1
Question
Plane passing through the points A(2, 1, 3), B(-1, 2, 4) and C(0, 2, 1). Determine its point of intersection with the line r=j+k+t(2i+k).
Plane passing through the points A(2, 1, 3), B(-1, 2, 4) and C(0, 2, 1). Determine its point of intersection with the line r=j+k+t(2i+k).
Open in App
Solution
The correct option is D (7, -1, 4).
If n be the normal of the plane ABC, thenn=→AB×→AC=−3i−8j−k
Plane is (r−a).n=0or r.n=a.nor r.(3i+8j+k)=17 ...(1)
The given line is r=(1+2t)i+(−1)j+(1+t)k=0 ...(2)
For point of intersection we have from (1) and (2) 3(1+2t)−8+(1+t)=17
∴7t=21 or t=3 putting for t in (2) the point of intersection is given by r=7i−j+4k.
∴ Its co-ordinates are (7,−1,4).
If n be the normal of the plane ABC, then
n=→AB×→AC=−3i−8j−k
Plane is (r−a).n=0or r.n=a.nor r.(3i+8j+k)=17 ...(1)
The given line is r=(1+2t)i+(−1)j+(1+t)k=0 ...(2)
For point of intersection we have from (1) and (2) 3(1+2t)−8+(1+t)=17
∴7t=21 or t=3 putting for t in (2) the point of intersection is given by r=7i−j+4k.
∴ Its co-ordinates are (7,−1,4).
Plane is (r−a).n=0or r.n=a.nor r.(3i+8j+k)=17 ...(1)
The given line is r=(1+2t)i+(−1)j+(1+t)k=0 ...(2)
For point of intersection we have from (1) and (2) 3(1+2t)−8+(1+t)=17
∴7t=21 or t=3 putting for t in (2) the point of intersection is given by r=7i−j+4k.
∴ Its co-ordinates are (7,−1,4).
Suggest Corrections
0
Similar questions
View More
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program
