Find the centre and radius of the sphere r- 2 -r- 4i- -2j- -6k- -11-0

Let r⃗ =xi⃗ +yj⃗ +zk⃗ , r=√( x ^ 2 + y ^ 2 + z ^ 2 ) ⇒ r ^ 2 = x ^ 2 + y ^ 2 + z ^ 2 ⇒r⃗·( 4i⃗ +2j⃗ 6k⃗ #160;) =4x+2y 6z So, r⃗ #1 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Equation of a Sphere : Radius-Centre Form

Find the cent...

Question

Find the centre and radius of the sphere ${\to r}^{2} - \to r \cdot (4 \to i + 2 \to j - 6 \to k) - 11 = 0$

Open in App

Solution

Suggest Corrections

Similar questions

{\to r}^{2} + \to r . (2 \to i - 2 \to j - 4 \to k) - 19 = 0

\to r . (\to i - 2 \to j + 4 \to k) + 8 = 0

\to r = (- \to i + 2 \to j + 3 \to k) + s (2 \to i + \to j + \to k)

\to r = (- 2 \to i + 3 \to j + 5 \to k) + t (\to i + 2 \to j + 3 \to k)

\to A = 2 \to i + \to k, \to B = \to i + \to j + \to k,

\to C = 4 \to i - 3 \to j + 7 \to k

\to R

\to R \times \to B = \to C \times \to B

\to R . \to A = 0

\to r = {\to r}_{1} + t {\to a}_{1}

\to r = {\to r}_{2} + t {\to a}_{2}

{\to r}_{1} = j, {\to a}_{1} = i + 2 j - k

{\to r}_{2} = i + j + k, {\to a}_{2} = - 2 i - 2 j

\to r = (\to i + 2 \to j - 5 \to k) + t (2 \to i - 3 \to j + 4 \to k)

\to r \cdot (2 \to i + 4 \to j - \to k) = 3

Join BYJU'S Learning Program