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  • Find the distance of a point (2, 4, -1) from the line frac{x+5}{1}=frac{y+3}{4}=frac{z-6}{-9}

 Find the distance of a point (2, 4, -1) from the line \frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}

 

Answers (1)

Given, the point P (2, 4, -1), the equation of the line is \frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}

We must find the distance of point P from this line.

Note, to find the distance between a point and a line, we should get foot of the perpendicular from the point on the line.

Let, P(2, 4, -1) be the given point and  be L:\frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}=\lambda the given line.

Direction ratio of the line L is (1, 4, -9) …(i)

Let us find any point on this line.

Taking L,

\frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}=\lambda

\frac{x+5}{1}=\lambda,\frac{y+3}{4}=\lambda,\frac{z-6}{-9}=\lambda

 Take \frac{x+5}{1}=\lambda\\ \Rightarrow x+5=\lambda\\ \Rightarrow x=\lambda - 5\\\\ Take \frac{y+3}{4}=\lambda\\ \Rightarrow y+3=4\lambda\\ \Rightarrow y=4\lambda-3\\ \\ Take \frac{z-6}{-9}=\lambda\\ \Rightarrow z-6=-9\lambda\\ \Rightarrow z=6-9\lambda

Therefore, any point on the line L is (\lambda - 5, 4\lambda - 3, 6 - 9\lambda)

Let this point be Q(\lambda - 5, 4\lambda - 3, 6 - 9\lambda), the foot of the perpendicular  from the point P (2, 4, -1) on the line L.

Hence, the direction ratio of PQ is given by

(\lambda - 5-2, 4\lambda - 3-4, 6 - 9\lambda-(-1))

=> Direction ratio of PQ=  (\lambda - 7, 4\lambda - 7, 7 - 9\lambda) …(ii)

Also, we know, if two lines are perpendicular to each other, then the dot product of their direction ratios should be 0.

Here, PQ is perpendicular to L. We have, from (i) and (ii),

Direction ratio of L = (1, 4, -9)

Direction ratio of PQ = (\lambda - 7, 4\lambda - 7, 7 - 9\lambda)

Therefore,

(1, 4, -9).(\lambda - 7, 4\lambda - 7, 7 - 9\lambda) = 0\\ \Rightarrow 1 (\lambda- 7) + 4 (4\lambda - 7) + (-9) (7 - 9\lambda) = 0\\ \Rightarrow \lambda - 7 + 16\lambda - 28 -63 + 81\lambda = 0\\ \Rightarrow \lambda + 16\lambda + 81\lambda - 7 - 28- 63 = 0\\ \Rightarrow 98\lambda - 98 = 0\\ \Rightarrow 98\lambda = 98\\ \Rightarrow \lambda = 1
Hence, the coordinate of Q, i.e. the foot of the perpendicular from the point on the given line is,
Q (\lambda - 5, 4\lambda - 3, 6 - 9\lambda) = Q (1 - 5, 4(1) - 3, 6 - 9)\\ \Rightarrow Q (\lambda - 5, 4\lambda - 3, 6 - 9\lambda) = Q (1 - 5, 4 - 3, 6 - 9)\\ \Rightarrow => Q(\lambda - 5, 4\lambda - 3, 6 - 9\lambda) = (-4, 1, -3)

Now, to find the perpendicular distance from P to the line, that is point   Q,

That is, to find \left | \vec{PQ} \right |

We know,

\left | \vec{PQ} \right |=(\lambda - 7, 4\lambda - 7, 7 - 9\lambda)

Substituting \lambda=1

\vec{PQ}=(1 - 7, 4(1)- 7, 7 - 9(1))\\ \Rightarrow \vec{PQ}=\left ( -6,4-7,7-9 \right )\\ \Rightarrow \vec{PQ}=\left ( -6,-3,-9 \right )

Now, to find

\left | \vec{PQ} \right |= \sqrt{(-6)^{2}+(-3)^{2}+(-2)^{2} }\\ \Rightarrow \left | \vec{PQ} \right |= \sqrt{36+9+4}\\ \Rightarrow \left | \vec{PQ} \right |= \sqrt{49}\\ \Rightarrow \left | \vec{PQ} \right |= 7

Therefore, the distance from the given point to the given line = 7 units.

 

Posted by

infoexpert24

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