a maths doubt

[anushey]

can anyone solve part 3 of this question?i don't understand how the anwer is 6.

10 Two planes, m and n, have equations x + 2y − 2s = 1 and 2x − 2y + s = 7 respectively. The line l has

equation r = i + j − k + l(2i + j + 2k).

(i) Show that l is parallel to m. [3]

(ii) Find the position vector of the point of intersection of l and n. [3]

(iii) A point P lying on l is such that its perpendicular distances from m and n are equal. Find the

position vectors of the two possible positions for P and calculate the distance between them.













Permission

 

[littlecloud11]

m: x +2y -2z =1
n: 2x- 2y +z =7
l: i + j − k + s(2i + j + 2k)

Any point on line l= (1+2s) i + (1+s)j + (-1 +2s) k
For the perpendicular distance, D, from a point to a plane the formula is-
D= |ax +by +cz +d|/√(a^2+ b^2 +c^2)
where x,y and z represent the coordinates of the point and a,b, c and d represent the plane

The distance between P and plane n=
n: 2x- 2y +z -7 =0

a=2, b= -2 c=1 d= -7
x= (1+2s), y= (1+s) and z= (-1 +2s)

D1= 2*(1+2s) -2*(1+s) +1*(-1 +2s) -7/ √(2^2+ (-2)^2 +1^2)
D1= |4s -8|/ √9
D1 = |(4s-8)|/3

The distance between P and m=
m: x +2y -2z -1=0

a= 1, b=2, c= -2, d= -1
x= (1+2s), y= (1+s) and z= (-1 +2s)

D2= |1*(1+2s) +2*(1+s) -2*(-1 +2s) -1|/ √(1^2+ (2)^2 +(-2)^2)
D2 = 4/3

Given in the question D1 = D2
so, |(4s-8)|/3 =4/3
As there is a modulus you can form 2 separate equations-

4s-8 = 4
s= 3
OR,
-4s + 8 = 4
s= 1

substitute the two values of s in the equation of P
P = (1+2s) i + (1+s)j + (-1 +2s) k
when s= 3
P= 7i + 4j + 5k

when s=1
P= 3i + 2j + k

Now find the perpendicular distance= √(7-3)^2 + (4-2)^2 +(5-1)^2 =√36 =6

 

[anushey]

m: x +2y -2z =1
n: 2x- 2y +z =7
l: i + j − k + s(2i + j + 2k)

Any point on line l= (1+2s) i + (1+s)j + (-1 +2s) k
For the perpendicular distance, D, from a point to a plane the formula is-
D= |ax +by +cz +d|/√(a^2+ b^2 +c^2)
where x,y and z represent the coordinates of the point and a,b, c and d represent the plane

The distance between P and plane n=
n: 2x- 2y +z -7 =0

a=2, b= -2 c=1 d= -7
x= (1+2s), y= (1+s) and z= (-1 +2s)

D1= 2*(1+2s) -2*(1+s) +1*(-1 +2s) -7/ √(2^2+ (-2)^2 +1^2)
D1= |4s -8|/ √9
D1 = |(4s-8)|/3

The distance between P and m=
m: x +2y -2z -1=0

a= 1, b=2, c= -2, d= -1
x= (1+2s), y= (1+s) and z= (-1 +2s)

D2= |1*(1+2s) +2*(1+s) -2*(-1 +2s) -1|/ √(1^2+ (2)^2 +(-2)^2)
D2 = 4/3

Given in the question D1 = D2
so, |(4s-8)|/3 =4/3
As there is a modulus you can form 2 separate equations-

4s-8 = 4
s= 3
OR,
-4s + 8 = 4
s= 1

substitute the two values of s in the equation of P
P = (1+2s) i + (1+s)j + (-1 +2s) k
when s= 3
P= 7i + 4j + 5k

when s=1
P= 3i + 2j + k

Now find the perpendicular distance= √(7-3)^2 + (4-2)^2 +(5-1)^2 =√36 =6

THANNKYOU for taking out time.You are gr8 in vectors.

 

[littlecloud11]

THANNKYOU for taking out time.You are gr8 in vectors.

No problem.