Question

Write the equation of the right circular cylinder whose axis is the line (x-x_(1 /(l =(y-y_(1 /(m =(z-z_(1 /(n and whose radius is r.`​

Answer 1

Step-by-step explanation:

$Formula to find the distance of a point from a straight line:</p><p>Let P be the point (x_{1},\:y_{1},\:z_{1}) and a straight line be given by</p><p>\quad \frac{x-\alpha}{l}=\frac{y-\beta}{m}=\frac{z-\gamma}{n}\quad ...(i)</p><p>where l,\:m,\:n are direction cosines of the straight line.</p><p>\therefore the distance of the point P from the straight line (i) is given by</p><p>\bigg({\left|\begin{array}{cc}x_{1}-\alpha&y_{1}-\beta\\l&m\end{array}\right|}^{2}+{\left|\begin{array}{cc}y_{1}-\beta&z_{1}-\gamma\\m&n\end{array}\right|}^{2}+{\left|\begin{array}{cc}z_{1}-\gamma&x_{1}-\alpha\\n&l\end{array}\right|}^{2}\bigg)^{\frac{1}{2}}</p><p>Finding equation of a right circular cylinder:</p><p>Let the cylinder be of radius r and let the axis be the straight line</p><p>\quad \frac{x-\alpha}{l}=\frac{y-\beta}{m}=\frac{z-\gamma}{n}</p><p>Let (x',\:y',\:z') be any point on the cylinder.</p><p>We have to use this point to determine r.</p><p>Then the equation of the right circular cylinder be</p><p>(x-\alpha)^{2}+(y-\beta)^{2}+(z-\gamma)^{2}-\bigg[\frac{l(x-\alpha)+m(y-\beta)+n(z-\gamma)}{\sqrt{l^{2}+m^{2}+n^{2}}}\bigg]^{2}=r^{2}</p><p>Solution:</p><p>The axis of the right circular cylinder is given by</p><p>\quad\quad x-2=z,\:y=0</p><p>This can be put in symmetrical form as follows,</p><p>\quad \frac{x-2}{\frac{1}{\sqrt{2}}}=\frac{y-0}{0}=\frac{z-0}{\frac{1}{\sqrt{2}}}\quad ...(i)</p><p>The point on the cylinder is A\:(3,\:0,\:0).</p><p>Then the distance of the point A from the straight line (i) is</p><p>=\bigg({\left|\begin{array}{cc}3-2&0-0\\ \frac{1}{\sqrt{2}}&0\end{array}\right|}^{2}+{\left|\begin{array}{cc}0-0&0-0\\0&\frac{1}{\sqrt{2}}\end{array}\right|}^{2}+{\left|\begin{array}{cc}0-0&3-2\\ \frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\end{array}\right|}^{2}\bigg)^{\frac{1}{2}}</p><p>=\big(0+0+\frac{1}{2}\big)^{\frac{1}{2}}</p><p>=\frac{1}{\sqrt{2}}</p><p>This distance is radius of the right circular cylinder. That is r=\frac{1}{\sqrt{2}}.$

$Formula to find the distance of a point from a straight line:</p><p>Let P be the point (x_{1},\:y_{1},\:z_{1}) and a straight line be given by</p><p>\quad \frac{x-\alpha}{l}=\frac{y-\beta}{m}=\frac{z-\gamma}{n}\quad ...(i)</p><p>where l,\:m,\:n are direction cosines of the straight line.</p><p>\therefore the distance of the point P from the straight line (i) is given by</p><p>\bigg({\left|\begin{array}{cc}x_{1}-\alpha&y_{1}-\beta\\l&m\end{array}\right|}^{2}+{\left|\begin{array}{cc}y_{1}-\beta&z_{1}-\gamma\\m&n\end{array}\right|}^{2}+{\left|\begin{array}{cc}z_{1}-\gamma&x_{1}-\alpha\\n&l\end{array}\right|}^{2}\bigg)^{\frac{1}{2}}</p><p>Finding equation of a right circular cylinder:</p><p>Let the cylinder be of radius r and let the axis be the straight line</p><p>\quad \frac{x-\alpha}{l}=\frac{y-\beta}{m}=\frac{z-\gamma}{n}</p><p>Let (x',\:y',\:z') be any point on the cylinder.</p><p>We have to use this point to determine r.</p><p>Then the equation of the right circular cylinder be</p><p>(x-\alpha)^{2}+(y-\beta)^{2}+(z-\gamma)^{2}-\bigg[\frac{l(x-\alpha)+m(y-\beta)+n(z-\gamma)}{\sqrt{l^{2}+m^{2}+n^{2}}}\bigg]^{2}=r^{2}</p><p>Solution:</p><p>The axis of the right circular cylinder is given by</p><p>\quad\quad x-2=z,\:y=0</p><p>This can be put in symmetrical form as follows,</p><p>\quad \frac{x-2}{\frac{1}{\sqrt{2}}}=\frac{y-0}{0}=\frac{z-0}{\frac{1}{\sqrt{2}}}\quad ...(i)</p><p>The point on the cylinder is A\:(3,\:0,\:0).</p><p>Then the distance of the point A from the straight line (i) is</p><p>=\bigg({\left|\begin{array}{cc}3-2&0-0\\ \frac{1}{\sqrt{2}}&0\end{array}\right|}^{2}+{\left|\begin{array}{cc}0-0&0-0\\0&\frac{1}{\sqrt{2}}\end{array}\right|}^{2}+{\left|\begin{array}{cc}0-0&3-2\\ \frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\end{array}\right|}^{2}\bigg)^{\frac{1}{2}}</p><p>=\big(0+0+\frac{1}{2}\big)^{\frac{1}{2}}</p><p>=\frac{1}{\sqrt{2}}</p><p>This distance is radius of the right circular cylinder. That is r=\frac{1}{\sqrt{2}}.$

please swipe right

Answer 2

Step-by-step explanation:

Claims outstanding is shown as insurance company under