Question

Find the equation of the plane which contains the line of intersection of the planes x+2y+3z-4=0 and 2x+y-z+5=0 and whose x intercepts is twice its z intercepts. Hence write the vector equation of the plane passing through the point (2,3,-1) and parallel to the plane obtained earlier
Plz immediately..

Answer 1

Answer:

7x + 11y +14z = 15.

7x + 11y +14z = 33.

Step-by-step explanation:

Equation of  any plane which contains the line of intersection of the planes P₁ and P₂ is given by P₁+λP₂.

Thus,the equation of the plane which contains the line of intersection of the planes x+2y+3z-4=0 and 2x+y-z+5=0 is given by

x+2y+3z-4 + λ(2x+y-z+5) = 0

(1+2λ)x + (2+λ)y + (3-λ)z = 4 - 5λ,

x/(4 - 5λ/1+2λ) + y/(4 - 5λ/2+λ) +z/(4 - 5λ/3-λ) =1 is the equation of plane in intercept form.

x-intercept, a is 4 - 5λ/1+2λ,

z-intercept,c is 4 - 5λ/3-λ

Given a = 2*c

=>4 - 5λ/1+2λ = 2*(4 - 5λ/3-λ)

=>3-λ=2*(1+2λ)=2 + 4λ

=>5λ = 1

=>λ=1/5

On substituting value of λ, we get

7x + 11y +14z = 15---(*).....Answer

Equation of plane parallel to 7x + 11y +14z = 15 will be in the form

7x + 11y +14z = k,---(1)

Given (1) passes through (2,3,-1)

=>k=7*2 +11*3+14*-1 =33

Thus equation of plane is 7x + 11y +14z = 33.

In vector form, since equation is of form 7x +11y +14x it is clear that normal to plane has dr's (7,11,14)

.Thus equation of plassing through a(2,3,-1) and having normal (n)

is (r-a).n = 0 If x = xi + yj +zk and a = 2i + 3j -k, n = 7i + 11j +14k

(Please note r,a ,n are vectors)