Advertisements
Advertisements
प्रश्न
Find the co-ordinates of the foot of perpendicular drawn from the point A(1, 8, 4) to the line joining the points B(0, –1, 3) and C(2, –3, –1).
Advertisements
उत्तर
Let L be the foot of perpendicular drawn from the points A (1, 8, 4) to the line passing through B and C as shown in the Figure.
The equation of line BC by using formula `vec"r" = vec"a" + lambda(vec"b" - vec"a")`, the equation of the line BC is
`vec"r" = (-hat"j" + 3hat"k") + lambda(2hat"i" - 2hat"j" - 4hat"k")`
⇒ `xhat"i" + yhat"j" + zhat"k" = 2lambdahat"i" - (2lambda + 1)hat"j" + lambda(3 - 4lambda)hat"k"`
Comparing both sides, we get
x = `2lambda, y = -(2lambda + 1), z = 3 - 4lambda` .....(1)
Thus, the co-ordinate of L are (2λ, – (2λ + 1), (3 – 4λ),
so that the direction ratios of the line AL are (1 – 2λ), 8 + (2λ + 1), 4 – (3 – 4λ),
i.e. 1 – 2λ, 2λ + 9, 1 + 4λ
Since AL is perpendicular to BC, we have,
(1 – 2λ)(2 – 0) + (2λ + 9)(–3 + 1) + (4λ + 1)(–1 –3) = 0

⇒ λ = `(-5)/6`
The required point is obtained by substituting the value of λ, in (1)
Which is `((-5)/3, 2/3, 19/3)`.
APPEARS IN
संबंधित प्रश्न
Find the equations of the planes that passes through three points.
(1, 1, 0), (1, 2, 1), (−2, 2, −1)
Find the equation of the plane passing through (a, b, c) and parallel to the plane `vecr.(hati + hatj + hatk) = 2`
Find the vector equation of the line passing through the point (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)`
Find the vector equation of a plane passing through a point with position vector \[2 \hat{i} - \hat{j} + \hat{k} \] and perpendicular to the vector \[4 \hat{i} + 2 \hat{j} - 3 \hat{k} .\]
Find the Cartesian form of the equation of a plane whose vector equation is
\[\vec{r} \cdot \left( 12 \hat{i} - 3 \hat{j} + 4 \hat{k} \right) + 5 = 0\]
Find the Cartesian form of the equation of a plane whose vector equation is
\[\vec{r} \cdot \left( - \hat{i} + \hat{j} + 2 \hat{k} \right) = 9\]
Find the vector and Cartesian equations of a plane passing through the point (1, −1, 1) and normal to the line joining the points (1, 2, 5) and (−1, 3, 1).
A plane passes through the point (1, −2, 5) and is perpendicular to the line joining the origin to the point
Show that the normals to the following pairs of planes are perpendicular to each other.
x − y + z − 2 = 0 and 3x + 2y − z + 4 = 0
Show that the normal vector to the plane 2x + 2y + 2z = 3 is equally inclined to the coordinate axes.
Find the vector equation of a plane which is at a distance of 5 units from the origin and which is normal to the vector \[\hat{i} - \text{2 } \hat{j} - \text{2 } \hat{k} .\]
Find the vector equation of the plane passing through the points P (2, 5, −3), Q (−2, −3, 5) and R (5, 3, −3).
Find the vector equation of the plane passing through points A (a, 0, 0), B (0, b, 0) and C(0, 0, c). Reduce it to normal form. If plane ABC is at a distance p from the origin, prove that \[\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} .\]
Find the vector equation of the plane passing through the points (1, 1, −1), (6, 4, −5) and (−4, −2, 3).
Find the equation of a plane passing through the point (−1, −1, 2) and perpendicular to the planes 3x + 2y − 3z = 1 and 5x − 4y + z = 5.
Find the equation of the plane passing through the points whose coordinates are (−1, 1, 1) and (1, −1, 1) and perpendicular to the plane x + 2y + 2z = 5.
