Advertisements
Advertisements
प्रश्न
Find the perpendicular distance of the point (3, −1, 11) from the line \[\frac{x}{2} = \frac{y - 2}{- 3} = \frac{z - 3}{4} .\]
Advertisements
उत्तर
Given equation of line AB is
`x/2=("y"-2)/3=("z"-3)/4=lambda`[say]
or `x/2=lambda,("y"-2)/3=lambda`
and `("z"-3)/4=lambda`
or x = 2λ, y = 3λ + 2
and z = 4λ + 3
∴ Any point P on the given line
= (2λ, 3λ + 2, 4λ + 3)
Let P be the foot of Perpendicular drawn from point
Q(3, -1, 11) on line AB. Now, DR's of line
QP = (2λ - 3, 3λ + 2 + 1, 4λ + 3 - 11)
= (2λ - 3, 3λ + 3, 4λ - 8)
Here, a1 = 2λ -3, b1 = 3λ + 3, c1 = 4λ - 8,
and a2 = 2, b2 = 3, c2 = 4
Since, QP ⊥ AB
∴ We have, a1a2 + b1b2 + c1c2 = 0 ...(i)
or 2(2λ - 3) + 3 (3λ + 3) + 4 (4λ - 8) = 0
or 4λ - 6 + 9λ + 9 + 16λ - 32 = 0
or 29λ - 29 = 0 or 29λ = 29 or λ = 1
∴ Foot of Perpendicular
P = (2, 3 + 2, 4 + 3)
= (2, 5, 7)
Now, equation of Perpendicular QP, Where Q (3, -1, 11) and P (2, 5, 7) is
`(x-3)/(2-3)=("y"+1)/(5+1)=("z"-11)/(7-11)`
`[("using two points form of equation of line"), ("i"."e". (x-x_1)/(x_2-x_1)=("y"-"y"_1)/("y"_2-"y"_1)=("z"-"z"_1)/("z"_2-"z"_1))]`
or `(x-3)/-1=("y"+1)/6=("z"-11)/-4`
Now, length of Perpendicular QP = distance between points Q(3, -1, 11) and P (2, 5, 7)
= `sqrt((2-3)^2+(5+1)^2+(7-11)^2)`
`[(∵ "distance"), (=sqrt((x_2-x_1)^2+("y"_2-"y"_1)^2+("z"_2-"z"_1)^2))]`
= `sqrt(1+36+16)`
= `sqrt53`
Hence, length of Perpendicular is `sqrt53`.
APPEARS IN
संबंधित प्रश्न
The Cartesian equations of line are 3x -1 = 6y + 2 = 1 - z. Find the vector equation of line.
Find the vector equation of the line passing through the points (−1, 0, 2) and (3, 4, 6).
Find the vector equation of a line which is parallel to the vector \[2 \hat{i} - \hat{j} + 3 \hat{k}\] and which passes through the point (5, −2, 4). Also, reduce it to cartesian form.
ABCD is a parallelogram. The position vectors of the points A, B and C are respectively, \[4 \hat{ i} + 5 \hat{j} -10 \hat{k} , 2 \hat{i} - 3 \hat{j} + 4 \hat{k} \text{ and } - \hat{i} + 2 \hat{j} + \hat{k} .\] Find the vector equation of the line BD. Also, reduce it to cartesian form.
Show that the three lines with direction cosines \[\frac{12}{13}, \frac{- 3}{13}, \frac{- 4}{13}; \frac{4}{13}, \frac{12}{13}, \frac{3}{13}; \frac{3}{13}, \frac{- 4}{13}, \frac{12}{13}\] are mutually perpendicular.
Find the angle between the following pair of line:
\[\overrightarrow{r} = \left( 4 \hat{i} - \hat{j} \right) + \lambda\left( \hat{i} + 2 \hat{j} - 2 \hat{k} \right) \text{ and }\overrightarrow{r} = \hat{i} - \hat{j} + 2 \hat{k} - \mu\left( 2 \hat{i} + 4 \hat{j} - 4 \hat{k} \right)\]
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 equation of the line passing through the point (1, 2, −4) and parallel to the line \[\frac{x - 3}{4} = \frac{y - 5}{2} = \frac{z + 1}{3} .\]
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 \[\frac{x - 1}{3} = \frac{y + 1}{2} = \frac{z - 1}{5} \text{ and } \frac{x + 2}{4} = \frac{y - 1}{3} = \frac{z + 1}{- 2}\] do not intersect.
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.
Find the foot of the perpendicular drawn from the point A (1, 0, 3) to the joint of the points B (4, 7, 1) and C (3, 5, 3).
A (1, 0, 4), B (0, −11, 3), C (2, −3, 1) are three points and D is the foot of perpendicular from A on BC. Find the coordinates of D.
Find the equation of the perpendicular drawn from the point P (−1, 3, 2) to the line \[\overrightarrow{r} = \left( 2 \hat{j} + 3 \hat{k} \right) + \lambda\left( 2 \hat{i} + \hat{j} + 3 \hat{k} \right) .\] Also, find the coordinates of the foot of the perpendicular from P.
