Advertisements
Advertisements
Question
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).
Solution
The coordinates of any point on the line \[\frac{x + 2}{3} = \frac{y + 1}{2} = \frac{z - 3}{2}\] are given by
\[\frac{x + 2}{3} = \frac{y + 1}{2} = \frac{z - 3}{2} = \lambda\]
\[ \Rightarrow x = 3\lambda - 2, y = 2\lambda - 1, z = 2\lambda + 3 . . . (1)\]
Let the coordinates of the desired point be \[\left( 3\lambda - 2, 2\lambda - 1, 2\lambda + 3 \right)\]
The distance between this point and (1, 3, 3) is 5 units.
\[\therefore \sqrt{\left( 3\lambda - 2 - 1 \right)^2 + \left( 2\lambda - 1 - 3 \right)^2 + \left( 2\lambda + 3 - 3 \right)^2} = 5\]
\[ \Rightarrow \left( 3\lambda - 3 \right)^2 + \left( 2\lambda - 4 \right)^2 + \left( 2\lambda \right)^2 = 25\]
\[ \Rightarrow 17 \lambda^2 - 34\lambda = 0\]
\[ \Rightarrow \lambda\left( \lambda - 2 \right) = 0\]
\[ \Rightarrow \lambda = 0 \text{ or } 2\]
Substituting the values of \[\lambda\] in (1) we get the coordinates of the desired point as (-2,-1,3) and (4, 3 , 7) .
APPEARS IN
RELATED QUESTIONS
The Cartestation equation of line is `(x-6)/2=(y+4)/7=(z-5)/3` find its vector equation.
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 line through the points (1, −1, 2) (3, 4, −2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6).
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 and the Cartesian equations of the lines that pass through the origin and (5, −2, 3).
Find the vector and the Cartesian equations of the line that passes through the points (3, −2, −5), (3, −2, 6).
Find the equation of a line parallel to x-axis and passing through the origin.
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 and cartesian equations of the line through the point (5, 2, −4) and which is parallel to the vector \[3 \hat{i} + 2 \hat{j} - 8 \hat{k} .\]
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 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:
\[\overrightarrow{r} = \left( 3 \hat{i} + 2 \hat{j} - 4 \hat{k} \right) + \lambda\left( \hat{i} + 2 \hat{j} + 2 \hat{k} \right) \text{ and } \overrightarrow{r} = \left( 5 \hat{j} - 2 \hat{k} \right) + \mu\left( 3 \hat{i} + 2 \hat{j} + 6 \hat{k} \right)\]
Find the angle between the following pair of line:
\[\frac{x + 4}{3} = \frac{y - 1}{5} = \frac{z + 3}{4} \text { and } \frac{x + 1}{1} = \frac{y - 4}{1} = \frac{z - 5}{2}\]
Find the angle between the following pair of line:
\[\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{- 3} \text { and } \frac{x + 3}{- 1} = \frac{y - 5}{8} = \frac{z - 1}{4}\]
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 equation of the line passing through the point \[\hat{i} + \hat{j} - 3 \hat{k} \] and perpendicular to the lines \[\overrightarrow{r} = \hat{i} + \lambda\left( 2 \hat{i} + \hat{j} - 3 \hat{k} \right) \text { and } \overrightarrow{r} = \left( 2 \hat{i} + \hat{j} - \hat{ k} \right) + \mu\left( \hat{i} + \hat{j} + \hat{k} \right) .\]
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 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:
\[\frac{x - 1}{2} = \frac{y + 1}{3} = z \text{ and } \frac{x + 1}{5} = \frac{y - 2}{1}; z = 2\]
Determine whether the following pair of lines intersect or not:
\[\frac{x - 5}{4} = \frac{y - 7}{4} = \frac{z + 3}{- 5} and \frac{x - 8}{7} = \frac{y - 4}{1} = \frac{3 - 5}{3}\]
Find the foot of the perpendicular from (1, 2, −3) to the line \[\frac{x + 1}{2} = \frac{y - 3}{- 2} = \frac{z}{- 1} .\]
Write the vector equations of the following lines and hence determine the distance between them \[\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z + 4}{6} \text{ and } \frac{x - 3}{4} = \frac{y - 3}{6} = \frac{z + 5}{12}\]
Find the shortest distance between the lines \[\overrightarrow{r} = \left( \hat{i} + 2 \hat{j} + \hat{k} \right) + \lambda\left( \hat{i} - \hat{j} + \hat{k} \right) \text{ and } , \overrightarrow{r} = 2 \hat{i} - \hat{j} - \hat{k} + \mu\left( 2 \hat{i} + \hat{j} + 2 \hat{k} \right)\]
Write the vector equation of a line passing through a point having position vector \[\overrightarrow{\alpha}\] and parallel to vector \[\overrightarrow{\beta}\] .
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 value of λ for which the lines \[\frac{x - 3}{- 3} = \frac{y + 2}{2\lambda} = \frac{z + 4}{2} \text{ and } \frac{x + 1}{3\lambda} = \frac{y - 2}{1} = \frac{z + 6}{- 5}\] are perpendicular to each other.
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.
Find the angle between the lines
\[\vec{r} = \left( 2 \hat{i} - 5 \hat{j} + \hat{k} \right) + \lambda\left( 3 \hat{i} + 2 \hat{j} + 6 \hat{k} \right)\] and \[\vec{r} = 7 \hat{i} - 6 \hat{k} + \mu\left( \hat{i} + 2 \hat{j} + 2 \hat{k} \right)\]
The angle between the lines
The equation of the line passing through the points \[a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k} \text{ and } b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k} \] is
The lines \[\frac{x}{1} = \frac{y}{2} = \frac{z}{3} \text { and } \frac{x - 1}{- 2} = \frac{y - 2}{- 4} = \frac{z - 3}{- 6}\]
Find the value of p for which the following lines are perpendicular :
`(1-x)/3 = (2y-14)/(2p) = (z-3)/2 ; (1-x)/(3p) = (y-5)/1 = (6-z)/5`
If y – 2x – k = 0 touches the conic 3x2 – 5y2 = 15, find the value of k.
If the lines represented by kx2 − 3xy + 6y2 = 0 are perpendicular to each other, then
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.