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.
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.
Find the separate equations of the lines represented by the equation 3x2 – 10xy – 8y2 = 0.
Find the vector and Cartesian equations of the line through the point (1, 2, −4) and perpendicular to the two lines.
`vecr=(8hati-19hatj+10hatk)+lambda(3hati-16hatj+7hatk) " and "vecr=(15hati+29hatj+5hatk)+mu(3hati+8hatj-5hatk)`
The Cartesian equation of a line is `(x-5)/3 = (y+4)/7 = (z-6)/2` Write its vector form.
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 cartesian equation of a line passing through (1, −1, 2) and parallel to the line whose equations are \[\frac{x - 3}{1} = \frac{y - 1}{2} = \frac{z + 1}{- 2}\] Also, reduce the equation obtained in vector form.
Find the direction cosines of the line \[\frac{4 - x}{2} = \frac{y}{6} = \frac{1 - z}{3} .\] Also, 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.
Find the points on the line \[\frac{x + 2}{3} = \frac{y + 1}{2} = \frac{z - 3}{2}\] at a distance of 5 units from the point P (1, 3, 3).
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.
Show that the line joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points (3, 5, −1) and (4, 3, −1).
Find the angle between the following pair of line:
\[\frac{x - 2}{3} = \frac{y + 3}{- 2}, z = 5 \text{ and } \frac{x + 1}{1} = \frac{2y - 3}{3} = \frac{z - 5}{2}\]
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 5, −12, 13 and −3, 4, 5
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 lines \[\frac{x - 1}{- 3} = \frac{y - 2}{2 \lambda} = \frac{z - 3}{2} \text{ and } \frac{x - 1}{3\lambda} = \frac{y - 1}{1} = \frac{z - 6}{- 5}\] are perpendicular, find the value of λ.
Find the value of λ so that the following lines are perpendicular to each other. \[\frac{x - 5}{5\lambda + 2} = \frac{2 - y}{5} = \frac{1 - z}{- 1}, \frac{x}{1} = \frac{2y + 1}{4\lambda} = \frac{1 - z}{- 3}\]
Find the direction cosines of the line
\[\frac{x + 2}{2} = \frac{2y - 7}{6} = \frac{5 - z}{6}\] Also, find the vector equation of the line through the point A(−1, 2, 3) and parallel to the given line.
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.
Determine whether the following pair of lines intersect or not:
\[\frac{x - 5}{4} = \frac{y - 7}{4} = \frac{z + 3}{- 5} and \frac{x - 8}{7} = \frac{y - 4}{1} = \frac{3 - 5}{3}\]
Show that the lines \[\vec{r} = 3 \hat{i} + 2 \hat{j} - 4 \hat{k} + \lambda\left( \hat{i} + 2 \hat{j} + 2 \hat{k} \right) \text{ and } \vec{r} = 5 \hat{i} - 2 \hat{j} + \mu\left( 3 \hat{i} + 2 \hat{j} + 6 \hat{k} \right)\] are intersecting. Hence, find their point of intersection.
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.
Find the length of the perpendicular drawn from the point (5, 4, −1) to the line \[\overrightarrow{r} = \hat{i} + \lambda\left( 2 \hat{i} + 9 \hat{j} + 5 \hat{k} \right) .\]
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: \[\overrightarrow{r} = \left( 2 \hat{i} - \hat{j} - \hat{k} \right) + \lambda\left( 2 \hat{i} - 5 \hat{j} + 2 \hat{k} \right) \text{ and }, \overrightarrow{r} = \left( \hat{i} + 2 \hat{j} + \hat{k} \right) + \mu\left( \hat{i} - \hat{j} + \hat{k} \right)\]
Find the shortest distance between the following pairs of lines whose vector equations are: \[\overrightarrow{r} = \left( 8 + 3\lambda \right) \hat{i} - \left( 9 + 16\lambda \right) \hat{j} + \left( 10 + 7\lambda \right) \hat{k} \]\[\overrightarrow{r} = 15 \hat{i} + 29 \hat{j} + 5 \hat{k} + \mu\left( 3 \hat{i} + 8 \hat{j} - 5 \hat{k} \right)\]
By computing the shortest distance determine whether the following pairs of lines intersect or not: \[\frac{x - 1}{2} = \frac{y + 1}{3} = z \text{ and } \frac{x + 1}{5} = \frac{y - 2}{1}; z = 2\]
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 coordinate axis to which the line \[\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 1}{0}\] is perpendicular.
Write the angle between the lines \[\frac{x - 5}{7} = \frac{y + 2}{- 5} = \frac{z - 2}{1} \text{ and } \frac{x - 1}{1} = \frac{y}{2} = \frac{z - 1}{3} .\]
Write the angle between the lines 2x = 3y = −z and 6x = −y = −4z.
Write the condition for the lines \[\vec{r} = \overrightarrow{a_1} + \lambda \overrightarrow{b_1} \text{ and } \overrightarrow{r} = \overrightarrow{a_2} + \mu \overrightarrow{b_2}\] to be intersecting.
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
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.
Choose correct alternatives:
If the equation 4x2 + hxy + y2 = 0 represents two coincident lines, then h = _______
Choose correct alternatives:
The difference between the slopes of the lines represented by 3x2 - 4xy + y2 = 0 is 2
Find the separate equations of the lines given by x2 + 2xy tan α − y2 = 0
Find the position vector of a point A in space such that `vec"OA"` is inclined at 60º to OX and at 45° to OY and `|vec"OA"|` = 10 units.