Find the shortest distance between the following pairs of lines whose vector equations are: \[\vec{r} = 3 \hat{i} + 8 \hat{j} + 3 \hat{k} + \lambda\left( 3 \hat{i} - \hat{j} + \hat{k} \right) \text{ and } \vec{r} = - 3 \hat{i} - 7 \hat{j} + 6 \hat{k} + \mu\left( - 3 \hat{i} + 2 \hat{j} + 4 \hat{k} \right)\]
Find the shortest distance between the following pairs of lines whose vector equations are: \[\overrightarrow{r} = \left( 3 \hat{i} + 5 \hat{j} + 7 \hat{k} \right) + \lambda\left( \hat{i} - 2 \hat{j} + 7 \hat{k} \right) \text{ and } \overrightarrow{r} = - \hat{i} - \hat{j} - \hat{k} + \mu\left( 7 \hat{i} - 6 \hat{j} + \hat{k} \right)\]
Find the shortest distance between the following pairs of lines whose vector equations are: \[\overrightarrow{r} = \left( 8 + 3\lambda \right) \hat{i} - \left( 9 + 16\lambda \right) \hat{j} + \left( 10 + 7\lambda \right) \hat{k} \]\[\overrightarrow{r} = 15 \hat{i} + 29 \hat{j} + 5 \hat{k} + \mu\left( 3 \hat{i} + 8 \hat{j} - 5 \hat{k} \right)\]
Find the shortest distance between the following pairs of lines whose cartesian equations are : \[\frac{x - 1}{2} = \frac{y + 1}{3} = z \text{ and } \frac{x + 1}{3} = \frac{y - 2}{1}; z = 2\]
By computing the shortest distance determine whether the following pairs of lines intersect or not: \[\overrightarrow{r} = \left( \hat{i} + \hat{j} - \hat{k} \right) + \lambda\left( 3 \hat{i} - \hat{j} \right) \text{ and } \overrightarrow{r} = \left( 4 \hat{i} - \hat{k} \right) + \mu\left( 2 \hat{i} + 3 \hat{k} \right)\]
By computing the shortest distance determine whether the following pairs of lines intersect or not: \[\frac{x - 1}{2} = \frac{y + 1}{3} = z \text{ and } \frac{x + 1}{5} = \frac{y - 2}{1}; z = 2\]
By computing the shortest distance determine whether the following pairs of lines intersect or not: \[\frac{x - 5}{4} = \frac{y - 7}{- 5} = \frac{z + 3}{- 5} \text{ and } \frac{x - 8}{7} = \frac{y - 7}{1} = \frac{z - 5}{3}\]
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 \[\overrightarrow{r} = \left( \hat{i} + 2 \hat{j} + \hat{k} \right) + \lambda\left( \hat{i} - \hat{j} + \hat{k} \right) \text{ and } , \overrightarrow{r} = 2 \hat{i} - \hat{j} - \hat{k} + \mu\left( 2 \hat{i} + \hat{j} + 2 \hat{k} \right)\]
Find the distance between the lines l1 and l2 given by \[\overrightarrow{r} = \hat{i} + 2 \hat{j} - 4 \hat{k} + \lambda\left( 2 \hat{i} + 3 \hat{j} + 6 \hat{k} \right) \text{ and } , \overrightarrow{r} = 3 \hat{i} + 3 \hat{j} - 5 \hat{k} + \mu\left( 2 \hat{i} + 3 \hat{j} + 6 \hat{k} \right)\]
Write the direction cosines of the line whose cartesian equations are 6x − 2 = 3y + 1 = 2z − 4.
Write the direction cosines of the line \[\frac{x - 2}{2} = \frac{2y - 5}{- 3}, z = 2 .\]
Write the formula for the shortest distance between the lines
\[\overrightarrow{r} = \overrightarrow{a_1} + \lambda \overrightarrow{b} \text{ and } \overrightarrow{r} = \overrightarrow{a_2} + \mu \overrightarrow{b} .\]
The cartesian equations of a line AB are \[\frac{2x - 1}{\sqrt{3}} = \frac{y + 2}{2} = \frac{z - 3}{3} .\] Find the direction cosines of a line parallel to AB.
The angle between the lines
If the direction ratios of a line are proportional to 1, −3, 2, then its direction cosines are
If a line makes angle \[\frac{\pi}{3} \text{ and } \frac{\pi}{4}\] with x-axis and y-axis respectively, then the angle made by the line with z-axis is
The shortest distance between the lines \[\frac{x - 3}{3} = \frac{y - 8}{- 1} = \frac{z - 3}{1} \text{ and }, \frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - 6}{4}\]
Show that the lines \[\frac{5 - x}{- 4} = \frac{y - 7}{4} = \frac{z + 3}{- 5} \text { and } \frac{x - 8}{7} = \frac{2y - 8}{2} = \frac{z - 5}{3}\] are coplanar.
Find the separate equations of the lines given by x2 + 2xy tan α − y2 = 0
Find the cartesian equation of the line which passes ·through the point (– 2, 4, – 5) and parallel to the line given by.
`(x + 3)/3 = (y - 4)/5 = (z + 8)/6`
