Advertisements
Advertisements
प्रश्न
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)`
Advertisements
उत्तर
Let us assume the required line
`vecr = hati + 2hatj - 4hatk + λ(b_1hati + b_2hatj + b_3hatk)` .....(i)
The lines `(x - 8)/3 = (y + 19)/-16 = (z -10)/7` and `vecr = hati + 2hatj - 4hatk + λ(b_1hati + b_2hatj + b_3hatk)` are perpendicular to each other.
The direction ratios of these lines are 3, −16, 7 and b1, b2, b3. These lines are perpendicular to each other if
3b1 − 16b2 + 7b3 = 0 .....(ii)
Similarly, the direction ratios of the lines `(x -15)/3 = (y - 29)/8 = (z - 5)/-5` and `vecr = hati + 2hatj - 4hatk + λ(b_1hati + b_2hatj + b_3hatk)` are 3, 8, −5 and b1, b2, b3 are mutually perpendicular.
∴ 3b1 + 8b2 − 5b3 = 0 .....(iii)
From equations (ii) and (iii),
`b_1/(80 - 56) = b_2/(21 + 15) = b_3/(24 + 18)`
and `b_1/24 = b_2/36 = b_3/72`
`"b"_1/2 = "b"_2/3 = "b"_3/6`
Putting the proportional values of b1, b2, b3 (i)
`vecr = hati + 2hatj - 4hatk + λ(2hati + 3hatj + 6hatk)`
This is the equation of the required line.
APPEARS IN
संबंधित प्रश्न
Find the equations of the planes that passes through three points.
(1, 1, −1), (6, 4, −5), (−4, −2, 3)
If the points (1, 1, p) and (−3, 0, 1) be equidistant from the plane `vecr.(3hati + 4hatj - 12hatk)+ 13 = 0`, then find the value of p.
Find the vector equations of the coordinate planes.
Find the vector equation of each one of following planes.
x + y − z = 5
\[\vec{n}\] is a vector of magnitude \[\sqrt{3}\] and is equally inclined to an acute angle with the coordinate axes. Find the vector and Cartesian forms of the equation of a plane which passes through (2, 1, −1) and is normal to \[\vec{n}\] .
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 normals to the following pairs of planes are perpendicular to each other.
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).
Obtain the equation of the plane passing through the point (1, −3, −2) and perpendicular to the planes x + 2y + 2z = 5 and 3x + 3y + 2z = 8.
Find the equation of the plane passing through the points (2, 2, 1) and (9, 3, 6) and perpendicular to the plane 2x + 6y + 6z = 1.
Find the equation of the plane passing through (a, b, c) and parallel to the plane \[\vec{r} \cdot \left( \hat{i} + \hat{j} + \hat{k} \right) = 2 .\]
Find the equation of the plane through (2, 3, −4) and (1, −1, 3) and parallel to x-axis.
Find the equation of a plane which passes through the point (3, 2, 0) and contains the line \[\frac{x - 3}{1} = \frac{y - 6}{5} = \frac{z - 4}{4}\] .
Find the image of the point (0, 0, 0) in the plane 3x + 4y − 6z + 1 = 0.
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 equation of the plane that contains the point (1, –1, 2) and is perpendicular to both the planes 2x + 3y – 2z = 5 and x + 2y – 3z = 8. Hence, find the distance of point P (–2, 5, 5) from the plane obtained
Write the equation of the plane parallel to XOY- plane and passing through the point (2, −3, 5).
Write the general equation of a plane parallel to X-axis.
Write the position vector of the point where the line \[\vec{r} = \vec{a} + \lambda \vec{b}\] meets the plane \[\vec{r} . \vec{n} = 0 .\]
Write the intercept cut off by the plane 2x + y − z = 5 on x-axis.
The vector equation of the plane containing the line \[\vec{r} = \left( - 2 \hat{i} - 3 \hat{j} + 4 \hat{k} \right) + \lambda\left( 3 \hat{i} - 2 \hat{j} - \hat{k} \right)\] and the point \[\hat{i} + 2 \hat{j} + 3 \hat{k} \] is
Find a vector of magnitude 26 units normal to the plane 12x − 3y + 4z = 1.
If the line drawn from (4, −1, 2) meets a plane at right angles at the point (−10, 5, 4), find the equation of the plane.
Find the equation of the plane which bisects the line segment joining the points (−1, 2, 3) and (3, −5, 6) at right angles.
Find the vector and Cartesian equations of the plane that passes through the point (5, 2, −4) and is perpendicular to the line with direction ratios 2, 3, −1.
Find the vector equation of the plane with intercepts 3, –4 and 2 on x, y and z-axis respectively.
Find the value of λ for which the following lines are perpendicular to each other `("x"-5)/(5λ+2) = (2 -"y")/(5) = (1 -"z")/(-1); ("x")/(1) = ("y"+1/2)/(2λ) = ("z" -1)/(3)`
hence, find whether the lines intersect or not
Find the equation of a plane which bisects perpendicularly the line joining the points A(2, 3, 4) and B(4, 5, 8) at right angles.
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.
Find the equation of the plane through the points (2, 1, –1) and (–1, 3, 4), and perpendicular to the plane x – 2y + 4z = 10.
The locus represented by xy + yz = 0 is ______.
Let A be the foot of the perpendicular from focus P of hyperbola `x^2/a^2 - y^2/b^2 = 1` on the line bx – ay = 0 and let C be the centre of hyperbola. Then the area of the rectangle whose sides are equal to that of PA and CA is,