Find the equation of the plane that contains the point (1, −1, 2) and is perpendicular to each of the planes 2x + 3y − 2z = 5 and x + 2y − 3z = 8.
Find the equation of the plane through the points (2, 2, −1) and (3, 4, 2) and parallel to the line whose direction ratios are 7, 0, 6.
Find the coordinates of the point where the line through (3, −4, −5) and (2, −3, 1) crosses the plane 2x + y + z = 7.
If the lines \[\frac{x - 1}{- 3} = \frac{y - 2}{- 2k} = \frac{z - 3}{2} \text{ and }\frac{x - 1}{k} = \frac{y - 2}{1} = \frac{z - 3}{5}\] are perpendicular, find the value of k and, hence, find the equation of the plane containing these lines.
Find the coordinates of the foot of the perpendicular drawn from the point (5, 4, 2) to the line \[\frac{x + 1}{2} = \frac{y - 3}{3} = \frac{z - 1}{- 1} .\]
Hence, or otherwise, deduce the length of the perpendicular.
Find the image of the point with position vector \[3 \hat{i} + \hat{j} + 2 \hat{k} \] in the plane \[\vec{r} \cdot \left( 2 \hat{i} - \hat{j} + \hat{k} \right) = 4 .\] Also, find the position vectors of the foot of the perpendicular and the equation of the perpendicular line through \[3 \hat{i} + \hat{j} + 2 \hat{k} .\]
Find the coordinates of the foot of the perpendicular from the point (1, 1, 2) to the plane 2x − 2y + 4z + 5 = 0. Also, find the length of the perpendicular.
Find the distance of the point (1, −2, 3) from the plane x − y + z = 5 measured along a line parallel to \[\frac{x}{2} = \frac{y}{3} = \frac{z}{- 6} .\]
Find the image of the point (1, 3, 4) in the plane 2x − y + z + 3 = 0.
Find the distance of the point P (–1, –5, –10) from the point of intersection of the line joining the points A (2, –1, 2) and B (5, 3, 4) with the plane x – y + z = 5.
Write the general equation of a plane parallel to X-axis.
Write the distance of the plane \[\vec{r} \cdot \left( 2 \hat{i} - \hat{j} + 2 \hat{k} \right) = 12\] from the origin.
Write the equation of the plane \[\vec{r} = \vec{a} + \lambda \vec{b} + \mu \vec{c}\] in scalar product form.
Find the vector equation of the plane with intercepts 3, –4 and 2 on x, y and z-axis respectively.
Find the vector equation of the plane which contains the line of intersection of the planes `vec("r").(hat"i"+2hat"j"+3hat"k"),-4=0, vec("r").(2hat"i"+hat"j"-hat"k")+5=0`and which is perpendicular to the plane`vec("r").(5hat"i"+3hat"j"-6hat"k"),+8=0`
Prove that the lines x = py + q, z = ry + s and x = p′y + q′, z = r′y + s′ are perpendicular if pp′ + rr′ + 1 = 0.
Find the foot of perpendicular from the point (2, 3, –8) to the line `(4 - x)/2 = y/6 = (1 - z)/3`. Also, find the perpendicular distance from the given point to the line.
Find the equations of the line passing through the point (3, 0, 1) and parallel to the planes x + 2y = 0 and 3y – z = 0.
Show that the points `(hat"i" - hat"j" + 3hat"k")` and `3(hat"i" + hat"j" + hat"k")` are equidistant from the plane `vec"r" * (5hat"i" + 2hat"j" - 7hat"k") + 9` = 0 and lies on opposite side of it.
The equation of a line, which is parallel to `2hat"i" + hat"j" + 3hat"k"` and which passes through the point (5, –2, 4), is `(x - 5)/2 = (y + 2)/(-1) = (z - 4)/3`.
If the foot of perpendicular drawn from the origin to a plane is (5, – 3, – 2), then the equation of plane is `vec"r".(5hat"i" - 3hat"j" - 2hat"k")` = 38.
