JEE MATHEMATICS MCQ: 3D GEOMETRY SYSTEM

1 Easy

1 of 1

Application +4 | -1

The equation of the sphere whose centre has the position vector

(3 \hat{i}+6 \hat{j}-4 \hat{k})

and which touches the plane

{r} \cdot(2 \hat{i}-2 \hat{j}-\hat{k})=10

View answer and explanation

Correct option: A

|{r}-(3 \hat{i}+6 \hat{j}-4 \hat{k})|=4

Option A (

|{r}-(3 \hat{i}+6 \hat{j}-4 \hat{k})|=4

) is correct because it represents the equation of the sphere with the given center and radius.

More Questions from 3D GEOMETRY SYSTEM

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\left( { - 10,\;8} \right)

\left( { - 6,\;12} \right)

\left( {4,\; - 2} \right)

\left( { - 2,\;4} \right)

y z

(1,-2,4)

(3,-4

5)

p

2 x+y-3=0

3 x-y-2=0

x-p y+5=0

O\left( {0,0} \right),A\left( {a,0} \right)

B\left( {0,a} \right).

A(1,0,3)

B(4,7,1)

C(3,5,3)

\frac{x}{1}=\frac{y}{2}=\frac{z}{3}, \frac{x}{2}=\frac{y}{1}=\frac{z}{3}

\frac{x-1}{1}=\frac{2-y}{2}, z=3