KVPY 2019 SB SX Exam Previous Papers

Question : 84 of 120

Marks: +1, -0

Let

a↖{→}=i↖{\^}+j↖{\^}+k↖{\^},b↖{→}=2i↖{\^}+2j↖{\^}+k↖{\^}

and

c↖{→}=5i↖{\^}+j↖{\^}-k↖{\^}

be the three vectors. The area of the region formed by the set of points whose position vectors

r↖{→}

satisfy the equations

r ↖{→}. a↖{→} = 5

and

| r↖{→} -b ↖{→}| +| r ↖{→} - c ↖{→} | = 4

is closest to the integer

Solution:

| r↖{→} -b ↖{→}| +| r ↖{→} - c ↖{→} | = 4

is an ellipsoid with foci

b ↖{→},c ↖{→}

.
When it is cut by plane

r ↖{→}. a↖{→} = 5

i.e. x + y + z = 5 then we get ellipse with 2a = 4 and

2\ae = |b ↖{→}-c ↖{→}|=√14.

Area of ellipse =

πab=2π√{a^2-a^2e^2}

=2π√{4-14/4}=π√2

Go to Question:

Prev Question

Next Question