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  • Find the foot of the perpendicular from the point (2, 3, -8) to the line frac{4-x}{2}=frac{y}{6}=frac{1-z}{3} Also, find the perpendicular distance from the given point to the line.

Q: Find the foot of the perpendicular from the point (2, 3, -8) to the line \frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}

Also, find the perpendicular distance from the given point to the line.

Answers (1)

Given, the perpendicular from the point (let) C (2, 3, -8) to the line of which the equation is,

\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}

This can be re-written as,

\frac{x-4}{-2}=\frac{y}{6}=\frac{z-1}{-3}

Hence, the vector equation of the line is, -2\hat{i}+6\hat{j}-3\hat{k}

We must find the foot of the perpendicular from the point C (2, 3, -8) to given line, as well as the perpendicular distance from the given point C to the line.

To start with, let us locate the point of intersection between the point and the line.

Let us take,

\frac{x-4}{-2}=\frac{y}{6}=\frac{z-1}{-3}=\lambda

\frac{x-4}{-2}=\lambda,\frac{y}{6}=\lambda,\frac{z-1}{-3}=\lambda

from\frac{x-4}{-2}=\lambda\\\Rightarrow x-4=-2\lambda \\\Rightarrow x=4-2\lambda\\ \\ from \frac{y}{6}=\lambda\\ \Rightarrow y=6\lambda\\ \\ from \frac{z-1}{-3}=\lambda\\ \Rightarrow z-1=-3\lambda\\\Rightarrow z=1-3\lambda

We have,x = 4 - 2\lambda , y = 6\lambda, z = 1 - 3\lambda

Therefore, the coordinates of any point on the given line is \left ( 4 - 2\lambda , 6\lambda, 1 - 3\lambda \right )

Let us consider the foot of the perpendicular from C(2, 3, -8) on line to beL\left ( 4 - 2\lambda , 6\lambda, 1 - 3\lambda \right )
Therefore, the direction ratios of CL\left ( 4 - 2\lambda-2 , 6\lambda-3, 1 - 3\lambda-(-8) \right )

=\left ( 4 - 2\lambda-2 , 6\lambda-3, 1+8 - 3\lambda \right )\\ =\left ( 2-2\lambda , 6\lambda-3, 9 - 3\lambda \right )

Also, the direction ratio of the line is,\frac{x-4}{-2}=\frac{y}{6}=\frac{z-1}{-3} (-2, 6, -3).

Since L is the foot of the perpendicular on the line,

Sum of the product of these direction ratios \left ( 2-2\lambda , 6\lambda-3, 9 - 3\lambda \right ) and     (-2, 6, -3) = 0.

-2\left ( 2-2\lambda\right ) +6\left (6\lambda-3 \right )+(-3)\left ( 9 - 3\lambda \right )\\\\ \Rightarrow -4+4\lambda+36\lambda-18-27+9\lambda=0\\ \Rightarrow \left ( 4\lambda+36\lambda+9\lambda \right )+\left ( -4-18-27 \right )=0\\ \Rightarrow 49\lambda-49=0\\ \Rightarrow 49\lambda=49\\ \Rightarrow \lambda= \frac{49}{49}\\Hence \: \: \lambda=1

If we substitute this value of λ in L\left ( 4 - 2\lambda , 6\lambda, 1 - 3\lambda \right ), we get

\Rightarrow L\left ( 4 - 2\lambda , 6\lambda, 1 - 3\lambda \right )=L(4 - 2(1), 6(1), 1 - 3(1))

\Rightarrow L\left ( 4 - 2\lambda , 6\lambda, 1 - 3\lambda \right )=L(4 - 2, 6, 1 - 3)

\Rightarrow L\left ( 4 - 2\lambda , 6\lambda, 1 - 3\lambda \right )=L(2, 6, -2)

Now, we must calculate the perpendicular distance of point C from the line, that is point L.

In other words, we need to find \left | \vec{CL} \right |

We know, \vec{CL} =\left ( 2-2\lambda,6\lambda-3,9-3\lambda \right )

Substituting λ = 1,

\vec{CL} =\left ( 2-2(1),6(1)-3,9-3(1) \right )\\ \Rightarrow \vec{CL} =\left ( 2-2,6-3,9-3 \right )\\ \Rightarrow \vec{CL} =\left ( 0,3,6 \right )

To find \left | \vec{CL} \right |

\left | \vec{CL} \right |=\sqrt{0^{2}+3^{2}+6^{2}}\\ \Rightarrow \left | \vec{CL} \right |=\sqrt{0+9+36} \Rightarrow \left | \vec{CL} \right |=\sqrt{45}\\ \Rightarrow \left | \vec{CL} \right |=3\sqrt{5}

Therefore, the foot of the perpendicular from the point C to the given line is (2, 6, -2) and the perpendicular distance is 3\sqrt{5} units.

Posted by

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