G&T 3D vectors(cartesian equation)

[Guest]

Hello i got this question that has three parts to it, i know how to do the first two parts but can u plz help me with the last. Ill write all the questions so u wont get confused.

Q: a) Find the vector and Cartesian equations of the sphere which has its centre at (-2,3,4) and radius of square root 59 units.

A: vector=[r-(-2i+3j+4k)]=root 59
Cartesian=(x+2)^2 + (y-3)^2 + (z-4)^2=59

Q: b)Find the vector equation of the line which passes through (-2,14,12) and is parallel to i-4j-5k

A: r=(&-2)i + (-4&+14)j + (-5&+12)k
*&=lamda*

Q: c)Find the coordinates of where the line and sphere intersects?

A:???? can u please help me with this

THANK YOU :wink:

 

[soroban]

Hello, barule22!

Q: a) Find the vector and Cartesian equations of the sphere which has centre (-2,3,4) and radius \(\displaystyle \sqrt{59.\)

A .Vector: .\(\displaystyle |r - (-2i + 3j + 4k)| \ =\ \sqrt{59}\)
. . Cartesian: .\(\displaystyle (x + 2)^2 + (y - 3)^2 + (z - 4)^2 \ =\ 59\)

Q: b)Find the vector equation of the line which passes through (-2,14,12) and is parallel to i - 4j - 5k.

A .\(\displaystyle r \ = \ (\lambda - 2)i + (-4\lambda + 14)j + (-5\lambda +12)k\)

Q: c) Find the coordinates of where the line and sphere intersect.

The line has parametric equations: .\(\displaystyle \begin{array}{ccc}x \ = \ \lambda - 2\qquad\qquad \\ \quad\; y \ = \ -4\lambda + 14\quad \\ z \ = \ -5\lambda + 12\end{array}\)

Substitute into the equation of the sphere:

. . . \(\displaystyle (\lambda - 2\ +\ 2)^2 + (-4\lambda + 14\ -\ 3)^2 + (-5\lambda + 12\ -\ 4)^2 \;= \;59\)

This simplifies to: .\(\displaystyle \lambda^2\ -\ 4\lambda\ +\ 3 \:= \:0\)

. . which factors: .\(\displaystyle (\lambda - 1)(\lambda - 3) \:= \:0\) . . . and has roots: .\(\displaystyle \lambda \,= \,1,\,3\)


Therefore, the intersections are: .\(\displaystyle (-1,\ 10,\ 7)\) and \(\displaystyle (1,\ 2,\ -3)\)