Advertisements
Advertisements
प्रश्न
Find the equation of the perpendicular drawn from the point P (2, 4, −1) to the line \[\frac{x + 5}{1} = \frac{y + 3}{4} = \frac{z - 6}{- 9} .\] Also, write down the coordinates of the foot of the perpendicular from P.
उत्तर
Let L be the foot of the perpendicular drawn from the point P (2, 4,-1) to the given line.
The coordinates of a general point on the line
\[\frac{x + 5}{1} = \frac{y + 3}{4} = \frac{z - 6}{- 9}\] are given by
\[\frac{x + 5}{1} = \frac{y + 3}{4} = \frac{z - 6}{- 9} = \lambda\]
\[ \Rightarrow x = \lambda - 5\]
\[ y = 4\lambda - 3\]
\[ z = - 9\lambda + 6\]
Let the coordinates of L be
\[\left( \lambda - 5, 4\lambda - 3, - 9\lambda + 6 \right)\]
The direction ratios of PL are proportional to
\[\lambda - 5 - 2, 4\lambda - 3 - 4, - 9\lambda + 6 + 1, i . e . \lambda - 7, 4\lambda - 7, - 9\lambda + 7\]
The direction ratios of the given line are proportional to 1, 4, - 9, but PL is perpendicular to the given line.
\[\therefore 1\left( \lambda - 7 \right) + 4\left( 4\lambda - 7 \right) - 9\left( - 9\lambda + 7 \right) = 0\]
\[ \Rightarrow \lambda = 1\]
Substituting
\[ \Rightarrow \lambda = 1\] in
\[\left( \lambda - 5, 4\lambda - 3, - 9\lambda + 6 \right)\] we get the coordinates of L as (-4,1,-3) Equation of the line PL is
\[\frac{x - 2}{- 4 - 2} = \frac{y - 4}{1 - 4} = \frac{z + 1}{- 3 + 1}\]
\[ = \frac{x - 2}{- 6} = \frac{y - 4}{- 3} = \frac{z + 1}{- 2}\]
APPEARS IN
संबंधित प्रश्न
If a line drawn from the point A( 1, 2, 1) is perpendicular to the line joining P(1, 4, 6) and Q(5, 4, 4) then find the co-ordinates of the foot of the perpendicular.
Find the separate equations of the lines represented by the equation 3x2 – 10xy – 8y2 = 0.
Let `A(bara)` and `B(barb)` be any two points in the space and `R(barr)` be a point on the line segment AB dividing it internally in the ratio m : n, then prove that `bar r=(mbarb+nbara)/(m+n)` . Hence find the position vector of R which divides the line segment joining the points A(1, –2, 1) and B(1, 4, –2) internally in the ratio 2 : 1.
Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector `3hati+2hatj-2hatk`.
Show that the lines `(x-5)/7 = (y + 2)/(-5) = z/1` and `x/1 = y/2 = z/3` are perpendicular to each other.
Find the vector and Cartesian equations of a line passing through (1, 2, –4) and perpendicular to the two lines `(x - 8)/3 = (y + 19)/(-16) = (z - 10)/7` and `(x - 15)/3 = (y - 29)/8 = (z - 5)/(-5)`
Find the vector equation of the lines which passes through the point with position vector `4hati - hatj +2hatk` and is in the direction of `-2hati + hatj + hatk`
Find the vector equation for the line which passes through the point (1, 2, 3) and parallel to the vector \[\hat{i} - 2 \hat{j} + 3 \hat{k} .\] Reduce the corresponding equation in cartesian from.
The cartesian equations of a line are x = ay + b, z = cy + d. Find its direction ratios and reduce it to vector form.
Find the vector equation of a line passing through the point with position vector \[\hat{i} - 2 \hat{j} - 3 \hat{k}\] and parallel to the line joining the points with position vectors \[\hat{i} - \hat{j} + 4 \hat{k} \text{ and } 2 \hat{i} + \hat{j} + 2 \hat{k} .\] Also, find the cartesian equivalent of this equation.
Show that the points whose position vectors are \[- 2 \hat{i} + 3 \hat{j} , \hat{i} + 2 \hat{j} + 3 \hat{k} \text{ and } 7 \text{ i} - \text{ k} \] are collinear.
Find the vector equation of the line passing through the point A(1, 2, –1) and parallel to the line 5x – 25 = 14 – 7y = 35z.
Show that the lines \[\frac{x - 5}{7} = \frac{y + 2}{- 5} = \frac{z}{1} \text { and }\frac{x}{1} = \frac{y}{2} = \frac{z}{3}\] are perpendicular to each other.
