Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
Let L be the foot of the perpendicular drawn from the point P ( -1, 3, 2) to the line \[\overrightarrow{r} = \left( 2 \hat{j} + 3 \hat{k} \right) + \lambda\left( 2 \hat{i} + \hat{j} + 3 \hat{k} \right) .\]

Let the position vector L be \[\overrightarrow{r} = \left( 2 \hat{j} + 3 \hat{k} \right) + \lambda\left( 2 \hat{i} + \hat{j} + 3 \hat{k} \right) = 2\lambda \hat{i} + \left( 2 + \lambda \right) \hat{j} + \left( 3 + 3\lambda \right) \hat{k} \] ...........(1)
Now,
\[\overrightarrow{PL} = \text{ Position vector of L - Position vector of P } \]
\[ \Rightarrow \overrightarrow{PL} = \left\{ 2\lambda \hat{i} + \left( 2 + \lambda \right) \hat{j} + \left( 3 + 3\lambda \right) \hat{k} \right\} - \left( - \hat{i} + 3 \hat{j} + 2 \hat{k} \right)\]
\[ \Rightarrow \overrightarrow{PL} = \left( 2\lambda + 1 \right) \hat{i} + \left( \lambda - 1 \right) \hat{j} + \left( 3\lambda + 1 \right) \hat{k} . . . (2)\]
Since \[\overrightarrow{PL}\] is perpendicular to the given line, which is parallel to \[\overrightarrow{b} = 2 \hat{i} + \hat{j} + 3 \hat{k} \]
we have ,
\[\overrightarrow{PL} . \overrightarrow{b} = 0\]
\[ \Rightarrow \left\{ \left( 2\lambda + 1 \right) \hat{i} + \left( \lambda - 1 \right) \hat{j} + \left( 3\lambda + 1 \right) \hat{k} \right\} . \left( 2 \hat{i} + \hat{j} + 3 \hat{k} \right) = 0 \]
\[ \Rightarrow 2\left( 2\lambda + 1 \right) + 1\left( \lambda - 1 \right) + 3\left( 3\lambda + 1 \right) = 0\]
\[ \Rightarrow \lambda = - \frac{2}{7}\]
Substituting \[ \Rightarrow \lambda = - \frac{2}{7}\] in (1),
we get the position vector of L as \[- \frac{4}{7} \hat{i} + \frac{12}{7} \hat{j} + \frac{15}{7} \hat{k} \] So, the coordinates of the foot of the perpendicular from P to the given line is L \[\left( - \frac{4}{7}, \frac{12}{7}, \frac{15}{7} \right)\]
Substituting \[\lambda = - \frac{2}{7}\] in (2), we get \[\overrightarrow{PL} = \frac{3}{7} \hat{i} - \frac{9}{7} \hat{j} + \frac{1}{7} \hat{k} \]
Equation of the perpendicular drawn from P to the given line is
\[\overrightarrow{r} = \text{ Position vector of P } + \lambda\left( \vec{PL} \right)\]
\[ = \left( - \hat{i} + 3 \hat{j} + 2 \hat{k} \right) + \lambda\left( 3 \hat{i} - 9 \hat{j} + \hat{k} \right)\]
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.
If the Cartesian equations of a line are ` (3-x)/5=(y+4)/7=(2z-6)/4` , write the vector equation for the line.
A line passes through (2, −1, 3) and is perpendicular to the lines `vecr=(hati+hatj-hatk)+lambda(2hati-2hatj+hatk) and vecr=(2hati-hatj-3hatk)+mu(hati+2hatj+2hatk)` . Obtain its equation in vector and Cartesian from.
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.
Show that the three lines with direction cosines `12/13, (-3)/13, (-4)/13; 4/13, 12/13, 3/13; 3/13, (-4)/13, 12/13 ` are mutually perpendicular.
Show that the line through the points (4, 7, 8) (2, 3, 4) is parallel to the line through the points (−1, −2, 1), (1, 2, 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`
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 equation of a line parallel to x-axis and passing through the origin.
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 equations of the line passing through the point (−1, 2, 1) and parallel to the line \[\frac{2x - 1}{4} = \frac{3y + 5}{2} = \frac{2 - z}{3} .\]
Find the equation of the line passing through the point (2, −1, 3) and parallel to the line \[\overrightarrow{r} = \left( \hat{i} - 2 \hat{j} + \hat{k} \right) + \lambda\left( 2 \hat{i} + 3 \hat{j} - 5 \hat{k} \right) .\]
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.
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:
\[\frac{x - 1}{2} = \frac{y + 1}{3} = z \text{ and } \frac{x + 1}{5} = \frac{y - 2}{1}; z = 2\]
Find the perpendicular distance of the point (3, −1, 11) from the line \[\frac{x}{2} = \frac{y - 2}{- 3} = \frac{z - 3}{4} .\]
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).
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: \[\overrightarrow{r} = \left( 3 \hat{i} + 5 \hat{j} + 7 \hat{k} \right) + \lambda\left( \hat{i} - 2 \hat{j} + 7 \hat{k} \right) \text{ and } \overrightarrow{r} = - \hat{i} - \hat{j} - \hat{k} + \mu\left( 7 \hat{i} - 6 \hat{j} + \hat{k} \right)\]
Find the shortest distance between the following pairs of lines whose cartesian equations are: \[\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} and \frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 5}{5}\]
By computing the shortest distance determine whether the following pairs of lines intersect or not: \[\frac{x - 5}{4} = \frac{y - 7}{- 5} = \frac{z + 3}{- 5} \text{ and } \frac{x - 8}{7} = \frac{y - 7}{1} = \frac{z - 5}{3}\]
Write the cartesian and vector equations of Y-axis.
Write the cartesian and vector equations of Z-axis.
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 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
If a line makes angles α, β and γ with the axes respectively, then cos 2 α + cos 2 β + cos 2 γ =
If the direction ratios of a line are proportional to 1, −3, 2, then its direction cosines are
The straight line \[\frac{x - 3}{3} = \frac{y - 2}{1} = \frac{z - 1}{0}\] is
The shortest distance between the lines \[\frac{x - 3}{3} = \frac{y - 8}{- 1} = \frac{z - 3}{1} \text{ and }, \frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - 6}{4}\]
The equation of a line is 2x -2 = 3y +1 = 6z -2 find the direction ratios and also find the vector equation of the line.
If y – 2x – k = 0 touches the conic 3x2 – 5y2 = 15, find the value of k.
The equation 4x2 + 4xy + y2 = 0 represents two ______
