CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

Find the vector equation of the plane passing through the intersection of the planes r(^i+^j+^k)=6 and r(2^i+3^j+4^k)=5 and the point (1,1,1)


Solution

Here, n1=^i+^j+^k and n2=^2i+3^j+4^k
d1=6 and d2=5
Hence, using the relation r.(n1+λn2)=d1+λd2 , we get 

r[(^i+^j+^k)+λ(^2i+3^j+4^k)]=65λ
r[(1+2λ)^i+(1+3λ)^j+(1+4λ)^k]=65λ   (1)
where, λ is some real number.

Taking r=x^i+y^j+z^k, we get
(x^i+y^j+z^k).[(1+2λ)^i+(1+3λ)^j+(1+4λ)^k]=65λ
(1+2λ)x+(1+3λ)y+(1+4λ)z=65λ
(x+y+z6)+λ(2x+3y+4z+5)=0   (2)

Given that the plane passes through the point (1,1,1), it must satisfy equation(2), i.e.,
(1+1+16)+λ(2+3+4+5)=0
3+λ(14)=0λ=314

Putting the value of λ in (1), we get 
 
r[(1+37)^i+(1+914)^j+(1+67)^k]=61514
r(107^i+2314^j+137^k)=6914
r(20^i+23^j+26^k)=69
Which is the required vector equation of the plane.

Mathematics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image