Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Mathematics
The vector equation of the sphere whose centre is the point (1, 0, 1) and radius is 4, is:
Q. The vector equation of the sphere whose centre is the point (1, 0, 1) and radius is 4, is:
1603
217
Bihar CECE
Bihar CECE 2006
Report Error
A
∣
r
−
(
i
^
+
k
^
)
∣
=
4
B
∣
r
+
(
i
^
+
k
^
)
∣
=
4
2
C
r
(
i
^
+
k
^
)
=
4
D
r
(
i
^
+
k
^
)
=
4
2
Solution:
Given, centre of sphere
=
(
1
,
0
,
1
)
and radius
=
4
∴
Vector equation of sphere is
∣
r
−
a
∣
=
R
and centre of sphere in vector form
=
i
^
+
0
j
^
+
k
^
=
i
^
+
k
^
Hence, the vector equation of sphere is
∣
r
−
(
i
^
+
k
^
)
∣
=
4