Advertisements
Advertisements
प्रश्न
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).
Advertisements
उत्तर
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
संबंधित प्रश्न
The Cartestation equation of line is `(x-6)/2=(y+4)/7=(z-5)/3` find its vector equation.
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 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 equation of the line in vector and in Cartesian form that passes through the point with position vector `2hati -hatj+4hatk` and is in the direction `hati + 2hatj - hatk`.
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 and the Cartesian equations of the lines that pass through the origin and (5, −2, 3).
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), (4, 3, – 1).
Find the vector and Cartesian equations of a line passing through (1, 2, –4) and perpendicular to the two lines `(x - 8)/3 = (y + 19)/(-16) = (z - 10)/7` and `(x - 15)/3 = (y - 29)/8 = (z - 5)/(-5)`
Show that the line through the points (1, −1, 2) and (3, 4, −2) is perpendicular to the through the points (0, 3, 2) and (3, 5, 6).
Find the equation of a line parallel to x-axis and passing through the origin.
Find the angle between the pairs of lines with direction ratios proportional to 1, 2, −2 and −2, 2, 1 .
Find the equations of the line passing through the point (2, 1, 3) and perpendicular to the lines \[\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{3} \text{ and } \frac{x}{- 3} = \frac{y}{2} = \frac{z}{5}\]
Find the equation of the line passing through the point (1, −1, 1) and perpendicular to the lines joining the points (4, 3, 2), (1, −1, 0) and (1, 2, −1), (2, 1, 1).
Show that the lines \[\frac{x - 5}{7} = \frac{y + 2}{- 5} = \frac{z}{1} \text{ and } \frac{x}{1} = \frac{y}{2} = \frac{z}{3}\] are perpendicular to each other.
Find the vector equation of the line passing through the point (2, −1, −1) which is parallel to the line 6x − 2 = 3y + 1 = 2z − 2.
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.
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.
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 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: \[\frac{x - 1}{2} = \frac{y + 1}{3} = z \text{ and } \frac{x + 1}{5} = \frac{y - 2}{1}; z = 2\]
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.
Find the Cartesian equations of the line which passes through the point (−2, 4 , −5) and is parallel to the line \[\frac{x + 3}{3} = \frac{4 - y}{5} = \frac{z + 8}{6} .\]
The lines `x/1 = y/2 = z/3 and (x - 1)/-2 = (y - 2)/-4 = (z - 3)/-6` are
The direction ratios of the line perpendicular to the lines \[\frac{x - 7}{2} = \frac{y + 17}{- 3} = \frac{z - 6}{1} \text{ and }, \frac{x + 5}{1} = \frac{y + 3}{2} = \frac{z - 4}{- 2}\] are proportional to
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
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 lines \[\frac{x}{1} = \frac{y}{2} = \frac{z}{3} \text { and } \frac{x - 1}{- 2} = \frac{y - 2}{- 4} = \frac{z - 3}{- 6}\]
The shortest distance between the lines \[\frac{x - 3}{3} = \frac{y - 8}{- 1} = \frac{z - 3}{1} \text{ and }, \frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - 6}{4}\]
If y – 2x – k = 0 touches the conic 3x2 – 5y2 = 15, find the value of k.
Find the value of λ, so that the lines `(1-"x")/(3) = (7"y" -14)/(λ) = (z -3)/(2) and (7 -7"x")/(3λ) = ("y" - 5)/(1) = (6 -z)/(5)` are at right angles. Also, find whether the lines are intersecting or not.
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`
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
Find the equations of the diagonals of the parallelogram PQRS whose vertices are P(4, 2, – 6), Q(5, – 3, 1), R(12, 4, 5) and S(11, 9, – 2). Use these equations to find the point of intersection of diagonals.
