Advertisements
Advertisements
प्रश्न
Find the foot of the perpendicular drawn from the point \[\hat{i} + 6 \hat{j} + 3 \hat{k} \] to the line \[\overrightarrow{r} = \hat{j} + 2 \hat{k} + \lambda\left( \hat{i} + 2 \hat{j} + 3 \hat{k} \right) .\] Also, find the length of the perpendicular
उत्तर
Let L be the foot of the perpendicular drawn from the point P ( \[\hat{i} + 6 \hat{j} + 3 \hat{k}\] ) to the line \[\overrightarrow{r} = \hat{j} + 2 \hat{k} + \lambda\left( \hat{i} + 2 \hat{j} + 3 \hat{k} \right) .\]
Let the position vector L be \[\overrightarrow{r} = \hat{j} + 2 \hat{k} + \lambda\left( \hat{i} + 2 \hat{j} + 3 \hat{k} \right) = \lambda \hat{i} + \left( 1 + 2\lambda \right) \hat{j} + \left( 2 + 3\lambda \right) \hat{k} \]
Now,
\[\overrightarrow{PL} = \text{ Position vector of L - Position vector of P } \]
\[ \Rightarrow \overrightarrow{PL} = \left\{ \lambda \hat{i} + \left( 1 + 2\lambda \right) \hat{j} + \left( 2 + 3\lambda \right) \hat{k} \right\} - \left( \hat{i} + 6 \hat{j} + 3 \hat{k} \right)\]
\[ \Rightarrow \overrightarrow{PL} = \left( \lambda - 1 \right) \hat{i} + \left( 2\lambda - 5 \right) \hat{j} + \left( 3\lambda - 1 \right) \hat{k} . . . (2)\]
Since
\[\overrightarrow{PL}\] is perpendicular to the given line, which is parallel to \[\overrightarrow{b} = \hat{i} + 2 \hat{j} + 3 \hat{k} \]
we have ,
\[\overrightarrow{PL} . \overrightarrow{b} = 0\]
\[ \Rightarrow \left\{ \left( \lambda - 1 \right) \hat{i} + \left( 2\lambda - 5 \right) \hat{j} + \left( 3\lambda - 1 \right) \hat{k} \right\} . \left( \hat{i} + 2 \hat{j} + 3 \hat{k} \right) = 0 \]
\[ \Rightarrow 1\left( \lambda - 1 \right) + 2\left( 2\lambda - 5 \right) + 3\left( 3\lambda - 1 \right) = 0\]
\[ \Rightarrow \lambda = 1\]
Substituting \[\lambda = 1\] in (1),
we get the position vector of L as \[\hat{i} + 3 \hat{j} + 5 \hat{k}\]
Substituting \[\lambda = 1\] in (2),
we get , \[\overrightarrow{PL} = - 3 \hat{j} + 2 \hat{k}\]
\[= \sqrt{\left( - 3 \right)^2 + 2^2}\]
\[ = \sqrt{13}\]
Hence, the length of the perpendicular from point P on PL is \[\sqrt{13} \text{ units} \] .
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.
The Cartestation equation of line is `(x-6)/2=(y+4)/7=(z-5)/3` find its vector equation.
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.
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.
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.
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 angle between the following pair of line:
\[\frac{- x + 2}{- 2} = \frac{y - 1}{7} = \frac{z + 3}{- 3} \text{ and } \frac{x + 2}{- 1} = \frac{2y - 8}{4} = \frac{z - 5}{4}\]
Find the angle between the pairs of lines with direction ratios proportional to 1, 2, −2 and −2, 2, 1 .
Find the angle between two lines, one of which has direction ratios 2, 2, 1 while the other one is obtained by joining the points (3, 1, 4) and (7, 2, 12).
Find the equations of the line passing through the point (−1, 2, 1) and parallel to the line \[\frac{2x - 1}{4} = \frac{3y + 5}{2} = \frac{2 - z}{3} .\]
Determine the equations of the line passing through the point (1, 2, −4) and perpendicular to the two lines \[\frac{x - 8}{8} = \frac{y + 9}{- 16} = \frac{z - 10}{7} \text{ and } \frac{x - 15}{3} = \frac{y - 29}{8} = \frac{z - 5}{- 5}\]
If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (−4, 3, −6) and (2, 9, 2) respectively, then find the angle between the lines AB and CD.
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 lines through A (0, −1, −1) and B (4, 5, 1) intersects the line through C (3, 9, 4) and D (−4, 4, 4). Also, find their point of intersection.
Determine whether the following pair of lines intersect or not:
\[\overrightarrow{r} = \left( \hat{i} - \hat{j} \right) + \lambda\left( 2 \hat{i} + \hat{k} \right) \text{ and } \overrightarrow{r} = \left( 2 \hat{i} - \hat{j} \right) + \mu\left( \hat{i} + \hat{j} - \hat{k} \right)\]
Determine whether the following pair of lines intersect or not:
\[\frac{x - 5}{4} = \frac{y - 7}{4} = \frac{z + 3}{- 5} and \frac{x - 8}{7} = \frac{y - 4}{1} = \frac{3 - 5}{3}\]
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 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 foot of the perpendicular from (1, 2, −3) to the line \[\frac{x + 1}{2} = \frac{y - 3}{- 2} = \frac{z}{- 1} .\]
Find the equation of line passing through the points A (0, 6, −9) and B (−3, −6, 3). If D is the foot of perpendicular drawn from a point C (7, 4, −1) on the line AB, then find the coordinates of the point D and the equation of line CD.
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: \[\overrightarrow{r} = \left( 8 + 3\lambda \right) \hat{i} - \left( 9 + 16\lambda \right) \hat{j} + \left( 10 + 7\lambda \right) \hat{k} \]\[\overrightarrow{r} = 15 \hat{i} + 29 \hat{j} + 5 \hat{k} + \mu\left( 3 \hat{i} + 8 \hat{j} - 5 \hat{k} \right)\]
By computing the shortest distance determine whether the following pairs of lines intersect or not: \[\overrightarrow{r} = \left( \hat{i} + \hat{j} - \hat{k} \right) + \lambda\left( 3 \hat{i} - \hat{j} \right) \text{ and } \overrightarrow{r} = \left( 4 \hat{i} - \hat{k} \right) + \mu\left( 2 \hat{i} + 3 \hat{k} \right)\]
Find the equations of the lines joining the following pairs of vertices and then find the shortest distance between the lines
(i) (0, 0, 0) and (1, 0, 2)
Write the vector equation of a line passing through a point having position vector \[\overrightarrow{\alpha}\] and parallel to vector \[\overrightarrow{\beta}\] .
Write the coordinate axis to which the line \[\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 1}{0}\] is perpendicular.
If the equations of a line AB are
\[\frac{3 - x}{1} = \frac{y + 2}{- 2} = \frac{z - 5}{4},\] write the direction ratios of a line parallel to AB.
The lines `x/1 = y/2 = z/3 and (x - 1)/-2 = (y - 2)/-4 = (z - 3)/-6` are
The perpendicular distance of the point P (1, 2, 3) from the line \[\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{- 2}\] is
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.
The separate equations of the lines represented by `3x^2 - 2sqrt(3)xy - 3y^2` = 0 are ______
Find the separate equations of the lines given by x2 + 2xy tan α − y2 = 0
The distance of the point (4, 3, 8) from the Y-axis is ______.
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`
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