Find the angle between the following pair of line:
\[\overrightarrow{r} = \left( 3 \hat{i} + 2 \hat{j} - 4 \hat{k} \right) + \lambda\left( \hat{i} + 2 \hat{j} + 2 \hat{k} \right) \text{ and } \overrightarrow{r} = \left( 5 \hat{j} - 2 \hat{k} \right) + \mu\left( 3 \hat{i} + 2 \hat{j} + 6 \hat{k} \right)\]
Find the angle between the following pair of line:
\[\frac{x - 5}{1} = \frac{2y + 6}{- 2} = \frac{z - 3}{1} \text{ and } \frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 6}{5}\]
Find the equation of the line passing through the point (2, −1, 3) and parallel to the line \[\overrightarrow{r} = \left( \hat{i} - 2 \hat{j} + \hat{k} \right) + \lambda\left( 2 \hat{i} + 3 \hat{j} - 5 \hat{k} \right) .\]
Find the equations of the line passing through the point (2, 1, 3) and perpendicular to the lines \[\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{3} \text{ and } \frac{x}{- 3} = \frac{y}{2} = \frac{z}{5}\]
Show that the lines \[\frac{x}{1} = \frac{y - 2}{2} = \frac{z + 3}{3} \text{ and } \frac{x - 2}{2} = \frac{y - 6}{3} = \frac{z - 3}{4}\] intersect and find their point of intersection.
Show that the lines \[\frac{x - 1}{3} = \frac{y + 1}{2} = \frac{z - 1}{5} \text{ and } \frac{x + 2}{4} = \frac{y - 1}{3} = \frac{z + 1}{- 2}\] do not intersect.
Show that the lines \[\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7} \text{ and } \frac{x - 2}{1} = \frac{y - 4}{3} = \frac{z - 6}{5}\] intersect. Find their point of intersection.
Prove that the line \[\vec{r} = \left( \hat{i }+ \hat{j }- \hat{k} \right) + \lambda\left( 3 \hat{i} - \hat{j} \right) \text{ and } \vec{r} = \left( 4 \hat{i} - \hat{k} \right) + \mu\left( 2 \hat{i} + 3 \hat{k} \right)\] intersect and find their point of intersection.
Determine whether the following pair of lines intersect or not:
\[\frac{x - 1}{3} = \frac{y - 1}{- 1} = \frac{z + 1}{0} and \frac{x - 4}{2} = \frac{y - 0}{0} = \frac{z + 1}{3}\]
Find the foot of the perpendicular from (0, 2, 7) on the line \[\frac{x + 2}{- 1} = \frac{y - 1}{3} = \frac{z - 3}{- 2} .\]
Find the shortest distance between the following pairs of lines whose vector equations are: \[\overrightarrow{r} = \left( 1 - t \right) \hat{i} + \left( t - 2 \right) \hat{j} + \left( 3 - t \right) \hat{k} \text{ and } \overrightarrow{r} = \left( s + 1 \right) \hat{i} + \left( 2s - 1 \right) \hat{j} - \left( 2s + 1 \right) \hat{k} \]
Find the shortest distance between the following pairs of lines whose cartesian equations are : \[\frac{x - 1}{2} = \frac{y + 1}{3} = z \text{ and } \frac{x + 1}{3} = \frac{y - 2}{1}; z = 2\]
Find the shortest distance between the following pairs of parallel lines whose equations are: \[\overrightarrow{r} = \left( \hat{i} + 2 \hat{j} + 3 \hat{k} \right) + \lambda\left( \hat{i} - \hat{j} + \hat{k} \right) \text{ and } \overrightarrow{r} = \left( 2 \hat{i} - \hat{j} - \hat{k} \right) + \mu\left( - \hat{i} + \hat{j} - \hat{k} \right)\]
Find the equations of the lines joining the following pairs of vertices and then find the shortest distance between the lines
(1, 3, 0) and (0, 3, 0)
Find the shortest distance between the lines \[\frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1} \text{ and } \frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1}\]
Find the shortest distance between the lines \[\overrightarrow{r} = 6 \hat{i} + 2 \hat{j} + 2 \hat{k} + \lambda\left( \hat{i} - 2 \hat{j} + 2 \hat{k} \right) \text{ and } \overrightarrow{r} = - 4 \hat{i} - \hat{k} + \mu\left( 3 \hat{i} - 2 \hat{j} - 2 \hat{k} \right)\]
Write the cartesian and vector equations of X-axis.
Write the direction cosines of the line whose cartesian equations are 6x − 2 = 3y + 1 = 2z − 4.
The angle between the straight lines \[\frac{x + 1}{2} = \frac{y - 2}{5} = \frac{z + 3}{4} and \frac{x - 1}{1} = \frac{y + 2}{2} = \frac{z - 3}{- 3}\] is
The direction ratios of the line x − y + z − 5 = 0 = x − 3y − 6 are proportional to
The straight line \[\frac{x - 3}{3} = \frac{y - 2}{1} = \frac{z - 1}{0}\] is
Find the vector equation of the lines passing through the point having position vector `(-hati - hatj + 2hatk)` and parallel to the line `vecr = (hati + 2hatj + 3hatk) + λ(3hati + 2hatj + hatk)`.
The lines `(x - 1)/2 = (y + 1)/2 = (z - 1)/4` and `(x - 3)/1 = (y - k)/2 = z/1` intersect each other at point
