this is the required option

$Let the plane P : 4x - y + z = 10 be rotated by an angle \frac{π}{2} about its line of intersection with the plane x + y - z = 4. If α is the distance of the point (2, 3, -4) from the new position of the plane P. then 35 α is equal to$

Solution

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