Question

The equation of right circular cylinder of radius \(2\ \text{cm}\) whose axis passes through the point \((1,2,3)\) and has direction cosines proportional to \((2,-3,6)\) is:

• A. \(45x^2+40y^2+13z^2+36yz+24zx+12xy+42x+280y+126z+294=0\)
• B. \(45x^2+40y^2+13z^2+36yz-24zx+12xy-42x-280y-126z+294=0\)
• C. \(45x^2+40y^2+13z^2+36yz-24zx-12xy+42x+280y+126z-294=0\)
• D. \(45x^2-40y^2-13z^2+36yz-24zx-12xy+42x+280y+126z-294=0\)

Hint:

For a cylinder with known axis, use distance of a point from a line equal to the radius.

Answer 1

The Correct Option is B

Concept:
For a cylinder, the perpendicular distance of any point \((x,y,z)\) from the axis is constant and equal to radius.

Step 1: Axis data.
Point on axis: \[ (1,2,3) \] Direction vector: \[ \vec{d}=2\hat{i}-3\hat{j}+6\hat{k} \] \[ |\vec{d}|^2=2^2+(-3)^2+6^2=49 \] Radius: \[ r=2 \]

Step 2: Use distance from point to line. \[ \frac{|(\vec{r}-\vec{r_0})\times \vec{d}|^2}{|\vec{d}|^2}=r^2 \] \[ |(\vec{r}-\vec{r_0})\times \vec{d}|^2=4\times49=196 \]

Step 3: Expanding the equation.
After expanding: \[ 45x^2+40y^2+13z^2+36yz-24zx+12xy-42x-280y-126z+294=0 \] \[ \therefore \text{Correct Answer is (B)} \]