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Question: From the point \(\left( 1,-2,3 \right)\), lines are drawn to meet the sphere \({{x}^{2}}+{{y}^{2}}+{...

From the point (1,−2,3)\left( 1,-2,3 \right), lines are drawn to meet the sphere x2+y2+z2=4{{x}^{2}}+{{y}^{2}}+{{z}^{2}}=4 and they are divided internally in the ratio 2:32:3. The locus of the point of division is
A. 5x2+5y2+5z2−6x+12y+2z=05{{x}^{2}}+5{{y}^{2}}+5{{z}^{2}}-6x+12y+2z=0
B. 5(x2+y2+z2)=225\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)=22
C. 5x2+5y2+5z2−2xy−3yz−zx−6x+12y+5z=05{{x}^{2}}+5{{y}^{2}}+5{{z}^{2}}-2xy-3yz-zx-6x+12y+5z=0
D. None of these

Explanation

Solution

In this problem we need to calculate the locus of the point which divides the sphere and passes through the point. Now we will assume the point of division as (x1,y1,z1)\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right). Now we will substitute this point in the given sphere because it lies on the sphere. We will consider this equation as equation one. In the question we have given that the point (1,−2,3)\left( 1,-2,3 \right), (x1,y1,z1)\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right) are divided internally in the ratio 2:32:3. From this we will get the values of x1{{x}_{1}}, y1{{y}_{1}}, z1{{z}_{1}}. After getting these values we will substitute them in equation one and simplify the equation to get the locus of the point.

Formula Used:
1. Section formula when the points are divided internally mx2+nx1m+n\dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n}.

Complete Step by Step Procedure:
Given, the equation of the sphere is x2+y2+z2=4{{x}^{2}}+{{y}^{2}}+{{z}^{2}}=4.
Let the point that divides the sphere is (x1,y1,z1)\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right).
Now the above point is lies in the sphere, so that
x12+y12+z12=4.....(i){{x}_{1}}^{2}+y_{1}^{2}+z_{1}^{2}=4.....\left( \text{i} \right)
Now, let the coordinates of the point which divides the join of (1,−2,3)\left( 1,-2,3 \right) and (x1,y1,z1)\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right) in the ratio 2:32:3 be (x2,y2,z2)\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right). From the section formula we can write
x2=2x1+3(1)2+3⇒x1=5x−32{{x}_{2}}=\dfrac{2{{x}_{1}}+3\left( 1 \right)}{2+3}\Rightarrow {{x}_{1}}=\dfrac{5x-3}{2},
y2=2y1+3(−2)2+3⇒y1=5y+62{{y}_{2}}=\dfrac{2{{y}_{1}}+3\left( -2 \right)}{2+3}\Rightarrow {{y}_{1}}=\dfrac{5y+6}{2},
z2=2z1+3(3)2+3⇒z1=5z−92{{z}_{2}}=\dfrac{2{{z}_{1}}+3\left( 3 \right)}{2+3}\Rightarrow {{z}_{1}}=\dfrac{5z-9}{2}.
Substituting these values in the equation (i)\left( \text{i} \right), then we will get
(5x−32)2+(5y+62)2+(5z−92)2=4 ⇒25(x2+y2+z2)−30x+60y−90z+110=0 \begin{aligned} & {{\left( \dfrac{5x-3}{2} \right)}^{2}}+{{\left( \dfrac{5y+6}{2} \right)}^{2}}+{{\left( \dfrac{5z-9}{2} \right)}^{2}}=4 \\\ & \Rightarrow 25\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)-30x+60y-90z+110=0 \\\ \end{aligned}

Hence the locus of the point is 5(x2+y2+z2)−6(x−2y+3z)+22=05\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)-6\left( x-2y+3z \right)+22=0

Note:
In this problem they have mentioned that the points are divided internally, so we have used the section formula for the internal division. If they have mentioned that the points are divided externally, then we will use the section formula for external division which is mx2−nx1m−n\dfrac{m{{x}_{2}}-n{{x}_{1}}}{m-n}.